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r Editor Tim Major and Editor Editor CalifCalifCalifCalifCalifororororornia Special Needs esa J visor::::: TTTTTerererereresa J esa J visor nia Special Needs AdAdAdAdAdvisor visor nia Special Needs nia Special Needs esa J..... Miller Miller Miller Miller esa J visor nia Special Needs Miller Teresa is a Mathematics Teacher in Capistrano Unified School District. She has a master’s degree in Education with an emphasis on Special Education. She holds two teaching credentials: Foundation Level Mathematics and Resource Specialist. ames Schierhierhierhierhierererererer ames Sc visor::::: J J J J James Sc ames Sc visor General al al al al AdAdAdAdAdvisor visor Gener Gener Gener ames Sc visor Gener James is a High School Mathematics Teacher for the King City Joint Union High School District. He has a bachelor’s degree in History. He holds two teaching credentials: Mathematics and History. visor::::: Daisy Lee Daisy Lee Daisy Lee visor General al al al al AdAdAdAdAdvisor visor Gener Gener Daisy Lee Daisy Lee visor Gener Gener Daisy is a High School Mathematics Teacher in Los Angeles Unified School District. She has a bachelor’s degree in Mathematics. terial: terial: ditional Ma AdAdAdAdAdditional Ma ditional Ma ence E. Da Da Da Da Davisvisvisvisvis,,,,, Ph.D Ph.D Ph.D ence E. ence E. Clar Clar terial: Clar ditional Material: Ph.D Clarence E. terial: ditional Ma Ph.D ence E. Clar Clarence is an educational consultant for Middle Grade Mathematics and was an Assistant Professor of Mathematics Education. AdAdAdAdAdditional Ma les Seraaaaaphinphinphinphinphin les Ser les Ser osemarie Ing terial: R R R R Rosemarie Ing osemarie Ing terial: terial: ditional Ma ditional Ma osemarie Ingles Ser ditional Material: les Ser osemarie Ing terial: ditional Ma Rosemarie is a Middle School and High School Mathematics Teacher, Fairfax County, Virginia. ting Editors:s:s:s:s: ting Editor ting Editor Suppor Suppor Supporting Editor Suppor ting Editor Suppor Chris Dennett Sarah Hilton Kate Houghton

Sharon Keeley Ali Palin Alan Rix Glenn Rogers Emma Stevens Ami Snelling Claire Thompson Julie Wakeling Tim Wakeling eading: Judith Curran Buck, Jennifer Heller, Amanda Jones, Rafiq Ladhani, Betsey McCarty and Cathy Podeszwa. eading: PrPrPrPrProofroofroofroofroofreading: eading: eading: phic Design: Caroline Batten, Russell Holden, Jane Ross and Ash Tyson. phic Design: GrGrGrGrGraaaaaphic Design: phic Design: phic Design: Mathematics Content Standards for California Public Schools reproduced by permission, California Department of Education, CDE Press, 1430 N Street, Suite 3207, Sacramento, CA 95814. ISBN 13: 978 1 60017 014 0 website: www.cgpeducation.com Wakeling Publishing Services: www.wakelingpublishingservices.co.uk Printed by Elanders Hindson Ltd, UK and Johnson Printing, Boulder, CO Clipart sources: CorelDRAW and VECTOR. Text, design, layout, and illustrations © CGP, Inc. 2007 All rights reserved. xxxxx Chapter 1 Working with Real Numbers Section 1.1 Sets and Expressions............................... 2 Section 1.2 The Real Number System....................... 14 Section 1.3 Exponents, Roots, and Fractions............ 37 Section 1.4 Mathematical Logic................................. 55 Investigation Counting Collections............................... 65 11111 TTTTTopicopicopicopicopic 1.1.11.1.1 1.1.11.1.1 1.1.1 Section 1.1 The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets The Basics of Sets California Standards: 1.0:1.0:1.0:1.0:1.0: Students identify Students identify Students identify Students identify and Students identify use the arithmetic properties integggggererererersssss and inte inte subsets of subsets of of subsets of subsets of inte inte subsets of rational, irrational, and real numbers, including closure properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about sets and subsets. Key words: set element subset universal set empty set Check it out: In this course

the universal set consists of all real numbers (see Topic 1.1.2), unless otherwise specified. Sets are a really useful way of being able to say whether numbers or variables have something in common. There’s some new notation here, but the math isn’t too hard at all. Elements Elements e Collections of e Collections of Sets ar Sets ar Elements e Collections of Elements Sets are Collections of Elements e Collections of Sets ar Sets ar A set is a collection of objects. Each object in the set is called an element or a member. You use the symbol Œ to show that something is a member of a set — you read it as “is an element of ” or “is a member of.” The symbol œ means the opposite — you read it as “is not an element of ” or “is not a member of.” Sets are usually named by capital letters. The elements of a set are enclosed in braces { }. For example, A = {1, 3, 5, 7} is a set containing 4 elements — the numbers 1, 3, 5, and 7. The universal set, denoted by the symbol U, is the set of all objects under consideration (so in a math course, the universal set often consists of all numbers). Example Example Example Example Example 11111 Given that set A = {1, x, 2, b}, determine whether each of the following statements is true or false. a) x Œ A b) 2 Œ A c) 1 œ A d) 3 Œ A Solution Solution Solution Solution Solution a) x is an element of A, so the statement is true. b) 2 is an element of A, so the statement is true. c) 1 is an element of A (1 Œ A), so the statement is false. d) 3 is not an element of A, so the statement is false. Guided Practice Given that set A = {x, 2, 4, y}, determine whether each of the following statements is true or false. 1. y Œ A 2. 2 Œ A 3. 6 œ A 4. 4 œ A 22222 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 The Empty (or Null) Set has No Elements The Empty (or Null) Set has No Elements The Empty (or Null) Set has No Elements The Empty (

or Null) Set has No Elements The Empty (or Null) Set has No Elements An empty set (or null set) is a set without any elements or members. It’s denoted by Δ or { }. e Contained WWWWWithin Other Sets ithin Other Sets ithin Other Sets e Contained e Contained Subsets ar Subsets ar ithin Other Sets Subsets are Contained ithin Other Sets e Contained Subsets ar Subsets ar A subset is a set whose elements are also contained in another set. The symbol à means “is a subset of.” Any set is a subset of itself — and the empty set is a subset of any set. Example Example Example Example Example 22222 Let A = {all odd numbers} and B = {1, 3, 5, 7}. Is B a subset of A? Solution Solution Solution Solution Solution Work through the elements of B one by one: 1 Œ A, 3 Œ A, 5 Œ A, and 7 Œ A, because they are all odd numbers. All elements of B are also elements of A — so B ÃÃÃÃà A. Example Example Example Example Example 33333 Let A = {0, 1, 2}. Determine all the subsets of set A. Solution Solution Solution Solution Solution List the empty set first because it’s a subset of any set. Then write all the subsets with 1 element, then all the subsets with 2 elements, and so on until finally you finish with the whole set. So the subsets of A are: ΔΔΔΔΔ, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, and {0, 1, 2} Guided Practice Let A = {a, 2, 3, b, c} and B = {1, 2, c, d}. Use sets A and B to answer questions 5 and 6: 5. Explain whether B à A. 6. List all the subsets of set B. Let C = {all prime numbers less than 13 but greater than 7}. 7. List set C and all its subsets. For exercises 8 and 9, let M = {all real numbers b such that b = 3x – 1}. 8. List the members of set M if x Œ {1, 2, 3, 5}. 9. List the subsets of set M if x

Œ {1, 2, 3, 5}. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 33333 Check it out: Another way of saying that set A and B are equal is that A is a subset of B......and B is a subset of A. Equality of Sets Equality of Sets Equality of Sets Equality of Sets Equality of Sets Two sets are equal if they have all of the same elements in them. So if A = {1, 3, 5} and B = {3, 5, 1}, then A and B are equal. Or more mathematically, two sets A and B are said to be equal if every element in set A is in set B, and every element in set B is in set A. Note also that all empty sets are equal — because they’re exactly the same. That’s why you say the empty set, not an empty set. Independent Practice 1. Determine the set H whose elements are all the multiples of both 2 and 3 that are less than 30 but greater than 12. 2. Let U = {all letters of the alphabet}. Determine set V, whose elements are the vowels. 3. Write down G = {all prime numbers greater than 11 but less than 13}. For exercises 4 and 5, let F = {red, yellow, blue, purple}, and let G à F and H à F. 4. Determine set G, whose elements are 3-letter words in F. 5. Determine set H, whose elements are 5-letter words in F. For exercises 6 and 7, let A = {all even numbers} and B = {2, 4, 6, 8}. 6. Explain whether B à A. 7. List all the subsets of B. For exercises 8 and 9, let M = {3, a, 9, b, 15} and N = {3, 6, 9}. 8. Explain whether N à M. 9. List all the subsets of N. 10. Let C = {all even numbers less than 10 but greater than 2}. List set C and all its subsets. 11. Let J = {all real numbers y such that y = 2x + 3}. List the members of set J if x Œ {0, 1, 2, 3}. 12. Let A = {1, 3, 5, 7, 9}

and B = {all odd numbers less than 10 but greater than 0}. Explain whether A = B. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Sets sound a little odd but they’re just a way of grouping together types of numbers or variables. You’ve seen a lot of the stuff in this Topic in earlier grades, but in Algebra I you’ve got to treat everything formally and give the proper names for things like the empty set. 44444 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 TTTTTopicopicopicopicopic 1.1.21.1.2 1.1.21.1.2 1.1.2 California Standards: 1.0:1.0:1.0:1.0:1.0: Students identify Students identify Students identify Students identify and Students identify use the arithmetic properties integggggererererers and s and s and inte inte subsets of subsets of subsets of inte of subsets of s and inte subsets of s and rrrrraaaaational, and realealealealeal and r and r tional, tional, tional, ir ir ir ir irrrrrraaaaational, tional, tional, tional, and r and r tional, tional, umbersssss, including closure umber umber nnnnnumber umber properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about the set of real numbers and its subsets. Key words: real numbers natural numbers whole numbers integers rational numbers irrational numbers set element subset Check it out: The natural numbers (N) are sometimes called the counting numbers, because they’re the numbers you use in everyday life to count things. Check it out: The symbol À means “is not a subset of.” eal Numbersssss eal Number eal Number the R the R Subsets of Subsets of the Real Number Subsets of the R eal Number the R Subsets of Subsets of eal Numbersssss eal Number eal Number the R the R Subsets of Subsets of the Real Number Subsets of the R the R Subsets of eal Number Subsets

of This Topic is about one set of numbers that is really important. You’ll be referring to the real numbers throughout Algebra I. s Has Subsets s Has Subsets eal Number he Set of R R R R Real Number eal Number he Set of TTTTThe Set of he Set of s Has Subsets eal Numbers Has Subsets s Has Subsets eal Number he Set of In Algebra I, the sets used are usually subsets of the real numbers. R — Real Numbers The set of real numbers consists of all positive numbers, zero, and all negative numbers. The following are subsets of R: N — Natural numbers W — Whole numbers Z — Integers N = {1, 2, 3, …} W = {0, 1, 2, 3, …} Z = {..., –2, –1, 0, 1, 2,...} Example Example Example Example Example 11111 Explain why N à W. Solution Solution Solution Solution Solution The set N of natural numbers is a subset of the set W of whole numbers because every member of set N is a member of set W. Example Example Example Example Example 22222 Explain why Z À N. Solution Solution Solution Solution Solution To show this, you need to find an element in Z that is not in N. (You can use any negative integer or zero.) Since –1 Œ Z, but –1 œ N, the set Z is not a subset of N (Z À N). Guided Practice 1. Determine the subset of N (natural numbers) whose elements are multiples of 2 that are greater than 10 but less than 20. 2. Determine the subset of Z (integers) whose elements are greater than –1 but less than 5. 3. Give an example of an element in R (real numbers) that is not in Z (integers). 4. Explain why R (real numbers) À Z (integers). Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 55555 Check it out: Notice that since q Œ N, q cannot equal zero. e Subsets of R e Subsets of e Subsets of o Mor tionals — TTTTTwwwwwo Mor o Mor tionals — tionals and Irrrrrraaaaationals — tionals — tionals and Ir RRRRRaaaaationals and Ir tionals and Ir o More

Subsets of e Subsets of o Mor tionals — tionals and Ir A rational number is a number that can be expressed in the form p q, where p is an integer and q is a natural number. Rational Numbers: = all numbers that can be expressed as fractions ⎧ ⎪⎪ ⎨ ⎩⎪⎪, where p q∈, ∈ p q ⎫ ⎪⎪ ⎬ ⎭⎪⎪ For example, 3.5 is a rational number — it can be expressed as 7 2. An irrational number is a number that cannot be expressed in the form p q, where p is an integer and q is a natural number. Irrational Numbers: I = all numbers that cannot be expressed as fractions ⎧ ⎪⎪ ⎨ ⎩⎪⎪, wher ee p q∈, ∈ p q ⎫ ⎪⎪ ⎬ ⎭⎪⎪ For example, ÷2 = 1.4142... is an irrational number — it can’t be expressed in the form p q. Both Q and I are subsets of the real numbers, R. Guided Practice 5. Explain why N Ã Q. 6. Explain why Z À I. 7. Determine the subset, C, of Q whose elements are also members of I. Independent Practice Remember that R = {real numbers}, N = {natural numbers}, W = {whole numbers}, Z = {integers}, Q = {rational numbers}, and I = {irrational numbers}. Classify each of the following statements as true or false. 2. Δ À N 1. N Ã W 3. R Ã Q { 4. Z À Q 5. 2 3. 0 4 7 25 8 6, }Ã Q.,,, ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up “Real numbers” is just a formal name for all positive and negative numbers and zero — there’s nothing more to it than a new name. You need to know all the common labels for the subsets. 66666 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.

1 Section 1.1 TTTTTopicopicopicopicopic 1.1.31.1.3 1.1.31.1.3 1.1.3 California Standards: 1.0:1.0:1.0:1.0:1.0: Students identify Students identify Students identify Students identify and Students identify use the arithmetic properties integggggererererers and s and s and inte inte subsets of subsets of subsets of inte of subsets of s and inte subsets of s and rrrrraaaaational, and realealealealeal and r and r tional, tional, tional, ir ir ir ir irrrrrraaaaational, tional, tional, tional, and r and r tional, tional, umbersssss, including closure umber umber nnnnnumber umber properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about unions and intersections of sets. Key words: union intersection set element Check it out: A » B is said “A union B.” sections sections Unions and Inter Unions and Inter sections Unions and Intersections sections Unions and Inter Unions and Inter sections sections Unions and Inter Unions and Inter Unions and Intersections sections Unions and Inter sections Unions and Inter This Topic is all about two symbols that represent ways of grouping elements in sets — unions and intersections. The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The Union of Sets is Anything in Either or Both Sets The union of sets A and B, denoted A »»»»» B, is the set of all elements in either A or B or both sets. Example Example Example Example Example 11111 Let A = {4, 6, 8, 20} and B = {6, 8, 9, 15}. Find A » B. Solution Solution Solution Solution Solution Elements are not counted twice — they are either “in or out.” So 8, which is in both A and B, only appears once in A » B. So A »»»»» B = {4, 6, 8, 9, 15, 20}. Guided Practice Let A = {1, 2, 4,

5}, B = {0, 3, 6}, C = {2, 4, 6, 8}, D = {1, 3, 5, 7}. 1. Find A » B. 2. Find C » D. 3. Find A » C. 4. Find B » D. Sets is AnAnAnAnAnything in Both Sets ything in Both Sets ything in Both Sets Sets is Sets is section of section of he Inter TTTTThe Inter he Inter ything in Both Sets section of Sets is he Intersection of ything in Both Sets Sets is section of he Inter The intersection of sets A and B, denoted A ««««« B, is the set of all elements that are in both set A and set B. Check it out: A « B is said “A intersect B” (or “A intersection B”). Example Example Example Example Example 22222 Let A = {4, 6, 8, 20} and B = {6, 9, 15}. Find A « B. Solution Solution Solution Solution Solution Only the number 6 is in both A and B. So A ««««« B = {6}. Example Example Example Example Example 33333 Let A = {all even numbers} and B = {all odd numbers}. Find A « B. Solution Solution Solution Solution Solution There are no elements in both A and B, so A ««««« B = ΔΔΔΔΔ. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 77777 Guided Practice Let A = {1, 2, 4, 5}, B = {0, 3, 6}, C = {2, 4, 6, 8}, D = {1, 3, 5, 7}. 5. Find A « B. 6. Find A « C. 7. Find C « D. 8. Find B « D. sections for or or or or TTTTTwwwwwo Sets o Sets o Sets sections f sections f ou Can Find Unions and Inter YYYYYou Can Find Unions and Inter ou Can Find Unions and Inter o Sets ou Can Find Unions and Intersections f o Sets sections f ou Can Find Unions and Inter If you have any two sets, you can always work out the union and intersection. Example Example Example Example Example 44444 Let A = {2, 4, 6

, 8, 10, 12} and B = {3, 6, 9, 12, 15}. Find A « B and A » B. Solution Solution Solution Solution Solution A « B is the set of all the elements that appear in both A and B. So A ««««« B = {6, 12}. A » B is the set of all the elements that appear in A or B, or both sets. So A »»»»» B = {2, 3, 4, 6, 8, 9, 10, 12, 15}. Independent Practice 1. Explain why A « B is a subset of A. If A = {1, 2, 3, 4, 5, 11, 12}, B = {2, 4, 6, 8, 10, 11, 20}, and C = {5, 10, 15, 20, 24}, write down the following sets. 2. A « B 3. (A « B) » C 4. (A « B) « C Let U = {all b Œ N such that b £ 20}, M = {all b in U such that b is a multiple of 2}, V = {all b in U such that b is a multiple of 3}, and H = {all b in U such that b is a multiple of 4}. 5. Determine the set M « V. 6. Determine the set (M « V) « H. 7. Determine the set (M » V) » H. Let A = {prime numbers}, B = {square numbers}, C = {even numbers}, D = {odd numbers}, and E = {natural numbers less than 10}. 8. Copy and complete the following expressions using sets A to E: a. A «...... = Δ, b....... «...... = {2}, c....... « D « E = {1, 9} d. A «...... «...... = {3, 5, 7} 9. Write down the number of subsets for each set a–d in exercise 8. 10. Use your answer to exercise 9 to write down a formula for the number of subsets of a set with n members. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The big thing to remember here is that the union sign is » and the intersection sign is «. Make sure you understand the definitions of

each way of grouping elements. 88888 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 TTTTTopicopicopicopicopic 1.1.41.1.4 1.1.41.1.4 1.1.4 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use prprprprproper umbersssss to umber ties of n n n n number umber ties of ties of oper oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll deal with simple expressions that contain numbers and unknowns. Key words: expression numeric expression algebraic expression variable aic and Numeric aic and Numeric AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic and Numeric aic and Numeric aic and Numeric AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic and Numeric aic and Numeric aic and Numeric aic and Numeric aic and Numeric essions essions ExprExprExprExprExpressions essions essions ExprExprExprExprExpressions essions essions essions essions Expressions are mathematical statements. This Topic is all about two kinds of expressions — ones that contain only numbers, and ones that contain both numbers and unknown values. lude Numbersssss lude Number lude Number ust Inc ust Inc essions J essions J Numeric Expr Numeric Expr ust Include Number essions Just Inc Numeric Expressions J lude Number ust Inc essions J Numeric Expr Numeric Expr A numeric expression contains only numbers — for example, 2.54 × 2 is a numeric expression. Since there are no unknown quantities, you can always work out the value of a numeric expression — the value of 2.54 × 2 is 5.08. Numeric expressions with the same value are like different names for the same thing. All these expressions have the same value: 2.54 × 2 5 + 0.08 6 – 0.92 The value of each of these numeric expressions is 55555.08.08.08.08.08. So if you saw (2.54 × 2) somewhere, you could

write (5 + 0.08) instead, since the two expressions have the same value — they’re describing the same thing. Guided Practice Calculate the value of each of these numeric expressions: 1. 18.4 + 8.23 2. 37.82 – 11.19 3. 716 ÷ 2 4. 1790 ÷ 5 5. 37 ÷ 37 6. 19284 ÷ 19284 For each of the following, write a numeric expression that contains the number 100 and has the same value as the expression given. 7. 465 – 253 8. 3850 ÷ 5 9. 15.6 × 5 10. 24.2 + 1.5 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 99999 Check it out: When you see a number and a letter written next to each other (like 2.54x), it means they are multiplied together. essions Contain VVVVVariaariaariaariaariabbbbbleslesleslesles essions Contain essions Contain aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr aic Expressions Contain essions Contain aic Expr A lot of the time in math, you can use letters to stand for unknown amounts. So if you were not sure how many inches long your desk was, you could just call the length x inches. When you see letters in equations, they’re just numbers that you don’t know the value of yet. Letters that represent unknown numbers are called variables. Example Example Example Example Example 11111 The length of a desk is x inches. How long is this in centimeters? To convert a length in inches to a length in centimeters, you multiply by 2.54. Solution Solution Solution Solution Solution You don’t know the length of the desk, so it’s been called x. If the desk was 10 inches long, it would be 2.54 × 10 = 25.4 cm long. If the desk was 50 inches long, it would be 2.54 × 50 = 127 cm. In the same way, a desk that is x inches long would be 2.54 × x = 2.54x cm. You don’t have enough information to write a numerical value, so leave x in the solution. 2.54x is called an algebraic

expression, as it contains an unknown quantity. The variable x represents the unknown quantity — the number of inches to be converted to centimeters. An algebraic expression always contains at least one variable (and very often it contains numbers as well). Example Example Example Example Example 22222 Write an algebraic or numeric expression for each of the following: a) The total cost of an item whose price is $15.75 plus sales tax of $1.30. b) The number of cents in d dollars. c) Three more than twice the length, l. Solution Solution Solution Solution Solution a) $15.75 + $1.30 = $17.05 (since there are no unknowns, this is a numeric expression) b) 100d (since the number of cents is always the number of dollars multiplied by 100) c) 2l + 3 (since twice the length l is 2 × l = 2l, and three more than this is 2l + 3) 1010101010 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 Guided Practice Write an algebraic expression representing each of the following. 11. The sum of twice m and 7. 12. The sum of three times x and 12. 13. The difference between the total cost c, and the down payment d. 14. The number of notebooks you can buy for $75 if each notebook costs $n. 15. The difference between (3x + 7) and t. 16. Two more than five times d. 17. Three less than the product of x and t. 18. The age of someone six years ago who is 2y years old now. Independent Practice Determine whether the following expressions are numeric or algebraic. 1. 3.14 × 2 2. 189x ÷ 2 3. 2w + 2l 4. 656.4 – 33 Write an algebraic or numeric expression for each of the following. 5. The difference between x and 3 6. The cost of 5 CDs at $x 7. The difference between the original price of a pair of jeans, p, and the sale price, s 8. The product of 2 and 4, plus 3 9. Three times the difference between t and 2 10. One-half of 200 times a number 11. The number of feet in x yards (3 feet = 1 yard) 12. The perimeter of a rectangle with length, l, and width, w. ound

Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Don’t worry if you come across an expression that contains variables. Variables are just letters or symbols that represent unknown numbers, and expressions containing variables follow all the same rules as numeric expressions. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 1111111111 TTTTTopicopicopicopicopic 1.1.51.1.5 1.1.51.1.5 1.1.5 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use prprprprproper umbersssss to umber ties of n n n n number umber ties of ties of oper oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll find out what coefficients are, and you’ll calculate the value of expressions. Key words: coefficient expression numeric expression algebraic expression variable ficients and Evaluaaluaaluaaluaaluatingtingtingtingting ficients and Ev ficients and Ev CoefCoefCoefCoefCoefficients and Ev ficients and Ev CoefCoefCoefCoefCoefficients and Ev ficients and Evaluaaluaaluaaluaaluatingtingtingtingting ficients and Ev ficients and Ev ficients and Ev You’ve just learned about variables — coefficients are just another element that make up a full expression. In this Topic you’ll learn about evaluating expressions, which just means calculating the value. y a y a VVVVVariaariaariaariaariabbbbblelelelele y a y a Amount to Multipl Amount to Multipl ficient is the ficient is the A Coef A Coef Amount to Multiply a ficient is the Amount to Multipl A Coefficient is the Amount to Multipl ficient is the A Coef A Coef In the algebraic expression 2.54x, the number 2.54 is the coefficient of x. A coefficient is a number that multiplies a variable. Example Example Example Example Example 11111 Find the coefficients in these expressions: a) the coefficient of

x in the algebraic expression 2x b) the coefficient of v in v + 5 c) the coefficient of y in 7 – 6y d) the coefficient of k in 5 – k e) the coefficient of m in 5(3 – 2m) Solution Solution Solution Solution Solution a) x is multiplied by 2, so the coefficient is 2. b) v is the same as 1v, so the coefficient is 1. c) The minus sign is part of the coefficient as well — so the coefficient is –6. d) There’s a minus sign here too — the coefficient is –1. e) First multiply out the parentheses: 5(3 – 2m) = 15 – 10m Now you can see that the coefficient is actually –10. Guided Practice Find the coefficient of the variable in each of these expressions: 1. 4a – 15 2. 15 – 4z 3. x – 3 4. 2 – k 5. 2(3b + 1) 6. 4(2 – m) 1212121212 Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 heir Simplest Fororororormmmmm heir Simplest F essions in TTTTTheir Simplest F heir Simplest F essions in essions in rite Expr AlAlAlAlAlwwwwwaaaaays ys ys ys ys WWWWWrite Expr rite Expr rite Expressions in heir Simplest F essions in rite Expr Different numeric expressions can have the same value. The expressions 4 + 3 and 14 ÷ 2 both have the same value of 7. You can think of these expressions as different names for the number 7. When you simplify an expression, you replace it with its simplest name. So you could simplify “4 + 3” or “14 ÷ 2” by writing “7” instead. Guided Practice Simplify each expression below. 9. 24 ÷ 6 7. 11 + 4 10. 10a + 4a 8. 10 × 3.14 11. 12a + 2 – 8a – 5 12. 28b ÷ 28 essions essions aic Expr ting AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr ting EvEvEvEvEvaluaaluaaluaaluaaluating ting essions aic Expressions

essions aic Expr ting You’ve already seen algebraic expressions — they’re expressions containing variables. But if you know the value of the letter, you can evaluate the expression — that means you calculate its value. For example, the algebraic expression 2.54x contains the variable x. When x = 33, the expression 2.54x is equal to 2.54 × 33 = 83.82. That’s what is meant by “evaluating an algebraic expression” — finding the value of the expression when any variables are replaced by specific numbers. Guided Practice Don’t forget: Variables are letters that represent unknown quantities. Evaluate each of these expressions when a = 2, b = 7, and c = –4. 13. a + b 14. b × a 15. c ÷ a 16. a + b + c 17. b – a 18. b – c Independent Practice 1. Find the coefficient of x in the algebraic expression 8 – 5x. 2. Evaluate 8 – 5x when x = 3. 3. Find the coefficient of k in 4.18k + 2. 4. Evaluate 10x + 2 when x = 0. 5. Evaluate 2x + 3y when x = 10 and y = –1. 6. Evaluate 2x + 3y – z when x = 5, y = 2, and z = 4. 7. The formula C = 5 9 degrees Fahrenheit (°F) to degrees Celsius (°C). What is 131 °F in °C? (F – 32) is used to convert temperatures from ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up A coefficient is just the number in front of a variable. It doesn’t matter whether an expression includes just numbers, or numbers and variables — you always follow the same rules when you simplify. Section 1.1 Section 1.1 Section 1.1 — Sets and Expressions Section 1.1 Section 1.1 1313131313 TTTTTopicopicopicopicopic 1.2.11.2.1 1.2.11.2.1 1.2.1 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0

: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n incincincincincluding c losureeeee losur losur luding c luding c luding closur losur luding c prprprprproper our basic our basic or the f or the f ties f ties f oper oper or the four basic our basic ties for the f operties f our basic or the f ties f oper tions wherherherherhereeeee tions w tions w arithmetic operaaaaations w arithmetic oper arithmetic oper arithmetic oper tions w arithmetic oper pplicabbbbblelelelele..... pplica aaaaapplica pplica pplica What it means for you: You’ll learn some of the rules that the real numbers always follow. Key words: binary operation equality real numbers substitution The Number System The Number System The Number System The Number System The Number System The Number System The Number System The Number System The Number System The Number System You’ve already met the real numbers in Section 1.1. Now it’s time to look at the properties of the real numbers in detail. eal Number System is Based on Simple Rulesulesulesulesules eal Number System is Based on Simple R TTTTThe Rhe Rhe Rhe Rhe Real Number System is Based on Simple R eal Number System is Based on Simple R eal Number System is Based on Simple R The rules of the real number system are based on the existence of a set of numbers, plus two binary operations. A binary operation allows you to combine two numbers in some way to produce a third number. In Algebra I, the set of numbers used is the real numbers, R, and the two binary operations are addition and multiplication. Equality AlAlAlAlAlwwwwwaaaaays Hold ys Hold ys Hold Equality Equality ties of ties of oper oper he Pr TTTTThe Pr he Pr ys Hold ties of Equality operties of he Proper ys Hold Equality ties of oper he Pr The following statements about equality hold true for any real numbers a, b, and c: For any number a Œ R: a =

a This is the reflexive property of equality. For any numbers a, b Œ R: if a = b then b = a This is the symmetric property of equality. For any numbers a, b, c Œ R: if a = b and b = c, then a = c This is the transitive property of equality. Guided Practice Name the property of equality being used in each statement. 1. If a = 3 then 3 = a. 2. If a = 3 and 3 = b, then a = b. 3. 3 = 3 4. If 3x = 2 and 2 = 2y, then 3x = 2y. 5. If 3x = 2y then 2y = 3x. Check it out: For addition, the inputs are called adadadadaddends dends dends dends, and for dends multiplication they are called fffffactor actor actor actorsssss. actor tions tions y Operaaaaations y Oper y Oper e Binar e Binar tion ar tion ar dition and Multiplica dition and Multiplica AdAdAdAdAddition and Multiplica tions e Binary Oper tion are Binar dition and Multiplication ar tions y Oper e Binar tion ar dition and Multiplica The operations of addition and multiplication are called binary operations. In order to carry either of them out you need to have two “inputs.” What this basically means is that you can combine two numbers to produce a third number. A set of numbers is said to be closed under a given binary operation if, when you perform that operation on any two members of the set, the result is also a member of the set. 1414141414 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out: Compare this to the set of irrational numbers, I, which is not closed under multiplication, since 2 × 2 = 2, where 2 Œ I, but 2 œ I. Check it out: Multiplication of a and b can be written as ab, a·b, or a × b. They all mean the same thing. The set of real numbers, R, is closed under both addition and multiplication — adding and multiplying real numbers always produces other real numbers. If a Œ R and b Œ R, then (a + b) ŒŒŒ�

�Œ R and (a × b) ŒŒŒŒŒ R. Example Example Example Example Example 11111 Use the fact that 10 Œ R and 6 Œ R to explain why 16 Œ R and 60 Œ R. Solution Solution Solution Solution Solution 10 Œ R and 6 Œ R, so you can add the two numbers to produce another number that is a member of R. So 10 + 6 = 16 Œ R. You can also multiply the two numbers to produce another number that is a member of R. So 10 × 6 = 60 Œ R. operty)ty)ty)ty)ty) oper oper he Substitution Principle (or Substitution Pr TTTTThe Substitution Principle (or Substitution Pr he Substitution Principle (or Substitution Pr he Substitution Principle (or Substitution Proper oper he Substitution Principle (or Substitution Pr For any real numbers a and b, the number a may be substituted for the number b if a = b. This principle means that if two expressions have the same value, then one expression can be substituted for (written instead of) the other. For example, the expression 23 can be replaced by the expression (10 + 13) or by (17 + 6). Similarly, the expression 2.3 × 4.5 can be replaced with 10.35. Independent Practice Determine whether each of the following statements is true or false. If false, rewrite the statement so that it is true. 1. If k = l then l = k. This is the symmetric property. 2. If k = l and l = 3, then k = 3. This is the reflexive property. In Exercises 3–6, demonstrate the closed nature of the set under both addition and multiplication using the numbers given. 3. R using 3 and 5 4. R using –10 and 11 5. N using 6 and 8 6. Q using 1.5 and 0.3 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up These rules are a bit abstract at the moment, but don’t worry. You’ll see how useful the real number properties are in the rest of the Section, and you’ll use them all the way through Algebra I. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1

.2 Section 1.2 1515151515 TTTTTopicopicopicopicopic 1.2.21.2.2 1.2.21.2.2 1.2.2 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under h operaaaaations as tions as tions as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding finding finding taking the opposite taking the opposite finding taking the opposite finding taking the opposite ocal ocal ecipr ecipr the r the r ocal, taking a root, eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll find out about identities, which don’t change your original value, and inverses, which take the opposite. Key words: identity inverse Check it out: The additive inverse is sometimes called a negative, or an opposite. Identities and Invvvvvererererersessessessesses Identities and In Identities and In Identities and In Identities and In Identities and Invvvvvererererersessessessesses Identities and In Identities and In Identities and In Identities and In The numbers 0 and 1 are special numbers, since they are the identities of addition and multiplication. dition and Multiplicationtiontiontiontion dition and Multiplica e Identities of AdAdAdAdAddition and Multiplica dition and Multiplica e Identities of e Identities of 0 and 1 ar 0 and 1 ar 0 and 1 are Identities of dition and Multiplica e Identities of 0 and 1 ar 0 and 1 ar In the real number system, there’s an additive identity. This means that there’s a number (called zero) which you can add to any number without changing the value of that number. For any number a Œ R There’s also a multiplicative identity. This is a number (called one) which any number can be multiplied by without its value being changed. For any number a Œ R to Make 0e 0e 0e 0e 0 d to Mak se is WW

WWWhahahahahat t t t t YYYYYou ou ou ou ou AdAdAdAdAdd to Mak d to Mak se is TTTTThe he he he he AdAdAdAdAdditiditiditiditiditivvvvve Ine Ine Ine Ine Invvvvverererererse is se is d to Mak se is Every real number has an additive inverse — “additive inverse” is just a more mathematical term for the “negative” of a number. When you add a number to its additive inverse, the result is 0 (the additive identity). Or, more formally: For every real number m, there is an additive inverse written –m. When you add a number to its additive inverse, you get 0 — which is the additive identity. So, for any number m Œ R: –m + m = 0 = m + (–m) The inverse of a negative number is a positive number — that is, –(–m) = m, for any m. And if you take the opposite of a number, you change it into its additive inverse. Guided Practice Find the additive inverse of the following: 1. 6 2. 81 3. –4 4. y 5. –k 6. a + 5 7. –5a – 2 8. 2ab 9. State the value of x if 51 + x = 0. 10. State the value of x if x – 5 = 0. 1616161616 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out: The multiplicative inverse is sometimes called the reciprocal. Check it out: See Topic 1.2.5 for more information about not being able to divide by zero. se to Make 1e 1e 1e 1e 1 se to Mak y the Multiplicatititititivvvvve Ine Ine Ine Ine Invvvvverererererse to Mak se to Mak y the Multiplica Multiply by by by by by the Multiplica y the Multiplica Multipl Multipl se to Mak y the Multiplica Multipl Multipl Multiplication has a property that’s similar to the property of addition: Every real number also has a multiplicative inverse (or “reciprocal”). When you multiply a number by

its multiplicative inverse, the result is 1 — the multiplicative identity. However, there’s an important exception: zero has no reciprocal — its reciprocal cannot be defined. That means that you can’t divide by zero. More formally this becomes: For every nonzero real number m, there is a multiplicative inverse written m–1. When you multiply a number by its multiplicative inverse, you get 1 — which is the multiplicative identity. So, for any number m Œ R, m π 0: m–1 × m = 1 = m × m–1 If m is a nonzero real number then its reciprocal is m–1, which is given by m–1 = 1 m. And the reciprocal of m–1 is m — that is, ( )m− − 1 1 = m, 1 − = m, or 1m ( 1 1 ) = m. m which means that Example Example Example Example Example 11111 Find the reciprocals of: a) 3 b) 12 c) x d) 3–1 Solution Solution Solution Solution Solution 3 1− = 1 3 a) b) 12 1− = 1 12 c) x− =1 1 x d) (3–1)–1 = 3 Guided Practice Find the multiplicative inverse: 12. –31 11. 2 a 14. − 1 b 16 15. 16. Find the value of x if 3x = 1. Independent Practice 13. 1 8 State the additive and multiplicative inverses of each expression below: 1. 6 1 8 4. 2. –11 5. x 3. 1 3 6. (a + b) ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Identity and inverse are just math names for quite simple things. The identity doesn’t change your value, and inverses just mean taking the opposite, or finding the reciprocal. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 1717171717 TTTTTopicopicopicopicopic 1.2.31.2.3 1.2.31.2.3 1.2.3 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use prprprprpro

per umbers s s s s to umber ties of n n n n number umber ties of ties of oper oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll use the number line to show real numbers, and you’ll describe them in terms of absolute values. Key words: real numbers absolute value The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line The Number Line alues alues Absolute VVVVValues Absolute Absolute and and alues and Absolute alues Absolute and and Absolute VVVVValues alues alues Absolute Absolute and and and Absolute alues Absolute and alues and The number line is a useful way of representing numbers visually. This Topic also includes information about absolute values, which show how far from zero numbers are on the number line. oints on a Line oints on a Line s as P s as P eal Number Number Lines Show Rw Rw Rw Rw Real Number eal Number Number Lines Sho Number Lines Sho oints on a Line s as Points on a Line eal Numbers as P oints on a Line s as P eal Number Number Lines Sho Number Lines Sho –6 –5 –4 –3 –2 –1 0 All real numbers can be found on the number line, no matter how big or small they are, and no matter whether they are rational or irrational. And for every point on the number line, there is a real number. 1,000,000,000 –4.5 ÷2 1 2 3 5 1 6 4 2 So there’s a one-to-one correspondence between the real numbers and the points on the number line. Check it out: A “graph” is just a type of “picture” of something mathematical. For each point on the number line, the corresponding real number is called the coordinate of the point. And for each real number, the corresponding point is called the graph of the number. Don’t forget: The “sign” of a number refers to whether it is positive or negative. The numbers to the left of zero on the number line are all negative — they’re less than zero. The numbers to the right of zero are positive — they’re greater than zero. Zero is neither negative nor positive. Guided Practice 1. “Only integers can be found on the

number line.” Is this statement true or false? 2. Identify the corresponding real numbers of points A–E on the number line below. A B C D E –5 –4 –3 –2 –. Draw the graph of 6 on a number line. 4. Draw the graph of –2 on a number line. 1818181818 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 tions tions or Calcula or Calcula he Number Line Can Be Useful f TTTTThe Number Line Can Be Useful f he Number Line Can Be Useful f tions or Calculations he Number Line Can Be Useful for Calcula tions or Calcula he Number Line Can Be Useful f If you think of the number line as a road, then you can think of coordinates as movement along the road — either to the left or to the right, depending on the coordinate’s sign. For example, –5 would indicate a movement of 5 units to the left, while 4 would mean 4 units to the right. Example Example Example Example Example 11111 Find 3 – 2 + 4. Solution Solution Solution Solution Solution Rewriting this as 3 + (–2) + 4, you can interpret this as: “Start at 3, move 2 to the left (to reach 1) and then 4 to the right (to reach 5).” So 3 – 2 + 4 = 5. om 0 om 0 a Number is its Distance fr a Number is its Distance fr alue of Absolute VVVVValue of alue of Absolute TTTTThe he he he he Absolute Absolute om 0 a Number is its Distance from 0 alue of a Number is its Distance fr om 0 a Number is its Distance fr alue of Absolute The opposite of a real number c (that is, –c) lies an equal distance from zero as c, but on the other side of zero. So the opposite of 4 (which lies 4 units to the right of zero) is –4 (which lies 4 units to the left of zero). And the opposite of –7 (which lies 7 units to the left of zero) is 7 (which lies 7 units to the right of zero). The distance from zero to a number is called the number’s absolute value. It doesn’t matter whether it is to the left or to the right of zero — so absolute value just means the “size”

of the number, ignoring its sign. The absolute value of c is written |c|. More algebraically... = c ⎧ c ⎪⎪⎪⎪ ⎪⎪⎪⎪ c if ⎨ c if 0 − c ⎩ c if > 0 = 0 < 0 The absolute value of a number can never be negative. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 1919191919 Example Example Example Example Example 22222 Find: a) |6| b) |0| c) |–23| Solution Solution Solution Solution Solution a) 6 is positive, so |6| = 6. b) |0| = 0 (by definition) c) –23 is negative, so |–23| = –(–23) = 23. Guided Practice Find the following absolute values. 5. |6| 7. |–3| 9. |0| 11. | 2 | 6. |15| 8. |–8| 10. | 1 2 | 12. − 1 3 Independent Practice 1. Choose the correct word from each pair to complete this sentence. On a number line, positive numbers are found to the (left/right) of zero and negative numbers are found to the (left/right) of zero. 2. On a single number line, draw the graphs with the following coordinates: 5 –3.5 0.5 In exercises 3–11, find x. 3. |3| = x 6. |3.14| = x 9. |465| = x 4. |–10.5| = x 7. |–2.17| = x 10. |–465| = x 11. |x| = 465 (Hint: Look at exercises 9 and 10) 12. What is wrong with the equation |x| = –1? 5. |–2| = x 8. |x| = 0 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You’ve seen the number line plenty of times in earlier grades, but it’s always useful. You don’t always need to draw it out, but you can imagine a number line to work out which direction an operation will move a number. 2020

202020 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 TTTTTopicopicopicopicopic 1.2.41.2.4 1.2.41.2.4 1.2.4 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n incincincincincluding c losureeeee losur losur luding c luding c luding closur losur luding c prprprprproper our basic our basic or the f or the f ties f ties f oper oper or the four basic our basic ties for the f operties f our basic or the f ties f oper tions wherherherherhereeeee tions w tions w arithmetic operaaaaations w arithmetic oper arithmetic oper arithmetic oper tions w arithmetic oper pplicabbbbblelelelele..... pplica aaaaapplica pplica pplica What it means for you: You’ll see how adding or multiplying positive and negative numbers affects the sign of the result. Key words: sum product addend Don’t forget: You don’t need to write the sign of a positive number — you can write +12 as 12. dition and Multiplicationtiontiontiontion dition and Multiplica dition and Multiplica AdAdAdAdAddition and Multiplica dition and Multiplica AdAdAdAdAddition and Multiplica dition and Multiplicationtiontiontiontion dition and Multiplica dition and Multiplica dition and Multiplica All real numbers fall into one of the following three categories: positive, zero, or negative. This is also true for the results of adding or multiplying real numbers (since R is closed under addition and multiplication). d Numbersssss hen YYYYYou ou ou ou ou AdAdAdAdAdd Number d Number d Number hen ou Get WW

WWWhen hen ou Get he Sum is WWWWWhahahahahat t t t t YYYYYou Get ou Get he Sum is TTTTThe Sum is he Sum is d Number hen ou Get he Sum is When you add two real numbers a and b (find the sum a + b), the result can be positive, zero, or negative, depending on a and b. The sum of any two real numbers a and b is: 1. Positive if both addends are positive. This means that (a + b) > 0 if a > 0 and b > 0. This is the same as saying that the set of positive real numbers is closed under addition — so if you add two positive real numbers, you always get another positive real number. 2. Negative if both addends are negative. This means that (a + b) < 0 if a < 0 and b < 0. This is the same as saying that the set of negative real numbers is closed under addition — so if you add two negative real numbers, you always get another negative real number. 3. Positive or Negative or Zero if a and b have opposite signs. In this case, the sign of (a + b) is the same as the sign of the addend with the larger absolute value. For example, (–5) + 7 = 7 + (–5) = 2 — positive, since |7| > |–5|. (The positive addend has a larger absolute value, so the sum is positive.) However, (–7) + 5 = 5 + (–7) = –2 — negative, since |–7| > |5|. (The negative addend has a larger absolute value, so the sum is negative.) Guided Practice State with a reason the sign of each sum, then find the sum. 1. 3 + 9 2. –3 + (–9) 3. –10 + (–5) 4. –3 + 9 5. 3 + (–9) 6. 10 + (–15) Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2121212121 Check it out: The numbers being multiplied together are called factors. ou Multiplyyyyy hen YYYYYou Multipl ou Multipl ou Multipl hen ou Get WWWWWhen hen ou Get oduct is WWWWWhahahahahat t t t

t YYYYYou Get ou Get oduct is oduct is he Pr TTTTThe Pr he Pr he Product is ou Multipl hen ou Get oduct is he Pr In a similar way, the sign of the product of any real numbers depends on the signs of the numbers being multiplied (the factors). The rules for the sign of the product of any two real numbers are as follows: Signs of the factors: Sign of the product: + + – + – – + + – For example, 5 × 2 = 10 — positive. The set of positive real numbers is closed under multiplication. For example, –5 × –2 = 10 — positive. The set of negative real numbers is not closed under multiplication. For example, 5 × –2 = –10 — negative. So the product of any real numbers is: 1. Positive if the expression contains an even number of negative factors (or only positive factors). For example, 5 × (–2) × 2 × (–3) × (–1) × (–4) = 240 This is positive, since the expression has four negative factors — an even number. 2. Negative if the expression contains an odd number of negative factors. For example, (–2) × 3 × (–4) × (–2) = –48 This is negative, since the expression has three negative factors — an odd number. 2222222222 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Guided Practice State with a reason the sign of each product, then find the product. 7. 2 × 6 8. 7 × (–7) 9. (–7) × (–7) 10. 9 × 10 11. (–11) × (–3) 12. (–1) × 3 × 4 Independent Practice State with a reason the signs of the following expressions. 1. (–5) + 3 2. 2 + (–8) 3. 5 × 3 4. (–10) × (–2) 5. (–2) × (–1) × (–8) × 6 × (–2) Evaluate the following numerical expressions. 6. –8 + (–2) 7. 8 + (–2) 8. 6 × (–4) 9. –8 + 2 10. –8 + 10 11. (–2) × (–5) × 4 12.

(–5) + 2 + 6 13. (–3) × (–5) × (–2) State the signs of the following expressions. 14. a3 where a < 0. 15. –3a2b2 where a < 0 and b > 0. 16. a2b3c7 where a > 0, b < 0, and c < 0 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You’ve done addition and multiplication a lot in previous grades — but a lot of Algebra I is about making formal rules for math methods. After a while you’ll carry out the rules of changing sign without really thinking about them — but for now you need to make sure you remember them. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2323232323 TTTTTopicopicopicopicopic 1.2.51.2.5 1.2.51.2.5 1.2.5 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n incincincincincluding c losureeeee losur losur luding c luding c luding closur losur luding c prprprprproper our basic our basic or the f or the f ties f ties f oper oper or the four basic our basic ties for the f operties f our basic or the f ties f oper tions wherherherherhereeeee tions w tions w arithmetic operaaaaations w arithmetic oper arithmetic oper arithmetic oper tions w arithmetic oper pplicabbbbblelelelele..... pplica aaaaapplica pplica pplica 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under

h operaaaaations as tions as tions as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding finding finding taking the opposite taking the opposite finding taking the opposite finding taking the opposite ocal ocal ecipr ecipr the r the r ocal, taking a root, eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll see how adding or multiplying positive and negative numbers affects the sign of the result. Key words: difference quotient reciprocal commutative Check it out: See Topic 1.2.7 for more on what “commutative” means. vision vision action and Di action and Di Subtr Subtr vision action and Division Subtraction and Di vision action and Di Subtr Subtr vision vision action and Di action and Di Subtr Subtr action and Division Subtraction and Di vision action and Di Subtr vision Subtr After addition and multiplication you can probably guess what’s next. This Topic gives more formal rules for subtraction and division that you’ve seen in earlier grades. ding the AdAdAdAdAdditiditiditiditiditivvvvve Ine Ine Ine Ine Invvvvvererererersesesesese ding the action Means AdAdAdAdAdding the ding the action Means action Means Subtr Subtr Subtraction Means ding the action Means Subtr Subtr Subtraction is the inverse operation of addition. It is defined as the addition of the opposite of a number. So to subtract a from b (that is, to find b – a), you add –a to b. For any a, b Œ R: b – a = b + (–a) Finding the difference between two numbers means you subtract them. For example, the difference between a and b would be a – b or b – a. Since subtraction is a type of addition, and R is closed under addition, R must be closed under subtraction. Subtraction is not commutative (meaning that a – b π b – a). Example Example Example Example Example 11111 Show that a – b π b – a for any a π b. Solution Solution Solution Solution Solution Write both subtractions

as additions: a – b = a + (–b) b – a = b + (–a) = (–a) + b The two subtractions “a – b” and “b – a” contain different addends when they’re written as additions — so a – b π b – a. Subtracting a negative number is the same as adding a positive number (since subtracting means adding the opposite, and the opposite of a negative number is a positive number). For any a, b Œ R: a – (–b) = a + b 2424242424 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Example Example Example Example Example 22222 a) Rewrite 11 – 7 as an addition. b) Simplify –7 – (–5). c) Find the sum –3 + (–8). Solution Solution Solution Solution Solution a) 11 – 7 = 11 + (–7) b) As the opposite of –5 is 5, subtracting –5 means adding 5. So –7 – (–5) = –7 + 5 = –2. c) –3 + (–8) = –(3 + 8) = –11 Guided Practice Rewrite the following subtractions as additions: 1. 9 – 6 2. 10 – 4 3. –4 – (–10) Evaluate the following: 4. 6 – (–5) 5. –3 + (–10) 6. –a – (–a) ocal ocal ecipr y a Ry a Recipr ecipr y a Ry a R ying b ying b vision Means Multipl DiDiDiDiDivision Means Multipl vision Means Multipl ocal eciprocal ying by a R vision Means Multiplying b ocal ecipr ying b vision Means Multipl Check it out: The reciprocal is another name for the multiplicative inverse. Division is the inverse operation of multiplication. It’s defined as multiplication by a reciprocal. So to divide b by a (that is, to find b ÷ a), you multiply b by a–1. For any a, b (a π 0) Œ R1 1 a One number divided by another is called a quotient. Division is not commutative (meaning that a ÷ b π b ÷ a

). Example Example Example Example Example 33333 Show that a ÷ b π b ÷ a for any a π b, a, b π 0. Solution Solution Solution Solution Solution Write both divisions as multiplications: a ÷ b = a × (b–1) b ÷ a = b × (a–1) The two divisions “a ÷ b” and “b ÷ a” contain different factors when they’re written as multiplications — so a ÷ b π b ÷ a. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2525252525 Dividing by a number’s reciprocal is the same as multiplying by the number itself. For any a, b (b π 0) Œ R ) And lastly, you cannot divide by 0 — since 0 has no reciprocal. Example Example Example Example Example 44444 a) Rewrite 25 ÷ 5 as a multiplication. b) Simplify 7 ÷ 5–1. Solution Solution Solution Solution Solution a) 25 ÷ 5 = 25 × 1 5 (or 25 × 0.2) b) The reciprocal of 5–1 is 5, so dividing by 5–1 means multiplying by 5. Therefore 7 ÷ 5–1 = 7 × 5 = 35. Guided Practice Rewrite these expressions as multiplications. 7. 24 ÷ 6 8. 36 ÷ 3 Evaluate the following expressions. 10. 10 ÷ 1 5 11. 16 ÷ 2–1 9. 2 ÷ 1 8 ⎛ 12. 16 ÷ 1 ⎜⎜⎜ ⎝ 8 − ⎞ 1 ⎟⎟⎟ ⎠ Independent Practice Simplify. 1. 12 – 15 4. –4 – 2 7. 15 ÷ 3–1 2. 18 + (–3) 3. 36 – (–4) 5. 20 ÷ (–2) 8. 18 ÷ ⎛ ⎜⎜⎜ ⎝ 1 6 − ⎞ 1 ⎟⎟⎟ ⎠ 6. 10 ÷ 1 2 9. Use examples to demonstrate that subtraction and division are not commutative. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up

RRRRRound Up ound Up ound Up Check over the part on reciprocals until you’re sure you understand it. Although the notation is tricky, the actual ideas behind it should make sense if you read through it carefully. 2626262626 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 TTTTTopicopicopicopicopic 1.2.61.2.6 1.2.61.2.6 1.2.6 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use prprprprproper umbers s s s s to umber ties of n n n n number umber ties of ties of oper oper operties of umber ties of oper demonstrate whether assertions are true or false. What it means for you: You’ll see why it’s important to have a set of rules about the order in which you have to deal with operations. Key words grouping symbols parentheses brackets braces exponents tions tions Operaaaaations Oper Oper der of der of OrOrOrOrOrder of tions der of Oper tions Oper der of Operaaaaations OrOrOrOrOrder of tions tions Oper Oper der of der of der of Oper tions Oper der of tions It’s important that all mathematicians write out expressions in the same way, so that anyone can reach the same solution by following a set of rules called the “order of operations.” k Out Firststststst k Out Fir t to WWWWWororororork Out Fir k Out Fir t to ouping Symbols Show w w w w YYYYYou ou ou ou ou WWWWWhahahahahat to t to ouping Symbols Sho GrGrGrGrGrouping Symbols Sho ouping Symbols Sho k Out Fir t to ouping Symbols Sho If you wanted to write a numerical expression representing “add 4 and 3, then multiply the answer by 2,” you might be tempted to write 4 + 3 × 2. But watch out — this expression contains an addition and a multiplication, and you get different answers depending on which you do first. If you do the addition first, you get the answer 7 × 2 = 14. If you do the multiplication first, you get

the answer 4 + 6 = 10. You might know the addition has to be done first, but somebody else might not. To be really clear which parts of a calculation have to be done first, you can use grouping symbols. Some common grouping symbols are: parentheses ( ), brackets [ ], and braces { }. Example Example Example Example Example 11111 Write an expression representing the phrase “add 4 and 3, then multiply the answer by 2.” Solution Solution Solution Solution Solution You need to show that the addition should be done first, so put that part inside grouping symbols: The expression should be (4 + 3) × 2. Guided Practice Write numeric expressions for these phrases: 1. Divide 4 by 8 then add 3. 2. Divide 4 by the sum of 8 and 3. 3. From 20, subtract the product of 8 and 2. 4. From 20, subtract 8 and multiply by 2. Evaluate the following sums and differences: 5. (3 – 2) – 5 6. 3 – (2 – 5) 7. 6 – (11 + 7) 8. (7 – 8) – (–3 – 8) – 11 9. (5 – 9) – (3 – 10) – 2 10. 9 + (5 – 3) – 4 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 2727272727 Nested grouping symbols are when you have grouping symbols inside other grouping symbols. When you see nested grouping symbols, you always start from the inside and work outwards. Example Example Example Example Example 22222 Evaluate {5 – [11 – (7 – 2)]} + 34. Solution Solution Solution Solution Solution Start from the inside and work outwards: {5 – [11 – (7 – 2)]} + 34 = {5 – [11 – 5]} + 34 = {5 – 6} + 34 = –1 + 34 = 33 Guided Practice Evaluate the following: 11. 8 + [10 + (6 – 9) + 7] 12. 9 – {[(–4) + 10] + 7} 13. [(13 – 12) + 6] – (4 – 2) 14. 14 – {8 + [5 – (–2)]} – 6 15. 13 + [10 – (4 + 5)] – (11 + 8) 16. 10 + {[7 – (–2)] – (3 – 1

)} + (–14) Check it out See Topic 1.3.1 for more about exponents. Check it out When you’re simplifying expressions within grouping symbols, follow steps 2–4 (see Example 3). te Firststststst te Fir t to Evaluaaluaaluaaluaaluate Fir te Fir t to Ev About WWWWWhahahahahat to Ev t to Ev About About ules ules e Other R TTTTTherherherherhere are are are are are Other R e Other R ules About e Other Rules te Fir t to Ev About ules e Other R This order of operations is used by all mathematicians, so that every mathematician in the world evaluates expressions in the same way. 1. First calculate expressions within grouping symbols — working from the innermost grouping symbols to the outermost. 2. Then calculate expressions involving exponents. 3. Next do all multiplication and division, working from left to right. Multiplication and division have equal priority, so do them in the order they appear from left to right. 4. Lastly, do any addition or subtraction, again from left to right. Addition and subtraction have the same priority, so do them in the order they appear from left to right too. 2828282828 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Example Example Example Example Example 33333 Simplify {4(10 – 3) + 32} × 5 – 11. Solution Solution Solution Solution Solution Work through the expression bit by bit: {4(10 – 3) + 32} × 5 – 11 = {4 × 7 + 32} × 5 – 11 inner ouping symbols firststststst ouping symbols fir ouping symbols fir most grrrrrouping symbols fir most g most g inner inner innermost g ouping symbols fir most g inner Now you have to calculate everything inside the remaining grouping symbols: = {4 × 7 + 9} × 5 – 11 = {28 + 9} × 5 – 11 = 37 × 5 – 11 xponent xponent k out the e k out the e firfirfirfirfirst wst wst wst wst wororororork out the e xponent k out the exponent xponent k out the e ultiplicationtiontiontiontion ultiplica ultiplica then do the m then do the m then do

the multiplica ultiplica then do the m then do the m dition dition then do the ad then do the ad dition then do the addition dition then do the ad then do the ad Now there are no grouping symbols left, so you can do the rest of the calculation: = 185 – 11 = 174 tion firststststst tion fir tion fir ultiplica ultiplica do the m do the m ultiplication fir do the multiplica tion fir ultiplica do the m do the m action action y the subtr y the subtr and finall and finall action y the subtraction and finally the subtr action y the subtr and finall and finall Guided Practice Evaluate the following: 17. 24 ÷ 8 – 2 18. 32 – 4 × 2 19. 21 – {–3[–5 × 4 + 32] – 9 × 23} + 17 ) 4 − − ( 2 3 3 ⎡ + ÷ + −( 4 10 ⎣ ⎡ 12 ⎣⎢ 3 ⎤ ) ⎦ ⎤ ⎦⎥ 3 20. 21. 8 + {10 ÷ [11 – 6] × (–4)} 22. [17 + {(–33 + 4) × 8 – 17}] ÷ 8 Independent Practice Evaluate the following. 1. 14 – [9 – 4 × 2] 2. 11 + (9 – 3) – (4 ÷ 2) 3. (–1) × (7 – 10 + 12) 4. 12 + 9 × [(1 + 2) – (6 – 14)] 5. [(11 – 8) + (7 – 2)] × 3 – 13 6. [(10 + 9 + 5) ÷ 2] – (6 – 12) Insert grouping symbols in each of the following statements so that each statement is true: 7. 12 + 4 2 × 24 – 18 ÷ 3 = 44 8. 20 + 3 2 – 14 – 12 × 6 = 517 9 = 1662 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up You really need to learn the order of operation rules. You’ll be using them again and again in Algebra I so you might as well make sure you remember them right now. Section 1.2 Section 1.2

Section 1.2 — The Real Number System Section 1.2 Section 1.2 2929292929 TTTTTopicopicopicopicopic 1.2.71.2.7 1.2.71.2.7 1.2.7 California Standards: 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and 1.0: Students identify and use the arithmetic use the arithmetic use the arithmetic use the arithmetic use the arithmetic prprprprproper ties of subsets of ties of ties of oper oper operties of ties of oper integers and rational, umbersssss,,,,, umber umber eal n eal n irrational, and rrrrreal n eal number umber eal n including closure properties for the four basic arithmetic operations where applicable. What it means for you: You’ll learn about the commutative, associative, and distributive laws of the real numbers. Key words commutative associative distributive binary operation eal Numbersssss eal Number eal Number ties of R R R R Real Number ties of ties of oper oper PrPrPrPrProper operties of eal Number ties of oper ties of R R R R Real Number PrPrPrPrProper eal Numbersssss eal Number eal Number ties of ties of oper oper operties of ties of oper eal Number In this Section you’ve already seen lots of real number properties — and now it’s time for some more. These rules are really important because they tell you exactly how to deal with real numbers. tter tter t Ma der Doesn’’’’’t Ma t Ma der Doesn der Doesn ws — the Or CommCommCommCommCommutautautautautatititititivvvvve Lae Lae Lae Lae Laws — the Or ws — the Or tter t Matter ws — the Order Doesn tter t Ma der Doesn ws — the Or It doesn’t matter which order you add two numbers in — the result is the same. This is called the commutative property (or law) of addition. For any a, b Œ R: a + b = b + a The same is true of multiplication — this is the commutative property (or law) of multiplication. For any a, b Œ R: a ×

b = b × a For example: 2 + 3 = 3 + 2 — commutative property of addition 2 × 3 = 3 × 2 — commutative property of multiplication oup Numbers s s s s AnAnAnAnAny y y y y WWWWWaaaaayyyyy oup Number oup Number ou Can Gr ws — YYYYYou Can Gr ou Can Gr ws — Associatititititivvvvve Lae Lae Lae Lae Laws — ws — Associa Associa ou Can Group Number oup Number ou Can Gr ws — Associa Associa Addition and multiplication are binary operations — you can only add or multiply two numbers at a time. So to add three numbers, for example, you add two of them first, and then add the third to the result. However, it doesn’t matter which two you add first: For any a, b, c Œ R: (a + b) + c = a + (b + c) Again, the same is true of multiplication: For any a, b, c Œ R: (a × b) × c = a × (b × c) These are called the associative properties (laws) of addition and multiplication. For example: (2 + 3) + 4 = 2 + (3 + 4) — associative property of addition (2 × 3) × 4 = 2 × (3 × 4) — associative property of multiplication 3030303030 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out In c), it’s fine to leave the parentheses out. It doesn’t matter if you do (3x + 3y) + 3z or 3x + (3y + 3z) — the result is the same. y Out Parararararentheses entheses entheses y Out P y Out P ws — Multipl Distributiutiutiutiutivvvvve Lae Lae Lae Lae Laws — Multipl ws — Multipl Distrib Distrib entheses ws — Multiply Out P entheses y Out P ws — Multipl Distrib Distrib The distributive law of multiplication over addition defines how multiplication and addition combine. When you multiply by a sum in parentheses, it’s the distributive law that you

’re using. For any a, b, c Œ R: a × (b + c) = (a × b) + (a × c) So a factor outside parentheses multiplies every term inside. Example Example Example Example Example 11111 Expand: a) 3(x + y) b) x(3 + y) c) 3[(x + y) + z] Solution Solution Solution Solution Solution a) 3(x + y) = 3x + 3y b) x(3 + y) = x·3 + xy = 3x + xy (using the commutative law of multiplication) c) 3[(x + y) + z] = 3(x + y) + 3z = (3x + 3y) + 3z = 3x + 3y + 3z Independent Practice Identify the property that makes each of the following statements true. 1. 8 × 3 = 3 × 8 2. 2(x + 3) = 2x + 6 3. 9 + 2 = 2 + 9 4. (9 + 2) + 1 = 9 + (2 + 1) 5. 4(p + 1) = 4p + 4 6. (9 × 4) × 5 = 9 × (4 × 5) Expand: 7. a(b + c) 8. 6(t + 5) 9. 12t(s – r) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The thing with these laws is that you’ve been using them for years without thinking about them, so it might seem strange to be taught them now. But when you rearrange equations, you’re using these laws. In Algebra I you can use these laws to justify anything you do to an equation. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 3131313131 TTTTTopicopicopicopicopic 1.2.81.2.8 1.2.81.2.8 1.2.8 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and

use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,,,,, finding taking the opposite taking the opposite taking the opposite taking the opposite the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll learn about multiplicative opposites, and numbers that don’t change a value when you multiply by them. Key words inverse identity reciprocal Check it out See Topic 1.2.5 for more on additive inverses. Check it out A minus sign outside parentheses changes the sign of everything inside the parentheses. e on Multiplicationtiontiontiontion e on Multiplica e on Multiplica MorMorMorMorMore on Multiplica e on Multiplica MorMorMorMorMore on Multiplica e on Multiplicationtiontiontiontion e on Multiplica e on Multiplica e on Multiplica You already dealt with changing signs during multiplication in Topic 1.2.4. This Topic is all about opposites in multiplication, and numbers that don’t change a value if you multiply by them. y –1 to Find the AdAdAdAdAdditiditiditiditiditivvvvve Ine Ine Ine Ine Invvvvverererererse or Opposite se or Opposite se or Opposite y –1 to Find the Multiply by by by by by –1 to Find the y –1 to Find the Multipl Multipl se or Opposite se or Opposite y –1 to Find the Multipl Multipl Every real number has an additive inverse — a number that will give zero when added to that number. So for every number m, there is an opposite number (or negative), –m. In particular, the number 1 has the additive inverse –1. And if you multiply any number by –1, you get that number’s inverse. For any m Œ R: –m = –1 × m = m × –1 So the opposite of 4 is –1 × 4 = –4. And the opposite of x is –1 × x = –x. Example Example Example Example Example 11111 Find the opposite of (2a + 3). Solution Solution Solution Solution Solution The question is asking you to find –(2a + 3). Find the opposite by multiplying by –1: –(2a + 3)

= –1 × (2a + 3) = [(–1) × (2a)] + [(–1) × (3)] = [(–1 × 2) × a)] + (–3) = –2a + (–3) = –2a – 3 Guided Practice Find the opposite of each of these expressions. 1. t 2. 3 3. t + 3 4. a – 2 5. –x + 1 6. –y – 4 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Anything Multiplied by 0 is 0 Zero is a special number — it’s the additive identity. This means that for any number m. But zero has another useful property as well... 3232323232 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 The product of any real number and 0 equals 0. For any m Œ R Check it out See Topic 1.2.5 for the definition of division. One result of this is that zero doesn’t have a reciprocal, since there is no number that you can possibly multiply zero by in order to get 1 — the multiplicative identity. What this means in practice is that you can’t divide by zero. Check it out See Topic 7.1.1 for more examples of solving equations using this rule. s and Get Zerooooo s and Get Zer s and Get Zer o Number o Nonzererererero Number o Number o Nonz t Multiply y y y y TTTTTwwwwwo Nonz o Nonz t Multipl ou Can’’’’’t Multipl t Multipl ou Can YYYYYou Can ou Can o Numbers and Get Zer s and Get Zer o Number o Nonz t Multipl ou Can If nonzero real numbers are multiplied together, the result is never zero. The product of two (or more) nonzero real numbers cannot be zero. If x, y Œ R, and xy = 0, then either x = 0 or y = 0 (or both x = 0 and y = 0). This has practical uses when it comes to solving equations. Example Example Example Example Example 22222 Find two possible solutions to the equation x(x – 1) = 0. Solution Solution Solution Solution Solution

There are two expressions multiplied together to give zero. Either one or the other must equal 0, so x = 0 or x – 1 = 0 — that is, x = 0 or x = 1. (You need to write “or,” since x can’t be both 0 and 1 at the same time.) Independent Practice Find the opposite. 1. –4 2. –a 3. g + 5 Solve each equation. 6. x + 1 = 0 7. y – 4 = 0 4. t – 6 5. –b + 8 8. y(y + 2) = 0 9. t(t – 3) = 0 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up These rules might just sound like common sense — but it’s important to write statements in formal math-speak to prove that they’re true. You’ll use these rules throughout Algebra I. Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 3333333333 TTTTTopicopicopicopicopic 1.2.91.2.9 1.2.91.2.9 1.2.9 California Standards: 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use 1.1: Students use ties of n n n n number prprprprproper umbers tos tos tos tos to umber umber ties of ties of oper oper operties of umber ties of oper hether hether demonstraaaaate wte wte wte wte whether demonstr demonstr hether demonstr hether demonstr ue or falsealsealsealsealse..... ue or f tions are tre tre tre tre true or f ue or f tions ar tions ar asser asser assertions ar ue or f tions ar asser asser What it means for you: You’ll see that the properties of real numbers that you learned in this section are useful to prove whether math statements are true. Key words theorem axiom postulate closure identity inverse commutative associative distributive Axioms Axioms eal Number eal Number TTTTThe Rhe Rhe Rhe Rhe Real Number Axioms eal Number Axioms Axi

oms eal Number TTTTThe Rhe Rhe Rhe Rhe Real Number Axioms Axioms eal Number eal Number eal Number Axioms Axioms eal Number Axioms Most of what has been covered in this Section has been about the axioms (or postulates) of the real number system. This Topic gives a summary of the axioms, and shows how you can use them. Assumptions Assumptions e Fundamental e Fundamental Axioms ar Axioms ar Assumptions e Fundamental Assumptions Axioms are Fundamental Assumptions e Fundamental Axioms ar Axioms ar A theorem is a statement that can be proved. An axiom, on the other hand, is a fundamental assumption — a statement that is accepted as true without having to be proved. You can use these axioms to justify solution steps when simplifying mathematical expressions, proving theorems, solving equations, and supporting mathematical arguments. You have to know all the axioms in this section, along with their names. Here’s a summary of them all together: Property Name Addition Multiplication Closure Property: a + b is a real number is a real number 3 × 5 = 15 Œ R Identity Property Inverse Property: a + (–a) = 0 = –a + a 3 + (–3) = 0 = (–3) + 3 a × a–1 = 1 = a–1 × a 3 × 3–1 = 1 = 3–1 × 3 Commutative Property Associative Property: (a + b) + c = a + (b + c) (3 + 5) + 6 = 3 + (5 + 6) (ab)c = a(bc) (3 × 5) × 6 = 3 × (5 × 6) Distributive Property of Multiplication over Addition: a(b + c) = ab + ac 3(5 + 6) = 3 × 5 + 3 × 6 and (b + c)a = ba + ca (5 + 6)3 = 5 × 3 + 6 × 3 3434343434 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 Check it out Any of the properties covered in the section as well as the axioms on the previous page can be used to justify steps. ps in Proofsoofsoofsoof

soofs ps in Pr ps in Pr ustify Ste ustify Ste Axioms to J Axioms to J Use the Use the ustify Steps in Pr Axioms to Justify Ste Use the Axioms to J ps in Pr ustify Ste Axioms to J Use the Use the These axioms can be used to justify steps in a mathematical proof. Example Example Example Example Example 11111 Show that (x + y) – x = y. Justify your steps. Solution Solution Solution Solution Solution (x + y) – x = (y + x) + (–x) dition, dition, ty of ad ad ad ad addition, ty of ty of oper oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper dition, operty of dition, ty of oper action action subtr subtr and the definition of and the definition of action subtraction and the definition of subtr action subtr and the definition of and the definition of = y + [x + (–x)] dition dition ty of ad ad ad ad addition ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa dition operty of dition ty of oper Associa Associa = y + 0 = y an additiditiditiditiditivvvvve ine ine ine ine invvvvvererererersesesesese an ad an ad Definition of Definition of Definition of an ad an ad Definition of Definition of dition dition ty of ad ad ad ad addition ty of ty of oper oper Identity pr Identity pr dition operty of Identity proper dition ty of oper Identity pr Identity pr Example Example Example Example Example 22222 Check it out You don’t usually need to write down how to justify each step — but you should know how to. Show that (x + y)(x – y) = x2 – y2. Justify your steps. Solution Solution Solution Solution Solution (x + y)(x – y) = (x + y)(x + (–y)) Definition of Definition of subtr subtr action action Definition of subtr subtraction action Definition of Definition of subtr action = (x + y)x + (x + y)(–y) Distrib Distrib Distributiutiuti

utiutivvvvve lae lae lae lae lawwwww Distrib Distrib = x2 + yx + x(–y) + y(–y) Distrib Distrib Distributiutiutiutiutivvvvve lae lae lae lae lawwwww Distrib Distrib = x2 + xy + x(–1·y) + y(–1·y) CommCommCommCommCommutautautautautatititititivvvvve lae lae lae lae law ofw ofw ofw ofw of ×, ×, ×, ×, ×, and m and mand m ultiplica ultiplica tion pr tion pr oper oper opertytytytyty and multiplica ultiplication pr tion proper and m ultiplica tion pr oper of –1 of –1 of –1 of –1 of –1 = x2 + xy + [x(–1)]y + [y(–1)]y Associa Associa Associa Associatititititivvvvve lae lae lae lae law ofw ofw ofw ofw of × × × × × Associa = x2 + xy + (–xy) + (–1)(y2) CommCommCommCommCommutautautautautatititititivvvvve lae lae lae lae law ofw ofw ofw ofw of ×, ×, ×, ×, ×, opertytytytyty oper oper tion pr tion pr ultiplica ultiplica and m and mand m tion proper ultiplication pr and multiplica oper tion pr ultiplica and m of –1 of –1 of –1 of –1 of –1 = x2 + 0 + (–y2) ty of + + + + + ty of ty of oper oper se pr se pr InInInInInvvvvverererererse pr operty of se proper ty of oper se pr = x2 – y2 and and +, +, ty of ty of oper oper Identity pr Identity pr and +, and ty of +, operty of Identity proper and +, ty of oper Identity pr Identity pr action action subtr subtr definition of definition of action subtraction definition of subtr action subtr definition of definition of Section 1.2 Section 1.2 Section 1.

2 — The Real Number System Section 1.2 Section 1.2 3535353535 Independent Practice State the real number property that justifies each statement below: 1. 3 + 5 is a real number. 2. 7 × 2 is a real number. 3. m + c = c + m for any real numbers m and c. 4. mc = cm for any real numbers m and c. 5. 12 × (7 × 4) = (12 × 7) × 4 6. 5(m – v) = 5m – 5v 7. 8 + (7 + 4) = (8 + 7) + 4 8. m–1 × m = m × m–1 = 1 9 10. 10 × 1 = 1 × 10 = 10 11. The following is a proof showing that 0c = 0, for any real c. Fill in the missing properties to support each step in the proof. 0c = 0c + 0 = 0c + [c + (–c)] = [0c + c] + (–c) = [0c + 1c] + (–c) = [0 + 1]c + (–c) = 1c + (–c) = c + (–c) = 0 Inverse property of + Identity property of × Identity property of × 12. Given real numbers m, c, and v, m(c – v) = mc – mv. Fill in the missing properties to support each step in the proof. m(c – v) = m[c + (–v)] = mc + m(–v) = mc + m(–1 × v) = mc + [m·(–1)] × v = mc + [(–1)·m] × v = mc + (–1)(mv) = mc + (–mv) = mc – mv Distributive property of × over + Associative property of × Associative property of × Definition of subtraction ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This Section is full of rules and properties. These aren’t just useless lists of abstract rules, though — you’ll be using them throughout Algebra I. You’ve been using most of these rules in previous grades without realizing it — but now you know all the proper names for them too. You

’ll sometimes have to state which rules you’re using when you’re doing math problems. 3636363636 Section 1.2 Section 1.2 Section 1.2 — The Real Number System Section 1.2 Section 1.2 TTTTTopicopicopicopicopic 1.3.11.3.1 1.3.11.3.1 1.3.1 Section 1.3 Exponent Lawswswswsws Exponent La Exponent La Exponent La Exponent La Exponent Lawswswswsws Exponent La Exponent La Exponent La Exponent La California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding taking a root,oot,oot,oot,oot, taking a r taking a r ocal, ocal, ecipr ecipr the r the r ocal, taking a r eciprocal, the recipr ecipr the r taking a r ocal, the r actional actional aising to a fr aising to a fr and r and r actional aising to a fractional and raising to a fr actional aising to a fr and r and r popopopopowwwwwererererer..... TTTTThehehehehey under stand stand y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of xponents..... eeeeexponents xponents xponents xponents What it means for you: You’ll learn about the rules of exponents. Key words: exponent base power product quotient Exponents have a whole set of rules to make sure that all mathematicians deal with them in the same way. There are a lot of rules written out in this Topic, so take care. tions tions ted Multiplica s are Re Re Re Re Reeeeepeapeapeapeapeated Multiplica ted Multiplica s ars ar s ar PPPPPooooowwwwwererererers ar tions ted Multiplications tions ted Multipl

ica A power is a multiplication in which all the factors are the same. For example, m2 = m × m and m3 = m × m × m are both powers of m. In this kind of expression, “m” is called the base and the “2” or “3” is called the exponent. Example Example Example Example Example 11111 a) Find the volume of the cube shown. Write your answer as a power of e. b) If the edges of the cube are 4 cm long, what is the volume? e e e Solution Solution Solution Solution Solution a) V = e × e × e = e3 b) V = e3 = (4 cm)3 = 43 cm3 = 64 cm3 Guided Practice Expand each expression and evaluate. 1. 23 2. 32 3. 52 × 32 4. 24y3 5. Find the area, A, of the square shown. Write your answer as a power of s. 6. If the sides of the square are 7 inches long, what is the area? s s 7. Find the volume of a cube if the edges are 2 feet long. (Volume V = e3, where e is the edge length.) Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 3737373737 Exponents Exponents ules of e Lots of R R R R Rules of ules of e Lots of TTTTTherherherherhere are are are are are Lots of e Lots of Exponents ules of Exponents Exponents ules of e Lots of 1) If you multiply m2 by m3, you get m5, since: m2 × m3 = (m × m) × (m × m × m = m5 The exponent of the product is the same as the exponents of the factors added together. This result always holds — to multiply powers with the same base, you simply add the exponents. ma × mb = ma+b 2) In a similar way, to divide powers, you subtract the exponents. ma ÷ mb = ma–b 3) When you raise a power to a power, you multiply the exponents — for example, (m3)2 = m3 × m3 = m6. 4) Raising a product or quotient to a power is the same as raising each of its elements to that

power. For example: (mb)3 = mb × mb × mb = (m × b) × (m × b) × (m × b = m3b3. (ma)b = mab (mb)a = maba ⎛ ⎜⎜⎜ ⎝ m b ⎞ a ⎟⎟⎟ = ⎠ a m a b 5) Using rule 1 above: ma × m0 = ma + 0 = ma. So m0 equals 1. m0 = 1 6) It’s also possible to make sense of a negative exponent: ma × m–a = ma–a = m0 = 1 (using rules 1 and 5 above) So the reciprocal of ma is m–a. − 1 a )m ( −= m a = 1 a m 7) And taking a root can be written using a fractional power. 1 n= a n a These rules always work, unless the base is 0. The exponents and the bases can be positive, negative, whole numbers, or fractions. The only exception is you cannot raise zero to a negative exponent — zero does not have a reciprocal. 3838383838 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 Independent Practice In exercises 1–6, write each expression using exponents. 1. 2 × 2 × 2 × 2 2. a × a × a × 4 3. a × b × a × b 6. Show that 6 4 = k2. k k Simplify the expressions in exercises 8–25 using rules of exponents. 8. 170 11. 6 3 4 3 14. ( 2 2 )3 3 3 9. 2–3 12. (23)2 · 22 10. 22 · 23 13. ⋅ 4 3 2 3 7 3 15. (x4 ÷ x2) · x3 16. (x2)3 ÷ x4 17. ⋅ 3 x ( ax 5 x 2 ) 18. ( − 3 3 )x − 4 x ⋅ 5 x 19. (2x–2)3 · 4x2 20. 3x0y–2 21. (3x)0xy–2 22. 5x–1 × 6(xy)0 23. y ( 2x )4 x 2 24.

y ( )2 3 2 x y− 2 ( 25. 32 26. An average baseball has a radius, r, of 1.45 inches. V Find the volume, V, of a baseball in cubic inches. ( = 4 3 = 1 2, 27. The kinetic energy of a ball (in joules) is given by E 2 where m is the ball's mass (in kilograms) and v is its velocity (in meters per second). If a ball weighs 1 kilogram and is traveling at 10 meters per second, what is its kinetic energy in joules? 3π r mv ) at= 1 2 2, where a is its acceleration (in meters per second 28. The speed of a ball (in meters per second) accelerating from rest is given by v squared) and t is its time of flight (in seconds). Calculate the speed of a ball in meters per second after 5 seconds of flight if it is accelerating at 5 meters per second squared. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That’s a lot of rules, but don’t worry — you’ll get plenty of practice using them later in the program. Exponents often turn up when you’re dealing with area and volume. The next Topic will deal just with square roots, which is a special case of Rule 7 from the previous page. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 3939393939 TTTTTopicopicopicopicopic 1.3.21.3.2 1.3.21.3.2 1.3.2 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite, finding taking a root,oot,oot,oot,oot, taking a r taking a r the reciprocal, taking a r taking a r actional actional aising to a fr aising to a fr and r and r actional aising to a fractional and raising to a fr action

al aising to a fr and r and r stand stand popopopopowwwwwererererer..... TTTTThehehehehey under y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of xponents..... eeeeexponents xponents xponents xponents What it means for you: You’ll look more closely at the rules of square roots. Key words: square root radical radicand principal square root minor square root Check it out: Technically, the square root of a number p can be written as p2, but you don’t usually write the 2. Square Re Re Re Re Rootsootsootsootsoots Squar Squar Squar Squar Square Re Re Re Re Rootsootsootsootsoots Squar Squar Squar Squar In the last Topic you learned about all the exponent rules — this Topic will look more closely at one rule in particular. Square roots are the type of root that you’ll come across most often in math problems — so it’s really important that you know how to deal with them. adical Sign adical Sign oot Sign is the R oot Sign is the R or the R or the R Another Name f Another Name f adical Sign oot Sign is the Radical Sign or the Root Sign is the R Another Name for the R adical Sign oot Sign is the R or the R Another Name f Another Name f The square root of p is written p. If you multiply p by itself, you get p — so p × p = p. Multiplying p by itself means you square it. The nth root of p is written pn. If you raise pn to the power n, you get p — so ( pn )n = p. The symbol is called the radical sign and shows the nonnegative root if more than one root exists. In the expression pn p is called the radicand. (the nth root of p), The square root of a number p is also written p 2. 1 You can show this using the rules of exponents For any real number p > 0, the square root is written as p (or p p 1 2 ). If r = p, then r2 = p and (–r)2 = p. r is called the principal square root of p

and –r is called the minor square root of p. Guided Practice Complete the following. 1. The radicand of 83 is......... 2. The 6th root of t is written........ in radical notation. 3. 9 ×........ = 9 4. b 2 =........ in radical notation. 1 4040404040 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 o Square Re Re Re Re Rootsootsootsootsoots o Squar s Havvvvve e e e e TTTTTwwwwwo Squar o Squar s Ha s Ha e Number ositivvvvve Number e Number ositiositi ositi PPPPPositi e Numbers Ha o Squar s Ha e Number In practice, this means that every positive number has two square roots — a positive one (the principal square root) and a negative one (the minor square root). Example Example Example Example Example 11111 Find the square roots of the following numbers: a) 100 b) n2 Solution Solution Solution Solution Solution a) 100 = 10, so the principal square root is 10, and the minor square root is –10. b) n 2 =, so the principal square root is n, and the minor square n root is – n. The principal square root of n is written as n. The minor square root of n is written as – n. To indicate both square roots you can write ± n. Guided Practice Find the principal square root and minor square root of these numbers: 5. 4 6. 100 7. 81 Use the “±” symbol to give the principal and minor square root of the following numbers: 8. 9 11. 352 14. t2 9. 16 12. x2 10. 144 13. 81 15. 9 × 9 16. (st)2 Evaluate the following, giving the principal and minor roots: 1 2 17. 4 1 18. 121 2 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4141414141 e Square Re Re Re Re Rootsootsootsootsoots e Squar Also Havvvvve Squar e Squar Also Ha Also Ha essions essions aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr

essions Also Ha aic Expressions e Squar Also Ha essions aic Expr You can also take the square root of an algebraic expression. Example Example Example Example Example 22222 Find the square root of ( x +1 2. ) Solution Solution Solution Solution Solution x +1 2 = |x + 1|, so the principal square root is x + 1 and the minor ) ( square root is – x + 1. Guided Practice Give the principal and minor square root of each of the following expressions. 19. t t× 20. t 2 2× t 21. a 2× a2 22 ) 23 ) 24. ( a b+ 2 ) 25. ( t +1 2 ) t a 26. [ ( b+ 2 )] 2 27. [ ( a b+ 2 )] Independent Practice 1. Is this statement true or false? “The radicand of 325 is 5.” Evaluate the following. 2. 64 5. 25 1 2 3. ( )49 6. 122 4. a2 7. j× j Find the square roots of the following. ( 8. a2 ) 2 11. ( m 2 2 2+ n ) 9. ( k −1 2 ) ) ( 12. 2 pq 2 10. ( ⎡ +( 13. a ⎣ b m n+ 2 ) )× +( c d 2 ⎤ ) ⎦ ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Remember that when you take the square root of a positive number, you always have two possible answers — a positive one and a negative one. You can give both answers neatly using the ± sign. 4242424242 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 TTTTTopicopicopicopicopic 1.3.31.3.3 1.3.31.3.3 1.3.3 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper and use suc tions as h oper and use suc taking the opposite,

finding taking a root,oot,oot,oot,oot, taking a r taking a r the reciprocal, taking a r taking a r actional actional aising to a fr aising to a fr and r and r actional aising to a fractional and raising to a fr actional aising to a fr and r and r stand stand popopopopowwwwwererererer..... TTTTThehehehehey under y under y under stand y understand stand y under and use the rules of and use the rules of and use the rules of and use the rules of and use the rules of xponents..... eeeeexponents xponents xponents xponents What it means for you: You’ll find out how to multiply and divide square roots. Key words: square root radical radicand numerator denominator Check it out: Remember, mc means the principal (nonnegative) square root of mc. ties of R R R R Rootsootsootsootsoots ties of ties of oper oper PrPrPrPrProper operties of ties of oper PrPrPrPrProper ties of R R R R Rootsootsootsootsoots ties of ties of oper oper operties of ties of oper In the last Topic you saw that positive numbers have two square roots. This Topic’s all about how to multiply and divide square roots, which is really important when you’re evaluating expressions involving more than one root sign. actorsssss actor actor s Can Be Split into F s Can Be Split into F Number Number s Can Be Split into Factor Numbers Can Be Split into F actor s Can Be Split into F Number Number A factor of a number is a number that divides into it without a remainder — for example, 1, 2, 5, and 10 are factors of 10. To factor a number or expression means to write it as a product of its factors — for example, 10 = 2 × 5. Factoring is a useful way of simplifying square roots — as you’ll see in the rest of this lesson. Square Re Re Re Re Rootsootsootsootsoots Squar Squar ty of ty of oper e Pre Proper oper e Pre Pr s a Multiplicatititititivvvvve Pr s a Multiplica TTTTTherherherherhere’e’e’e’e’s a Multiplica s a Multipl

ica ty of Squar operty of Squar ty of oper s a Multiplica mc = ⋅ m c This means that to make finding a square root easier, you can try to factor the radicand first. Example Example Example Example Example 11111 Find the following: a) 400 b) 8 c ) Solution Solution Solution Solution Solution a) 400 = × 4 100 b) 8 = × 4 2 = × 4 100 × 2 10 2 = 2 = × = 20 2 10 = × 4 2 2 = × 2 = 2 2 The same technique can work if you have an algebraic expression. You need to find factors that are squares of other expressions: c) +( x 2 ) 1 −( 3 x 2 ) = 1 +( x 2 ) 1 −( 3 x 2 ) = ( 1 x )( + 1 3 x ) 1– Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4343434343 Guided Practice Simplify the following square roots. 1. 900 2. 225 3. 20 4. 200 5. 32 6. 4 2x 7. 81 2t 8. 4 t +( 2 ) 2 9. 36 ( j − 3 2 ) 10. 64 ( k + 4 2 ) oots TTTTToooooooooo oots Square Re Re Re Re Roots oots Squar Squar ty of ty of oper oper vision Pr vision Pr s a Di TTTTTherherherherhere’e’e’e’e’s a Di s a Di ty of Squar operty of vision Proper s a Division Pr oots Squar ty of oper vision Pr s a Di = m c m c Again, the idea is to look for any factors in the numerator or denominator that are squares. Example Example Example Example Example 22222 Find the following: b) 3 16 c) 49 225 Solution Solution Solution Solution Solution a) 49 225 = = 49 225 2 7 2 15 = 7 15 b) = 3 16 = 3 16 3 42 = 3 4 Check it out: In Example 1 c), the numerator and the denominator are both squares. c) 2 +( 2 x −( 3 x ) 1 ) 1 2 = +( 2 x 2 ) 1 −( 4444444444 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section

1.3 Section 1.3 Guided Practice Find the following. 11. 12. 13. 14. 15. 16 4 25 9 125 16 50 4 x 2 36 Independent Practice Find the following. 1. 49 4× 2. 25 2m 3. 4. 5. 64 9 121 144 t 2 81 6. 48 2t 7. 72 2x 8. 8 2m m, ≠ 0 16. 200 2x, x π 0 17. 242 2a a, ≠ 0 18 19. 27 2y 10. 11. 12. 13. 14. 15. 300 2t, t π 0 2 x y2 81 36 2y 49 2 100 t 2 a 16 a, ≠ 0 2 x 2 16 y y, ≠ 0 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Root signs are really tricky, so it’s a good idea to get rid of them whenever you can. When you multiply or divide roots, you’re left with simpler expressions, which makes math problems a lot easier. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4545454545 TTTTTopicopicopicopicopic 1.3.41.3.4 1.3.41.3.4 1.3.4 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding ocal, taking a root, ocal ocal ecipr ecipr the r the r eciprocal the recipr ocal ecipr the r the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll simplify fractions to their lowest terms. Key words: equivalent numerator denominator greatest common factor simplify actions actions alent Frrrrractions alent F alent F EquiEquiEquiEquiEquivvvvvalent F actions actions alent F al

ent Frrrrractions EquiEquiEquiEquiEquivvvvvalent F actions actions alent F alent F actions alent F actions Here’s another Algebra I Topic that you’ve seen in earlier grades. You’ve used fractions a lot before, but in Algebra I you’ll treat them more formally. This Topic goes over stuff on simplifying fractions that should feel quite familiar to you. tor and a Denominatortortortortor tor and a Denomina e a Numeraaaaator and a Denomina tor and a Denomina e a Numer actions Havvvvve a Numer e a Numer actions Ha FFFFFrrrrractions Ha actions Ha tor and a Denomina e a Numer actions Ha A fraction is any number expressed as one integer divided by another integer. Fractions are written in the form p q, where p and q are integers, and q π 0. The top number (p) is called the numerator and the bottom number (q) is the denominator. A fraction with a denominator of zero is undefined, because you can’t divide by zero. e the Same VVVVValuealuealuealuealue e the Same actions Havvvvve the Same e the Same actions Ha alent Frrrrractions Ha actions Ha alent F EquiEquiEquiEquiEquivvvvvalent F alent F e the Same actions Ha alent F Equivalent fractions are fractions of the same value — for example, 2 3 and 4 6, 1 2 and. 4 8 To simplify 4 6 to 2 3, you can rewrite the numerator and denominator as products of factors (factor them). You can then cancel any common factors by dividing both the numerator and denominator by those factors to produce an equivalent fraction. Example Example Example Example Example 11111 Convert 4 6 to 2 3. Solution Solution Solution Solution Solution First factor the numerator and denominator. You can cancel factors that are common to both the numerator and denominator, so in this case, parts of the fractions cancel, leaving 4646464646 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 Guided Practice 1. Identify the numerator and the denominator in the fraction 9 10. In Exercises 2-4, convert the fractions

to 2. 12 16 3. 9 12 3 4. In exercises 5-7, convert the fractions to 5 8. 5. 25 40 6. 15 24 4. 21 28 7. 40 64 8. Show that 3 5 and 9 15 are equivalent fractions. 9. Show that 5 10 and 3 6 are equivalent fractions. actions TTTTToooooooooo actions alent Frrrrractions actions alent F te Equivvvvvalent F alent F te Equi ou Can Creaeaeaeaeate Equi te Equi ou Can Cr YYYYYou Can Cr ou Can Cr actions alent F te Equi ou Can Cr The greatest common factor of two numbers a and b is the largest possible number that will divide exactly into both a and b. To simplify a fraction (or reduce it to its lowest terms) is to convert it to an equivalent fraction in which the numerator and denominator have a greatest common factor of 1. Example Example Example Example Example 22222 Reduce 56 64 Solution Solution Solution Solution Solution = 56 64 ⋅ 7 8 ⋅ 8 8 to its lowest terms. FFFFFactor the n actor the n actor the n umer umer umeraaaaator and denomina tor and denomina tor and denomina tor and denominatortortortortor actor the numer actor the n umer tor and denomina actors frs frs frs frs fromomomomom actor actor Cancel common f Cancel common f Cancel common factor actor Cancel common f Cancel common f tor and denominatortortortortor tor and denomina tor and denomina umeraaaaator and denomina umer umer the n the n the numer tor and denomina umer the n the n The greatest common factor (GCF) of 7 and 8 is 1, so this is the simplest form of 56 64. Similarly, you can produce an equivalent fraction by multiplying both the numerator and denominator by the same number. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4747474747 Guided Practice Complete these statements. 10. The largest possible number that will divide exactly into two numbers a and b is called the........................................... of a and b. 11. A fraction is expressed in its lowest terms if the numerator and denominator have a GCF of........... Find the greatest common

factor of each of these pairs. 12. 81 and 90 13. 56 and 77 14. 42 and 54 15. 13 and 19 Simplify these fractions. 12 14 4 12 16. 17. 18. 30 33 19. 9 24 Independent Practice 1. Identify the numerator in the fraction 11 13. 2. Identify the denominator in the fraction 14 15. In exercises 3–5, show how to simplify each fraction to 18 27 10 15 8 12 3. 4. 5. 2 3. 6. Show that 10 16 and 5 8 are equivalent fractions. Find the greatest common factor of each of these pairs of numbers. 7. 15 and 20 8. 21 and 33 9. 26 and 39 Simplify these fractions. 12 20 11. 12 15 10. 13. 44 48 14. 81 90 12. 15. 21 39 56 77 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Hopefully you recognized a lot of the stuff in this Topic from earlier grades. In the next couple of Topics you’ll go over multiplying, dividing, adding, and subtracting fractions — and you’ll need to be happy with simplifying fractions each time. 4848484848 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 TTTTTopicopicopicopicopic 1.3.51.3.5 1.3.51.3.5 1.3.5 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding ocal, taking a root, ocal ocal ecipr ecipr the r the r eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll multiply and divide fractions, then simplify them. Key words: numerator denominator reciprocal Multiplying and Multiplying and Multiplying and Multiplying and Multip

lying and Multiplying and Multiplying and Multiplying and Multiplying and Multiplying and actions actions viding Frrrrractions viding F viding F DiDiDiDiDividing F actions actions viding F viding Frrrrractions DiDiDiDiDividing F actions actions viding F viding F actions viding F actions You did lots of work on multiplying and dividing fractions in grade 7. This Topic is mainly a reminder of those techniques, because you’re going to be using them a lot in later math problems in Algebra I. ying the TTTTTop and Bottom op and Bottom op and Bottom ying the ying the y Multipl y Multipl actions b Multiply Fy Fy Fy Fy Frrrrractions b actions b Multipl Multipl op and Bottom y Multiplying the actions by Multipl op and Bottom ying the y Multipl actions b Multipl Multipl To multiply two fractions, find the product of the numerators and divide that by the product of the denominators. Given any numbers a, b, c, and d Œ R (b π 0, d π 0): ⋅ a c ⋅ b d c ⋅ = d ac bd a b = Example Example Example Example Example 11111 a) Multiply 2 5 and 7 3. b) Multiply 3 4 and 9 2. Solution Solution Solution Solution Solution a = 14 15 b = 27 8 Guided Practice Find each of these products. 1. ⋅ 2 3 4 5 4. 3 10 ⋅ 1 8 7. ⋅ 8 11 3 5 10. 3 8 ⋅ 15 2 2. ⋅ 3 5 1 2 5. ⋅ 7 8 3 4 8. 9 10 ⋅ 3 5 11. ⋅ 5 9 4 7 3. 6. 2 7 ⋅ 5 11 12. ⋅ 8 5 7 3 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 4949494949 Check it out: If there are two fractions multiplied together, you can cancel a factor from anywhere in the numerator with a matching factor in the denominator. They don’t have to have come from the same fraction. e Solutions in the Simplest Fororororormmmmm e Solutions in the Simplest F

ys Givvvvve Solutions in the Simplest F e Solutions in the Simplest F ys Gi AlAlAlAlAlwwwwwaaaaays Gi ys Gi e Solutions in the Simplest F ys Gi You should always give your answer in its simplest form — so with more complicated examples, factor the numerators and denominators and cancel common factors. It’ll save time if you do this before you compute the products. Example Example Example Example Example 22222 Multiply and simplify ⋅ 56 64 8 28. Solution Solution Solution Solution Solution 56 64 ⋅ ⋅ = 28 ⋅ and denominatortortortortorsssss s and denomina s and denomina umeraaaaatortortortortors and denomina umer umer actor the n actor the n F F F F Factor the n actor the numer s and denomina umer actor the n actorsssss actor actor Cancel all the common f Cancel all the common f Cancel all the common factor actor Cancel all the common f Cancel all the common f Guided Practice Multiply and simplify these expressions. 13. 16. 3 5 4 9 ⋅ 7 12 ⋅ 36 37 19. 40 15 ⋅ 24 64 14. 17. ⋅ 8 21 3 16 ⋅ 22 15 75 26 20. 81 90 ⋅ 10 90 15. 18. 14 25 36 55 ⋅ ⋅ 40 63 11 12 21. ⋅ 25 21 28 40 ocal ocal ecipr ecipr y the R y the R ying b ying b y Multipl y Multipl actions b vide Frrrrractions b actions b vide F DiDiDiDiDivide F vide F ocal eciprocal y the Recipr ying by the R y Multiplying b actions by Multipl ocal ecipr y the R ying b y Multipl actions b vide F The reciprocal of a fraction c d is d c, since c d ⋅ d c = 1. To divide by a fraction, you multiply by its reciprocal. Given any nonzero numbers a, b, c, and d Œ R: a b 5050505050 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1. = ad bc Example Example Example Example Example 33333 Divide 72 96 by 9 144. Solution Solution

Solution Solution Solution 72 96 ÷ 9 144 72 = ⋅ 96 144 9 72 96 ⋅ 144 9 = ⋅ 8 9 ⋅ 12 8 ⋅ ⋅ 12 12 9 ocal ocal ecipr ecipr y the r y the r tion b tion b ultiplica ultiplica write as a m write as a m RRRRReeeeewrite as a m ocal eciprocal y the recipr tion by the r ultiplication b write as a multiplica ocal ecipr y the r tion b ultiplica write as a m s and denominatortortortortorsssss s and denomina s and denomina umeraaaaatortortortortors and denomina umer umer actor the n actor the n FFFFFactor the n actor the numer s and denomina umer actor the n 1 1 ⋅ 8 9 ⋅ 12 8 1 1 ⋅ 1 ⋅ 12 12 9 1 = 12 actorsssss actor actor Cancel all the common f Cancel all the common f Cancel all the common factor actor Cancel all the common f Cancel all the common f Guided Practice In Exercises 1–6, divide and simplify. 22. 4 9 ÷ 16 15 25. 28. 48 7 24 32 ÷ ÷ 16 35 9 56 23. 26. 29. 5 21 13 27 40 72 ÷ ÷ 25 14 2 63 ÷ 24 36 24. 27. 30. ÷ ÷ ÷ 34 39 15 56 15 70 42 13 3 32 36 56 Independent Practice Evaluate the following. Simplify your answer where appropriate. 1. 42 4 ⋅ 10 28 4. 24 6 ⋅ 30 8 7. 21 22 ÷ 21 14 10. ÷ 36 24 28 48 2. 18 20 ⋅ 16 24 5. 8. ÷ ÷ 30 50 21 32 45 55 24 36 11. 64 72 ⋅ 88 40 3. ⋅ 14 12 2 21 6. 9. 18 12 35 14 ÷ 8 6 ÷ 21 44 12. 12 50 ⋅ 88 44 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Unless you’re told otherwise, you should always cancel fraction solutions down to the most simple form. In the next Topic you�

�ll look at adding and subtracting fractions, which is a bit tougher. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 5151515151 TTTTTopicopicopicopicopic 1.3.61.3.6 1.3.61.3.6 1.3.6 California Standards: 2.0:2.0:2.0:2.0:2.0: Students under stand stand Students under Students under Students understand stand stand Students under tions as tions as h operaaaaations as h oper h oper and use suc and use suc tions as and use such oper tions as and use suc h oper and use suc finding finding taking the opposite, finding finding finding ocal, taking a root, ocal ocal ecipr ecipr the r the r eciprocal the recipr the r ocal ecipr the r and raising to a fractional power. They understand and use the rules of exponents. What it means for you: You’ll add and subtract fractions with the same denominator, then fractions with different denominators. Key words: numerator denominator least common multiple prime factor ding and ding and AdAdAdAdAdding and ding and ding and ding and ding and AdAdAdAdAdding and ding and ding and actions actions acting Frrrrractions acting F acting F Subtr Subtr actions Subtracting F actions acting F Subtr Subtr acting Frrrrractions actions actions acting F acting F Subtr Subtr Subtracting F actions acting F Subtr actions Subtr You dealt with multiplying and dividing fractions in Topic 1.3.5. This Topic deals with adding and subtracting, which is a little bit harder if the denominators are different in each of the fractions. actions with the Same Denominatortortortortor actions with the Same Denomina + and – with Frrrrractions with the Same Denomina actions with the Same Denomina + and – with F + and – with F actions with the Same Denomina + and – with F + and – with F To find the sum (or the difference) of two fractions with the same denominator, just add (or subtract) the numerators, then divide by the common denominator. Example Example Example Example Example 11111 a) Calculate 1 7 5 +. 7 b) Calculate 5 7

1 −. 7 Solution Solution Solution Solution Solution a) Add the numerators and divide the answer by the common denominator, 7) Subtract the second numerator from the first and divide the answer by the common denominator, 7 Guided Practice Perform the indicated operations and simplify each expression in exercises 1–9. 1. + 5 6 7 6 4. 7. 15 32 53 32 + 21 32 − 15 32 2. 2 15 + 8 15 5. 8. 12 27 32 45 + 11 27 − 37 45 3. 13 18 + 11 18 6. − 24 49 11 49 9. − 21 16 15 16 5252525252 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 ent Denominatortortortortorsssss ent Denomina actions with Difffffferererererent Denomina ent Denomina actions with Dif ding Frrrrractions with Dif actions with Dif ding F AdAdAdAdAdding F ding F ent Denomina actions with Dif ding F You can find the sum (or difference) of two fractions with different denominators by first converting them into equivalent fractions with the same denominator. This common denominator should be the least common multiple (LCM) of the two denominators. The LCM of two numbers a and b is the smallest possible number that is divisible by both a and b. Check it out: See Topic 2.3.1 for more tions tions actoriza actoriza prime f prime f tions. about prime f actorizations prime factoriza tions actoriza prime f The prime factorizations of a and b can be used to calculate the LCM — the LCM is the product of the highest power of each prime factor that appears in either factorization. Example Example Example Example Example 22222 Calculate 2 3 7 +. 12 Solution Solution Solution Solution Solution To find the LCM of the denominators, write 3 and 12 as products of prime factors: 3 = 3, 12 = 22 × 3 So the LCM of 3 and 12 is 3 × 22 = 12 2 3 = 8 12 ConConConConConvvvvvererererert t t t t 2 3 to an equi to an equi to an equivvvvvalent fr alent fr alent fr action o action o action ovvvvver 12 er 12 er 12 alent fraction o er

12 to an equi to an equi alent fr action o er 12 8 12 7 + = 12 15 12 = 15 12 ⋅ 5 3 ⋅ 4 3 = 5 4 AdAdAdAdAdd frd frd frd frd fractions with the same denomina actions with the same denomina actions with the same denomina actions with the same denominatortortortortorsssss actions with the same denomina FFFFFactor the n actor the n actor the n umer umer umeraaaaator and denomina tor and denomina tor and denomina tor and denominatortortortortor actor the numer actor the n umer tor and denomina and cancel an and cancel an y common f y common f actor actor actorsssss and cancel any common f y common factor and cancel an and cancel an y common f actor So, to add or subtract fractions with different denominators: 1. Find the least common multiple of the denominators. 2. Convert each fraction into an equivalent fraction with the LCM as the denominator. 3. Add or subtract the fractions with the same denominators, then simplify the resultant fraction if possible. Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 5353535353 Example Example Example Example Example 33333 Calculate 5 8 5 −. 12 Solution Solution Solution Solution Solution The LCM of 8 and 12 is 24. = 5 8 15 24, 5 12 = 10 24 alent alent action into an equivvvvvalent action into an equi action into an equi t each frh frh frh frh fraction into an equi t eac t eac ConConConConConvvvvvererererert eac alent alent action into an equi t eac tor of 24 24 24 24 24 tor of tor of action with a denomina action with a denomina frfrfrfrfraction with a denomina action with a denominator of tor of action with a denomina 15 24 − = 5 10 24 24 Guided Practice actions with the same denominatortortortortorsssss actions with the same denomina actions with the same denomina act fr act fr Subtr Subtr act fractions with the same denomina Subtract fr actions with the same denomina act fr Subtr Subtr

Find the least common multiple of each pair of numbers. 11. 5 and 6 10. 4 and 6 14. 21 and 49 13. 14 and 18 17. 16 and 40 16. 12 and 42 20. 24 and 32 19. 36 and 52 12. 9 and 12 15. 18 and 27 18. 15 and 36 21. 25 and 60 Work out and simplify the following. 22. 5 9 + 8 3 25. 28. 13 18 9 15 + 11 27 − 17 10 23. + 11 12 3 8 26. + 7 16 3 40 29. 9 8 − 19 20 24. 27. 30 12 Independent Practice Perform the indicated operations and simplify each expression. 1. 4. 7. 6 20 48 60 11 18 + 8 20 − + 18 60 4 12 2. 5. 8. 18 24 8 50 7 30 + + − 7 24 3 10 3 24 3. 6 16 − 4 16 6. 4 8 − 5 12 9. + 2 21 7 11 10. Steve and Joe are collecting money to donate to charity. They set a goal of collecting $100. Steve has collected $50, while Joe collected only 1/5 of the money needed to meet the goal. What fraction of their goal do they still need to collect? Simplify the following expressions as far as possible. 11. 3 7 7 + − 11 2 13 12 )( 1 + + 2 x 6 ) 1 1 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Fractions will be popping up in all sorts of places throughout Algebra I, so it’s a good idea to make sure you really know the rules for them now. 5454545454 Section 1.3 Section 1.3 Section 1.3 — Exponents, Roots, and Fractions Section 1.3 Section 1.3 TTTTTopicopicopicopicopic 1.4.11.4.1 1.4.11.4.1 1.4.1 Section 1.4 tical Proofsoofsoofsoofsoofs tical Pr tical Pr thema thema MaMaMaMaMathema thematical Pr tical Pr thema MaMaMaMaMathema tical Proofsoofsoofsoofsoofs tical Pr tical Pr thema thema thematical Pr tical Pr thema California Standards: 24.2: Students identify the

24.2: Students identify the 24.2: Students identify the 24.2: Students identify the 24.2: Students identify the hhhhhypothesis and conc lusion lusion ypothesis and conc ypothesis and conc lusion ypothesis and conclusion lusion ypothesis and conc in logical deduction. in logical deduction. in logical deduction. in logical deduction. in logical deduction. 25.2: Students judge the 25.2: Students judge the 25.2: Students judge the 25.2: Students judge the 25.2: Students judge the gument gument an ar an ar alidity of alidity of vvvvvalidity of gument an argument alidity of an ar gument an ar alidity of hether the hether the ding to w ding to w accor accor ding to whether the according to w hether the ding to w accor hether the accor prprprprproper the realealealealeal the r the r ties of ties of oper oper ties of the r operties of oper the r ties of nnnnnumber system and the umber system and the umber system and the umber system and the umber system and the tions havvvvveeeee tions ha operaaaaations ha tions ha oper oper der of der of orororororder of der of oper tions ha oper der of pplied corrrrrrectlectlectlectlectly ay ay ay ay attttt pplied cor pplied cor been a been a been applied cor been a pplied cor been a h steppppp..... h ste eaceaceaceaceach ste h ste h ste What it means for you: You’ll justify each step of mathematical proofs, and you’ll learn about the hypothesis and conclusion of “if... then” statements. Key words: justify hypothesis conclusion proof Check it out: ties of ties of oper There are four prprprprproper oper ties of operties of ties of oper equality — adadadadaddition dition dition equality equality dition, dition equality equality ultiplicationtiontiontiontion, ultiplica ultiplica action, mmmmmultiplica action action subtr subtr subtraction action subtr ultiplica subtr and dididididivision vision vision vision. They basically vision say that you can do the same thing to both sides of an equation

. For a formal definition of each, refer to Topic 2.2.1. A lot of Algebra I asks you to give formal proofs for stuff that you covered in earlier grades. You’re sometimes asked to state exactly which property you’re using for every step of a math problem. tical Proofoofoofoofoof tical Pr tical Pr thema thema a Ma h Step ofp ofp ofp ofp of a Ma a Ma h Ste h Ste ustify Eac ustify Eac ou Must J YYYYYou Must J ou Must J thematical Pr a Mathema ustify Each Ste ou Must Justify Eac tical Pr thema a Ma h Ste ustify Eac ou Must J A mathematical proof is a logical argument. When you write a mathematical proof, you have to justify each step in a logical way. In Algebra I, you do this using the axioms covered earlier in this chapter. You’ve seen lots of proofs already in this chapter — although some of them weren’t described as proofs at the time. Solving an equation to find the value of a variable is a form of mathematical proof. Look at the example below. It shows a mathematical proof written in two columns — with each step of the logical argument written on the left, and the justification for it written on the right. Example Example Example Example Example 11111 If 6x + 4 = 22, what is the value of x? Solution Solution Solution Solution Solution 6x + 4 = 22 (6x + 4) – 4 = 22 – 4 (6x + 4) + (–4) = 22 – 4 6x + (4 + (–4)) = 22 – 4 6x + (4 + (–4)) = 18 6x + 0 = 18 6x = 18 1 6 ⎛ ⎜⎜⎜ ⎝ × ( 6 )x 1 = × 6 18 × 6 ⎞ 1 ⎟⎟⎟⋅ = × ⎠ 6 x 18 1 6 ⎛ ⎜⎜⎜ ⎝ ⎛ ⎜⎜⎜ ⎝ 1 6 1 6 × 6 ⎞ ⎟⎟⎟⋅ =x ⎠ 18 6 × 6 ⎞ ⎟⎟�

��⋅ =x ⎠ 3 1·x = 3 x = 3 GiGiGiGiGivvvvven equa en equa en equa en equationtiontiontiontion en equa Subtr Subtr action pr action pr oper oper ty of ty of equality equality Subtraction pr action proper operty of ty of equality equality Subtr Subtr action pr oper ty of equality Definition of Definition of subtr subtr action action Definition of subtr subtraction action Definition of Definition of subtr action Associa Associa Associatititititivvvvve pr e pre proper e pre pr oper oper ty of ty of ty of ad ad ad ad addition dition dition operty of dition Associa Associa oper ty of dition Subtr Subtr acting acting Subtracting acting Subtr Subtr acting InInInInInvvvvverererererse pr se pr se pr oper oper ty of ty of ty of ad ad ad ad addition dition dition se proper operty of dition se pr oper ty of dition Identity pr Identity pr oper oper ty of ty of ty of ad ad ad ad addition dition dition Identity proper operty of dition Identity pr Identity pr oper ty of dition Multiplica Multiplica tion pr tion pr oper oper ty of ty of equality equality Multiplication pr tion proper operty of ty of equality equality Multiplica Multiplica tion pr oper ty of equality ultiplicationtiontiontiontion ultiplica ultiplica ty of m m m m multiplica ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa operty of ultiplica ty of oper Associa Associa vision vision di di Definition of Definition of vision division Definition of di vision di Definition of Definition of viding viding DiDiDiDiDividing viding viding InInInInInvvvvverererererse pr oper oper se pr se pr ty of m m m m multiplica ty of ty of ultiplicationtiontiontiontion ultiplica ultiplica se proper operty of se pr ty of oper ultiplica Identity pr Identity pr oper oper ty of ty of ty of m m m m multiplica ultiplica ultiplica ultiplicationtiontiontiontion Identity proper operty of Identity pr Identity pr oper ty of ultiplica Section

1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 5555555555 Guided Practice Complete these statements: 1. A mathematical proof is called a................. each step in a logical way using mathematical................. 2. Mathematical proofs can be written in two columns, with the..................... on the left and the..................... on the right...................... because you have to y Combining Stepspspspsps y Combining Ste y Combining Ste tened b tened b oofs Can Often be Shor PrPrPrPrProofs Can Often be Shor oofs Can Often be Shor tened by Combining Ste oofs Can Often be Shortened b y Combining Ste tened b oofs Can Often be Shor Proofs can very often be written in the kind of two-column format used in the last example. The next statement in your argument goes on the left, and the justification for it goes on the right. Usually the justification will be something from earlier in this chapter. However, it’s not likely that you’d often need to include every single possible stage in a proof. Usually you’d solve an equation in a few lines, as shown below. Check it out: In Example 2 you’re doing several steps at once — but it’s the same thing as in Example 1. Example Example Example Example Example 22222 If 6x + 4 = 22, what is the value of x? Solution Solution Solution Solution Solution 6x + 4 = 22 6x = 18 x = 3 Usually it’s quicker (and a much better idea) to solve an equation the short way, like in Example 2. But you must be able to do it the long way if you need to, justifying each step using the real number axioms. lusion lusion es a Hypothesis and a Conc Givvvvves a Hypothesis and a Conc es a Hypothesis and a Conc Gi hen...””””” Gi Gi hen......, TTTTThen... hen......, “If“If“If“If“If...,..., lusion es a Hypothesis and a Conclusion lusion es a Hypothesis and a Conc Gi hen......, Mathematical statements can often be written in the form: “If..., then...” For example

, when you solve an equation like the one in Example 2, what you are really saying is: “If 6x + 4 = 22, then the value of x is 3.” A sentence like this can be broken down into two basic parts — a hypothesis and a conclusion. The hypothesis is the part of the sentence that follows “if” — here, it is 6x + 4 = 22. The conclusion is the part of the sentence that follows “then” — here, it is x = 3. IF hypothesis, THEN conclusion. 5656565656 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 Check it out: Watch out — a true conclusion doesn’t imply a true hypothesis, and a false hypothesis doesn’t imply that the conclusion is false. This doesn’t just apply to mathematical statements — it’s true for non-mathematical “If..., then...” sentences as well. For example: If an animal is an insect, then it has six legs. If you are in California, then you are in the United States. Now, both the hypothesis and the conclusion can be either true or false. For example, an animal may or may not be an insect, and it may or may not have six legs. However, the conclusion has to be a logical consequence of the hypothesis. Using the example above, this just means that if it is an insect, then it will have six legs. Once you’ve figured out a hypothesis and a conclusion, you can apply the following logical rules: If the hypothesis is true, then the conclusion will also be true. If the conclusion is false, then the hypothesis will also be false. So if an animal doesn’t have six legs, then it isn’t an insect. If you aren’t in the United States, then you’re not in California. And if x is not 3, then 6x + 4 π 22. 4x = 12 means that x = 3. x + y = 1 means that x = 1 – y. b + 4 = 17 – y means that b = 13 – y. Rewrite the following in “If..., then...” format. 1. 2. 3. Identify the hypothesis and conclusion in the following statements. 4. If 5y = 30, then y = 6. 5. If

x2 + y2 = 16, then x2 = 16 – y2. 6. If d – 12 = 23z, then d = 23z + 12. 7. An animal has four legs if it is a dog. 8. Complete this proof by adding the missing justification steps. x – 7 = 17 (x – 7) + 7 = 17 + 7 [x + (–7)] + 7 = 17 + 7 [x + (–7)] + 7 = 24 x + [(–7) + 7] = 24 x + 0 = 24 x = 24 Given equation................................... Definition of subtraction Adding................................... Inverse property of +................................... ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The important thing with mathematical proofs is to take each line of the math problem step by step. If you’re asked to justify your steps, make sure that you state exactly which property you’re using. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 5757575757 TTTTTopicopicopicopicopic 1.4.21.4.2 1.4.21.4.2 1.4.2 California Standards: xplain the xplain the Students e Students e 24.1: 24.1: 24.1: Students e xplain the Students explain the Students e 24.1: xplain the 24.1: difdifdifdifdifffffferererererence betw ence betweeneeneeneeneen ence betw ence betw ence betw e and deductivvvvveeeee e and deducti inductivvvvve and deducti e and deducti inducti inducti inducti e and deducti inducti rrrrreasoning and identify and easoning and identify and easoning and identify and easoning and identify and easoning and identify and eac eac xamples of xamples of vide e prprprprprooooovide e vide e each.h.h.h.h. xamples of eac vide examples of eac xamples of vide e 24.3: Students use 24.3: Students use 24.3: Students use 24.3: Students use 24.3: Students use countere

eeeexamples to sho xamples to showwwww xamples to sho xamples to sho counter counter xamples to sho counter counter tion is falsealsealsealsealse thathathathathat an asser tion is f tion is f t an asser t an asser t an assertion is f tion is f t an asser t a singlelelelele t a sing t a sing e tha e tha and recoecoecoecoecognizgnizgnizgnizgnize tha and r and r e that a sing t a sing e tha and r and r countereeeeexample is xample is xample is counter counter xample is counter xample is counter sufsufsufsufsufficient to r efute an efute an ficient to r ficient to r efute an ficient to refute an ficient to r efute an tion. tion. asser asser tion. assertion. tion. asser asser What it means for you: You’ll identify inductive and deductive reasoning in math problems, and you’ll find how to use counterexamples to disprove a rule. Key words: inductive reasoning deductive reasoning counterexample e and e and Inductivvvvve and Inducti Inducti e and e and Inducti Inducti e and e and Inductivvvvve and Inducti Inducti e and e and Inducti Inducti easoning easoning Deductivvvvve Re Re Re Re Reasoning Deducti Deducti easoning easoning Deducti Deducti Deductivvvvve Re Re Re Re Reasoning easoning easoning Deducti Deducti easoning Deducti easoning Deducti There are different types of mathematical reasoning. Two types mentioned in the California math standards are inductive reasoning and deductive reasoning. easoning Means Finding a General Ral Ral Ral Ral Ruleuleuleuleule easoning Means Finding a Gener Inductivvvvve Re Re Re Re Reasoning Means Finding a Gener easoning Means Finding a Gener Inducti Inducti easoning Means Finding a Gener Inducti Inducti Inductive reasoning means finding a general rule by considering a few specific cases. For example, look at this sequence of square numbers: 1, 4, 9

, 16, 25, 36... If you look at the differences between successive terms, you find this: The difference between the first and second terms is 4 – 1 = 3. The difference between the second and third terms is 9 – 4 = 5. The difference between the third and fourth terms is 16 – 9 = 7. The difference between the fourth and fifth terms is 25 – 16 = 9. If you look at these differences, there’s a pattern — each difference is an odd number, and each one is 2 greater than the previous difference. So using inductive reasoning, you might conclude that: The difference between successive square numbers is always odd, and each difference is 2 greater than the previous one. Watch out though — this doesn’t actually prove the rule. This rule does look believable, but to prove it you’d have to use algebra. Guided Practice Use inductive reasoning to work out an expression for the nth term (xn) of these sequences. For example, the formula for the nth term of the sequence 1, 2, 3, 4,... is xn = n. 1. 2, 3, 4, 5,... 2. 11, 12, 13, 14,... 3. 2, 4, 6, 8,... 4. –1, –2, –3, –4,... In exercises 5-6, predict the next number in each pattern. 5. 1 = 12, 1 + 3 = 22, 1 + 3 + 5 = 32. 1, 1, 2, 3, 5, 8, 13, … 5858585858 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 ule Doesn’’’’’t t t t t WWWWWorororororkkkkk ule Doesn xample Prooooovvvvves es es es es TTTTThahahahahat a Rt a Rt a Rt a Rt a Rule Doesn ule Doesn xample Pr One Countereeeeexample Pr xample Pr One Counter One Counter ule Doesn xample Pr One Counter One Counter If you are testing a rule, you only need to find one counterexample (an example that does not work) to prove that the rule is not true. Once you have found one counterexample, you don’t need to look for any more — one is enough. Example Example Example Example

Example 11111 Decide whether the following statement is always true: “2n + 1 is always a prime number, where n is a natural number.” Solution Solution Solution Solution Solution At first, the rule looks believable. If n = 1: 2n + 1 = 21 + 1 = 2 + 1 = 3. This is a prime number, so the rule holds for n = 1. If n = 2: 2n + 1 = 22 + 1 = 4 + 1 = 5. This is a prime number, so the rule holds for n = 2. If n = 3: 2n + 1 = 23 + 1 = 8 + 1 = 9. This is not a prime number, so the rule doesn’t hold for n = 3. So a counterexample is n = 3, and this proves that the rule is not always true. Once you’ve found one counterexample, you don’t need to find any more. Guided Practice Give a counterexample to disprove each of the following statements. 7. All odd numbers are of the form 4n + 1, where n is a natural number. 8. If a number is divisible by both 6 and 3, then it is divisible by 12. 9. |x + 4| ≥ 4 10. The difference between any two square numbers is always odd. 11. The difference between any two prime numbers is always even. 12. The difference between two consecutive cube numbers is always prime. 13. All quadrilaterals are squares. 14. All angles are right angles. 15. All prime numbers are odd. 16. A number is always greater than its multiplicative inverse. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 5959595959 ying a General Ral Ral Ral Ral Ruleuleuleuleule ying a Gener easoning Means AAAAApplpplpplpplpplying a Gener ying a Gener easoning Means Deductivvvvve Re Re Re Re Reasoning Means easoning Means Deducti Deducti ying a Gener easoning Means Deducti Deducti Deductive reasoning is almost the opposite of inductive reasoning. In deductive reasoning, you use a general rule to find out a specific fact. Example Example Example Example Example 22222 A number is a multiple of 3 if the sum of its

digits is a multiple of 3. Use this information to decide whether 96 is a multiple of 3. Solution Solution Solution Solution Solution The sum of the digits is 9 + 6 = 15, which is divisible by 3. The statement says that a number is a multiple of 3 if the sum of its digits is a multiple of 3. Using deductive reasoning, that means that you can say that 96 is divisible by 3. Guided Practice Use deductive reasoning to work out the 10th term of these sequences: 18. xn = 6n 21. xn = 20n + 1 17. xn = n 20. xn = 3n – 1 Use deductive reasoning to reach a conclusion: 23. Ivy is older than Peter. Stephen is younger than Peter. 24. Lily lives in Maryland. Maryland is in the United States. 19. xn = 2n – 1 22. xn = n(n + 1) Independent Practice Use inductive reasoning in Exercises 1–3. 1. Give the next three numbers of the sequence: 25, 29, 34, 40, … 2. Write an expression for the nth term (xn) of the sequence: 24, 72, 216, 648, … 3. Audrey needs $650 to buy a digital camera. Her savings account shows the following balances: If the pattern continues, at the start of which month will she be able to buy the digital camera? Date Balance Jan 1 $100.00 Feb 1 $150.00 March 1 $250.00 April 1 $400.00 Use deductive reasoning in exercises 4-5. 4. Find the first five terms of the sequence xn = 3n. 5. Find the first five terms of the sequence xn = n(n – 1). ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Inductive reasoning means that you can make a general rule without having to check every single value — so it saves you a lot of work. 6060606060 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 TTTTTopicopicopicopicopic 1.4.31.4.3 1.4.31.4.3 1.4.3 California Standards: Givvvvven a specific en a specific en a specific Gi Gi 25.3: 25

.3: en a specific 25.3: Gi 25.3: en a specific Gi 25.3: algalgalgalgalgeeeeebrbrbrbrbraic sta tement tement aic sta aic sta aic statement tement aic sta tement inininininvvvvvolving linear quadraaaaatictictictictic,,,,, quadr olving linear,,,,, quadr quadr olving linear olving linear quadr olving linear or aor aor aor aor absolute v bsolute valuealuealuealuealue bsolute v bsolute v bsolute v tions or tions or essions or equa essions or equa eeeeexprxprxprxprxpressions or equa tions or essions or equations or tions or essions or equa inequalities,,,,, students students students inequalities inequalities students inequalities students inequalities hether the hether the mine w mine w deter deter hether the mine whether the determine w mine w deter hether the deter stastastastastatement is tr tement is trueueueueue tement is tr tement is tr tement is tr alwwwwwaaaaaysysysysys,,,,, or al al sometimes,,,,, al sometimes sometimes or or or or al sometimes sometimes nenenenenevvvvvererererer..... What it means for you: You’ll learn how to recognise whether statements are true, and if they’re true sometimes or all the time. Key words: hypothesis conclusion Check it out: See Topic 1.4.1 for more on hypotheses and conclusions. Using the logic from that Topic, if x is not 3 or –3, then x2 is not equal to 9. tements tements aic Sta aic Sta AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Sta tements aic Statements tements aic Sta AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Sta tements tements aic Sta aic Sta aic Statements tements aic Sta tements When you’re trying to solve a mathematical problem, sometimes the solution is a single value. This Topic is about other situations, when things can be a bit more complicated. e than One Parararararttttt e than One P e than One P Mor Mor Solutions Can Consist of

Solutions Can Consist of More than One P Solutions Can Consist of Mor e than One P Mor Solutions Can Consist of Solutions Can Consist of Example Example Example Example Example 11111 Find x given that x2 = 9. Solution Solution Solution Solution Solution You can think of the equation as a hypothesis — then you need to find a logical conclusion. One value that satisfies the equation is x = 3. However, the statement “If x2 = 9, then x = 3” is not true — because x = –3 also satisfies the equation. There are two values satisfying the equation, and you need to include both of them in your answer. So the solution is actually: “x = 3 or x = –3.” Or in “If..., then...” form: If x2 = 9, then x = 3 or x = –3. tions and Inequalities are e e e e AlAlAlAlAlwwwwwaaaaays ys ys ys ys TTTTTrrrrrueueueueue tions and Inequalities ar tions and Inequalities ar Some Equa Some Equa Some Equations and Inequalities ar tions and Inequalities ar Some Equa Some Equa If you’re given an algebraic statement (such as an equation or an inequality), there won’t always be a single value (or even two values) that satisfy the statement. Example Example Example Example Example 22222 What values of x satisfy x2 – 9 = (x + 3)(x – 3)? Solution Solution Solution Solution Solution If you look at the problem above, you might see that x = 0 satisfies the equation — because if you put x = 0, then both sides equal –9. But you might also realize that x = 3 satisfies the equation — since if x = 3, both sides equal 0. But that’s not all. If x = –3, both sides also equal 0. So x = –3 also satisfies the equation. In fact, the above equation is always true — no matter what value you pick for x. So to say “either x = 0, x = 3, or x = –3” is incorrect. You need to say that it is always true. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 6161616161 Check it out: See Topic 1.

2.3 for more about absolute value equations. Check it out: If |a| < |b|, then: 0 < a < b, or b < a < 0 or a < 0 < b, or b < 0 < a. tement is Nevvvvver er er er er TTTTTrrrrrueueueueue tement is Ne tement is Ne Sometimes a Sta Sometimes a Sta Sometimes a Statement is Ne tement is Ne Sometimes a Sta Sometimes a Sta There is another possibility. Look at the following absolute value equation: Example Example Example Example Example 33333 Find x given that |x| + 3 = 0. Solution Solution Solution Solution Solution You need to find values for x that satisfy the above equation. However, if you subtract 3 from both sides, you form the equivalent equation: |x| = –3 But the absolute value of a number is its distance from 0 on the number line — and a distance cannot be negative. That means that there are no values of x that satisfy the equation. So the equation is never true. Guided Practice 1. Find x given that x2 = 25. 2. What values of x satisfy x2 – 25 = (x + 5)(x – 5)? 3. Find x given that |x| + 5 = 0 4. Find x given that x2 = –16, x Œ R 5. What values of x satisfy x2 > 0? 6. Find x given that x2 + 1 = 17. tical Logicgicgicgicgic tical Lo tical Lo thema e Examples of Ma Ma Ma Ma Mathema thema e Examples of MorMorMorMorMore Examples of e Examples of thematical Lo tical Lo thema e Examples of You can combine the ideas from this section. Look at the following examples. Example Example Example Example Example 44444 Is the following statement true? If |a| < |b|, then a < b. Solution Solution Solution Solution Solution This is an example where you can find a counterexample to disprove the rule. For example, if a = –1 and b = –2, then |a| < |b|, but a > b. Here, the conclusion isn’t a logical consequence of the hypothesis — so you can’t apply the logical rules from earlier. That means that the statement is not true. 6262626262 Section 1.4 Section 1.4 Section 1.4

— Mathematical Logic Section 1.4 Section 1.4 Example Example Example Example Example 55555 If x is a real number, find the possible values for which the following is true: x2 + 1 < 2x Solution Solution Solution Solution Solution You can use the inequality as a hypothesis. Now you need to find a suitable conclusion using a series of logical steps. x2 + 1 < 2x fi x2 – 2x + 1 < 0 fi (x – 1)2 < 0 So your “If..., then...” statement is: If x2 + 1 < 2x, then (x – 1)2 < 0. Your hypothesis is: x2 + 1 < 2x Your conclusion is: (x – 1)2 < 0 When you square a real number, you can never get a negative number — so (x – 1)2 cannot be negative. This means that your conclusion is false. And using the logic from earlier, this means that your hypothesis is also false. So you’ve proved that there is no real number x for which x2 + 1 < 2x. The statement is never true. Example Example Example Example Example 66666 “If a number is odd, its square is odd.” The number 71774784 is a perfect square. Use the statement above to say whether the square roots of 71774784 are odd. Solution Solution Solution Solution Solution Hypothesis: a number is odd Conclusion: its square is odd The square of a number is 71774784 — which is an even number. So the conclusion is false. This means that the hypothesis is also false, and the square root of 71774784 is not odd. Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 6363636363 Independent Practice 1. If x is a real number, find the possible values for which the following is true: x2 – 4x + 4 < 0 2. Is this statement true? “If |x| < 3, then x < 3.” 3. Find x given that x2 = 64 4. What values of x satisfy x2 – 100 = (x + 10)(x – 10)? 5. Find x given that |x| + 5 = 0. 6. If x is a real number, find the possible values for which the following is true: x2 + 9 < 6x In exercises 7

-11, say whether the algebraic statements are true sometimes, always, or never. 7. 5 + x = 10 10. x2 < 0 8. x2 = –9 11. x2 > x 9. x2 = 49 12. Is this statement true? “If x > 0, then x3 > x2” 13. Find x given that |x| + 7 = 0. 14. What values of x satisfy x2 – 25 = (x + 5)(x – 5)? 15. The area of a rectangle is given by A = l × w where l and w are the length and width of the rectangle, respectively. Can the area of a rectangle ever be less than zero? 16. Find x given that x2 = 81 17. Is the following statement true? If x > 0 then x2 > x 18. Prove that the following statement is not true: “the square root of a number is always smaller than the number itself.” 19. Prove that the difference between successive square numbers is always odd, and each difference is 2 greater than the previous difference. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up There won’t always be a one-part solution to an algebraic statement such as an equation or inequality. If the statement is always true, it’s no good just giving one value the statement holds for. And if it’s never true, you have to state that too. 6464646464 Section 1.4 Section 1.4 Section 1.4 — Mathematical Logic Section 1.4 Section 1.4 Chapter 1 Investigation Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Counting Collections Sets and subsets aren’t only useful in Math class — they can be used to describe everyday situations. A cereal company is giving away baseball erasers free in their boxes of cereal. There are 7 erasers to collect. All of the children at an elementary school want to collect the whole set. At the moment, they all have different collections and none have more than one of any one eraser. What is the maximum number of children there could be at the school? Things to think about: How many children could there be if there were only two erasers to collect? How many children

could there be if there were three erasers to collect? How many children could there be if there were four erasers to collect? Look at your answers — what do you notice? The collections can overlap — for example, one child could have erasers a, b, and d, while another could have erasers b, c, and g. Extension 1) If there were 8 erasers in the set, how many different collections could there be? What if there were 20 erasers to collect? Try to find a general rule for the number of different collections for sets with n items. 2) A set of trading cards consists of 78 numbered cards. • How many people could have different collections of cards? The cards come in sealed packs that cost $1.80 per pack. Each pack contains 8 randomly selected cards. Your friend says that it would only cost $18 to get the full set. Is your friend right? 3) Set G is the set of two-digit prime numbers. How many subsets of G are there? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This Investigation shows that you can use sets and subsets to model real-life situations. In fact, you probably divide things into subsets without even realizing it — for example, sorting out your favourite types of candy from a mixed box. estigaaaaationtiontiontiontion — Counting Collections estigestig estig pter 1 Invvvvvestig pter 1 In ChaChaChaChaChapter 1 In pter 1 In pter 1 In 6565656565 Chapter 2 Single Variable Linear Equations Section 2.1 Algebra Basics.......................................................... 67 Section 2.2 Manipulating Equations............................................. 76 Section 2.3 More Equations......................................................... 81 Section 2.4 Using Equations........................................................ 89 Section 2.5 Consecutive Integer Tasks, Time and Rate Tasks.... 95 Section 2.6 Investment and Mixture Tasks................................. 108 Section 2.7 Work-Related Tasks................................................ 121 Section 2.8 Absolute Value Equations....................................... 128 Investigation Wildlife Park Trains.................................................. 135 6666666666 TTTTTopicopicopicopicopic 2.1.12.1.1 2.1.12.1.1 2.1.1 California Standards: 4.0: Students simplify 4.

0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions essions essions essions before solving essions linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll combine like terms to simplify expressions. Key words: algebraic expression simplify like terms Don’t forget: See Topic 1.3.1 if you’re not sure what exponents are. Section 2.1 Simplifying AlgAlgAlgAlgAlgeeeeebrbrbrbrbraicaicaicaicaic Simplifying Simplifying Simplifying Simplifying Simplifying AlgAlgAlgAlgAlgeeeeebrbrbrbrbraicaicaicaicaic Simplifying Simplifying Simplifying Simplifying essions essions ExprExprExprExprExpressions essions essions essions essions ExprExprExprExprExpressions essions essions Unless you’re told otherwise, you always need to give algebraic solutions in the simplest form possible. essions Contain VVVVVariaariaariaariaariabbbbbleslesleslesles essions Contain essions Contain aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr aic Expr aic Expressions Contain essions Contain aic Expr Algebraic expressions are made up of terms. A term can be a product of numbers and variables, like 4x2 or 5x, or just a number. For example, the algebraic expression 4x2 – 5x + 7 – 2x2 + 2x – 3 has six terms. The terms are separated by plus and minus signs. Each sign “belongs” to the term that it’s in front of. 2 8 + 2 x xy Invisible + sign Guided Practice – 6 + 7 – 2 y y 2 This minus sign belongs to the 6y term. Write the number of terms in the expressions in Exercises 1–4. 1. 3x2 + 4x – 2 3. 3x2 2. 8x4 + 7x3 + 2x2 – 8 4. 8 + 7xy – 2xy9 + 4x7 + 3y2 + 55y3 5.

Which variable is multiplied by –4 in the algebraic expression 4x2 – 4y + 8 + 4xy? 6. Counting from left to right, which term is the fourth term in the algebraic expression 8x2 + 2xy – 6y + 9xy3 – 4? LikLikLikLikLike e e e e TTTTTerererererms Can Be Combined ms Can Be Combined ms Can Be Combined ms Can Be Combined ms Can Be Combined Like terms are terms with identical variables that have identical exponents. The terms 4x2 and –2x2 are like terms because they have the same variable, x, with the same exponent, 2. Likewise, –5x and 2x are like terms, and 7 and –3 are like terms. Like terms can be combined using the distributive, commutative, and associative properties. Example Example Example Example Example 11111 Simplify 4x2 – 2x2. Solution Solution Solution Solution Solution 4x2 – 2x2 = (4 – 2)x2 = 2x2 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 6767676767 Don’t forget: See Lesson 1.2.9 for more on number properties. Check it out: Subtracting a number is just the same as adding a negative number. So, 4x – 5 is the same as 4x + (–5). Don’t forget: When there are no more like terms to collect, the expression is simplified as much as possible. essions essions ms to Simplify Expr Combine Like e e e e TTTTTerererererms to Simplify Expr ms to Simplify Expr Combine Lik Combine Lik essions ms to Simplify Expressions essions ms to Simplify Expr Combine Lik Combine Lik To simplify an algebraic expression, use number properties to first group and then combine like terms. Example Example Example Example Example 22222 Simplify the following: a) (4x – 5) – 2x b) 5x2 – 3x + 7x2 – 4x + 9 Solution Solution Solution Solution Solution a) (4x – 5) – 2x = 4x + (–5 – 2x) = 4x + (–2x – 5) = (4x + –2x) – 5 = [(4 – 2

)x] – 5 = 2x – 5 dition dition ty of ad ad ad ad addition ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa dition operty of dition ty of oper Associa Associa dition dition ty of ad ad ad ad addition ty of ty of oper oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper dition operty of dition ty of oper dition dition ty of ad ad ad ad addition ty of ty of oper oper e pre proper e pre pr Associatititititivvvvve pr Associa Associa dition operty of dition ty of oper Associa Associa opertytytytyty oper oper e pre proper e pre pr Distributiutiutiutiutivvvvve pr Distrib Distrib oper Distrib Distrib b) 5x2 – 3x + 7x2 – 4x + 9 = 5x2 + 7x2 – 3x – 4x + 9 = (5x2 + 7x2) + (–3x – 4x) + 9 = (5 + 7)x2 + (–3 – 4)x + 9 = 12x2 + (–7)x + 9 = 12x2 – 7x + 9 Guided Practice Simplify the following expressions in Exercises 7–14. 7. 7a + 12a – 4 9. 9x + (20 – 5x) 11. 7x2 + 7 + 20x2 + 3x 13. 5a – 4 × 3a + 7 × 2 – 3 × 6 8. 7a +3b – 5a 10. 5 – 10x – 2x – 7 12. 3 – 8x2 + 4x2 + 6x2 – 10 14. 3 × 4a – 2 × 5a2 + 2 × 2a2 In Exercises 15–18, simplify the expressions and determine the number of terms in each simplified form. 15. (4a – 9) + (2a – 18) 16. 15n + 3n + 8 – 2 – 6 17. 6a2 + 3a – 9a2 + 2a + 7 + 6 18. 6a + 3 × 7b – 2 × 5c + 7 – 9 +

2 × 4c 6868686868 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 Independent Practice In Exercises 1–5, determine the number of terms in each algebraic expression: 1. 7b + 14a – 4 3. (27x2 + 4x) – 13 5. 2x + 4xy + 4x2 – (10 + 12y + 19y2) 2. 2a 4. 5 +10x + 20x2 + 3a In Exercises 6–9, simplify each algebraic expression: 6. 12m – 7 + 3c – 7m – 8c 7. 4a + 3b + 11a – 8b 8. 2 × 6x – 3 × 5x – 3x × 4 + 5 × 2x + 12x 9. 5m × 3 + 2 × 7m – 4m × 4 + 7 × 2m – 17m In Exercises 10–13, simplify each algebraic expression: 10. 3x + 1 2 a – 3 11. 1 10 8 y – 1 x 2 b + 8 10 12. (17.8n + 13.08q) – 3.8q – 9.9n 13. 0.4x2 + 3 8 x2 – 0.14 – 5 8 b – 6 12 x – 1 2 a 14. Which expression below simplifies to 2x + 1? i. x + 4x + 4 – 3 – 3x + x ii. 7 + 5x – 6 + 5 – 4x + x – 5 15. Which expression below simplifies to 3x? i. 5 + 4x2 – 3x – 2 + 4x2 – 8x2 – 3 ii. –2x2 + 6 + 4x2 – 3 + 3x – 2x2 – 3 In Exercises 16–17, find a simplified expression for the perimeter of the figure. 16. 2x x 4 + 1 17. 3 2a 2a 2x 2x 7 18. Juan bought 3 baseball cards for b dollars each and 2 baseball cards for c dollars each. He has bought 4 comic books for $5.00 each. Write and simplify an algebraic expression showing the total money Juan spent on baseball cards and comic books. 19. Three friends Tom, Leo, and Maria have several pieces of candy to eat. Tom has (2x + 4

) pieces of candy, Leo has (8 – 2x) pieces of candy, and Maria has 8 pieces of candy. Write and simplify an algebraic expression showing the total number of pieces of candy the three friends have to eat. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up You’ve combined like terms before, in earlier grades — so this Topic should feel like good practice. It’s always important to give your final answers to algebraic problems in the simplest form. Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 6969696969 TTTTTopicopicopicopicopic 2.1.22.1.2 2.1.22.1.2 2.1.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions essions essions essions before solving essions linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll use the distributive property to simplify expressions. Key words: distributive property commutative property Don’t forget: After getting rid of the grouping symbols, collect the like terms together. Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of Getting Rid of ouping Symbols ouping Symbols GrGrGrGrGrouping Symbols ouping Symbols ouping Symbols GrGrGrGrGrouping Symbols ouping Symbols ouping Symbols ouping Symbols ouping Symbols You already saw the distributive property in Topic 1.2.7. In this Topic you’ll simplify expressions by using the distributive property to get rid of grouping symbols. ouping Symbols ouping Symbols es Gr operty Rty Rty Rty Rty Remoemoemoemoemovvvvves Gr es Gr oper oper e Pr he Distributiutiutiutiutivvvvve Pr e Pr he Distrib TTTTThe Distrib he Distrib ouping Symbols es Grouping

Symbols e Proper ouping Symbols es Gr oper e Pr he Distrib The expression 5(3x + 2) + 2(2x – 1) can be simplified — both parts have an “x” term and a constant term. To simplify an expression like this, you first need to get rid of the grouping symbols. The way to do this is to use the distributive property of multiplication over addition: a(b + c) = ab + ac. Example Example Example Example Example 11111 Simplify 5(3x + 2) + 2(2x – 1). Solution Solution Solution Solution Solution 5(3x + 2) + 2(2x – 1) = 15x + 10 + 4x – 2 = 15x + 4x + 10 – 2 = 19x + 8 Guided Practice Distrib Distrib Distributiutiutiutiutivvvvve pr e pre proper e pre pr oper oper opertytytytyty Distrib Distrib oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper oper oper ty of ty of ty of ad ad ad ad addition dition dition operty of dition oper ty of dition In Exercises 1–7, simplify the following expressions: 1. 2(4x + 5) + 8 2. 12(5a – 8) + 4x + 3 3. 6(2j + 3c) + 8(5c + 4z) 4. 10(x + 2) + 7(3 – 4x) 5. 6(a – b) + 4(2b – 3) 6. 5(3x + 4) + 3(4x + 10) + 2(8x + 9) 7. 8(2n – 3) + 9(4n – 5) + 4(3n + 7) 7070707070 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 y a Negggggaaaaatititititivvvvve Number e Number e Number y a Ne y a Ne ying b ying b hen Multipl e Care we we we we when Multipl hen Multipl e Car TTTTTakakakakake Car e Car e Number ying by a Ne hen Multiplying b e Number y a Ne ying b hen Multipl e Car If

a number outside a grouping symbol is negative, like in –7(2x + 1), you have to remember to use the multiplicative property of –1. This means that the signs of the terms within the grouping symbols will change: “+” signs will change to “–” signs and vice versa. Example Example Example Example Example 22222 Simplify the following: a) –7(2x + 1) b) –6(–x – 3) c) –3(5x – 4) Solution Solution Solution Solution Solution a) The +2x and +1 become negative. –7(2x + 1) = –14x – 7 b) The two negative terms inside the grouping symbols are multiplied by the negative term outside. They both become positive. –6(–x – 3) = 6x + 18 c) –3(5x – 4) = –15x + 12 Example Example Example Example Example 33333 Simplify the expression 4(2x – 1) – 5(x – 2). Show your steps. Check it out 3x + 6 has no more like terms, so it can’t be simplified any further. Solution Solution Solution Solution Solution 4(2x – 1) – 5(x – 2) = 8x – 4 – 5x + 10 = 8x – 5x – 4 + 10 = 3x + 6 Guided Practice GiGiGiGiGivvvvven e en een exprxprxprxprxpression en een e ession ession ession ession Distrib Distrib Distributiutiutiutiutivvvvve pr e pre proper e pre pr oper oper opertytytytyty Distrib Distrib oper CommCommCommCommCommutautautautautatititititivvvvve pre pre pre pre proper oper oper ty of ty of ty of ad ad ad ad addition dition dition operty of dition oper ty of dition In Exercises 8–13, simplify each algebraic expression: 8. –2(5a – 3c) 9. –8(3c – 2) 10. –2(–3x – 4) + 4(6 – 2x) 11. 7(2a + 9) – 4(a + 11) 12. –8(2y + 4) – 5(y + 4) 13. − 2 ⎛ ⎜⎜

⎜ ⎝ 1 2 + n 2 ⎞ ⎟⎟⎟− ⎠ ⎛ ⎜⎜⎜ 3 ⎝ 1 3 − n 4 ⎞ ⎟⎟⎟ ⎠ 14. Simplify 12(2n – 7) – 9(3 – 4n) + 6(4x – 9). 15. Simplify 5(x – 2) – 7(–4x + 3) – 3(–2x). Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 7171717171 Independent Practice In Exercises 1–6, simplify the algebraic expressions: 1. –4(a + 2b) 3. 5(3b – 2q) – (3q + 4) 5. –17(3a – 5b) – 4(x + 3) 2. 3(2n + 4) + n(–4) 4. –9(2 + 3b) + 3(3 – 2b) 6. –2(10n – 4x) + 2(n + 6x) – 3(7x – 2n) 7. Simplify –7(3p – 9q) – 4(25q – 3r) + 12(5p + 8). In Exercises 8–12, find and simplify an expression for the perimeter of the shape shown. 8 11. 9. 10. - 2 x- - – 1 4 – 3x 12. 3 n – 4 - - - 2 – x1 20 – 3n (3x + 4) – 1 2 In Exercises 13–15, simplify the algebraic expressions: 13. 3 8 14. –0.1(25n + 46) – 0.8(16n – 5) + 0.4(3 – 4n) 15. – 1 3 (2x – 4) – 0.2(–7x – 10) (5 – 9x) – 1 8 (4x – 8) 16. Mia has 4(x + 1) dolls, where x is Mia's age. Madeline has 5(2x + 3) dolls. Write and simplify an algebraic expression showing the total number of dolls in Mia and Madeline's collection. 17. Ruby, Sara, and Keisha are

counting stamps. If x represents Ruby's age, she has 4(x – 4) stamps, Sara has 2(8 – x) stamps, and Keisha has 8(7 + 2x) stamps. Write and simplify an algebraic expression showing the total number of stamps owned by the three friends. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The distributive property is really useful — it’s always good to get rid of confusing grouping symbols whenever you can. The main thing you need to watch out for is if you’re multiplying the contents of parentheses by a negative number — it will change the sign of everything in the parentheses. 7272727272 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 TTTTTopicopicopicopicopic 2.1.32.1.3 2.1.32.1.3 2.1.3 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions essions essions essions before solving essions linear equations and inequalities in one variable, such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll use the rules of exponents to simplify expressions, and you’ll show that proposed solutions to equations are correct. Key words: exponents commutative property equation Don’t forget: After getting rid of the grouping symbols, collect any like terms together. Check it out: 2xy and yx are like terms. e Simplifying and e Simplifying and MorMorMorMorMore Simplifying and e Simplifying and e Simplifying and e Simplifying and e Simplifying and MorMorMorMorMore Simplifying and e Simplifying and e Simplifying and king AnsAnsAnsAnsAnswwwwwererererersssss king king ChecChecChecChecChecking king ChecChecChecChecChecking king AnsAnsAnsAnsAnswwwwwererererersssss king king king This Topic gives another method of simplifying expressions — using the rules of exponents. Exponents to Multip

ly y y y y VVVVVariaariaariaariaariabbbbbleslesleslesles Exponents to Multipl Exponents to Multipl ules of ules of Use R Use R ules of Exponents to Multipl Use Rules of Exponents to Multipl ules of Use R Use R To simplify expressions like 4x(x2 – 2x + 1), you need to apply the distributive property as well as rules of exponents, such as xa × xb = xa+b. Example Example Example Example Example 11111 Simplify –2x(x2 – 2y + 1) – x(–4xy + y). Solution Solution Solution Solution Solution –2x(x2 – 2y + 1) – x(–4xy + y) = –2x3 + 4xy – 2x + 4x2y – xy = –2x3 + 3xy – 2x + 4x2y Get rid of Get rid of Get rid of g g g g grrrrrouping symbols fir ouping symbols fir ouping symbols fir ouping symbols firststststst Get rid of Get rid of ouping symbols fir TTTTThen collect lik hen collect lik hen collect lik e ter e ter e termsmsmsmsms hen collect like ter hen collect lik e ter If you have two or more different variables multiplied together, it doesn’t matter what order they’re in. For example, xy is the same as yx, and ab is the same as ba. This is because of the commutative property of multiplication. Example Example Example Example Example 22222 Simplify 2x(y + 1) – y(x + 3). Solution Solution Solution Solution Solution 2x(y + 1) – y(x + 3) = 2xy + 2x – yx – 3y = 2xy – yx + 2x – 3y = xy + 2x – 3y Guided Practice In Exercises 1–9, simplify the algebraic expressions: 1. 2x(3x – 4) 2. –4x(x – 4) 3. –2y(3yx + 2) 4. 2x(x + 5y) + 3y(y + 3) 5. 2y(2x + 2) – 4(2x + 2) 6. 6y(yx – 4) + 5(yx –

4) 7. 7n(3a + b) – 4a(7n + 2b) 8. 2x(2x2 – x) + x(2x – 8) + 3x(x – 4) 9. 2x(k – 9) – k(x – 7) + xk(4 – 3x) Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 7373737373 Don’t forget: See Topic 2.2.1 for the difference between expressions and equations. Check it out: Continue simplifying until you end up with the same number on each side of the equation. ritten as an Equationtiontiontiontion ritten as an Equa essions Can Be WWWWWritten as an Equa ritten as an Equa essions Can Be essions Can Be o Equal Expr TTTTTwwwwwo Equal Expr o Equal Expr o Equal Expressions Can Be ritten as an Equa essions Can Be o Equal Expr For any equation: Left-hand side = Right-hand side When you think you know what numbers the variables in an equation represent, you should always put them into the equation to check that both sides of the equation are equal. Replace the variables with the numbers and make sure that the left side of the equation is equal to the right side of the equation. You do exactly the same thing if a question asks you to prove or show that an equation is true. Example Example Example Example Example 33333 Show that x = –1 is a solution of 4(x – 2) + 2 = 4x – 2(2 – x). Solution Solution Solution Solution Solution 4(x – 2) + 2 = 4x – 2(2 – x) You’re told that x = –1 is a solution, so replace each x with –1. fi 4(–1 – 2) + 2 = 4(–1) – 2[2 – (–1)] fi 4(–3) + 2 = 4(–1) – 2(3) fi –12 + 2 = –4 – 6 fi –10 = –10 This shows that 4(x – 2) + 2 = 4x – 2(2 – x) is a true statement for x = –1. Guided Practice 10. Given x = –1, show that 2(5

x + 3) – 3(3x + 2) = 4x + 3. 11. Given b = –1, show that ( 12. Given x = – 5 3, show that 4(2x – 1) – 5(x – 2) = 1. 13. Show whether x = –2 is or is not a solution of –6x – 15 = –17 – 9x. − − 14. If b = –10, show that. 15. Show whether x = – 2 3 is or is not a solution of –6x – 15 = –17 – 9x. 16. Verify that x = 0 is a solution of 3 8 (3x + 4) – 1 2 (4x – 8) = 5.5. 17. Verify that x = –1 is a solution of (4x – 9)2 = 169. 18. Verify that x = –7 is a solution of 0.2( 4 9 x + 3) – 3 29 45. 7474747474 Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 o or More e e e e VVVVVariaariaariaariaariabbbbbleslesleslesles o or Mor y Contain TTTTTwwwwwo or Mor o or Mor y Contain y Contain tions ma EquaEquaEquaEquaEquations ma tions ma tions may Contain o or Mor y Contain tions ma Sometimes your equation will have two or more different variables. Example Example Example Example Example 44444 Given that a = 3, b = –2, and c = 0, show that –2a(4 + b) = b(5 – c) – 2. Solution Solution Solution Solution Solution Again, just substitute the numbers for the variables and simplify until you show that both sides of the equation are equal. –2a(4 + b) = b(5 – c) – 2 fi –2(3)(4 – 2) = –2(5 – 0) – 2 fi –6(2) = –2(5) – 2 fi –12 = –10 – 2 fi –12 = –12 \ –2a(4 + b) = b(5 – c) – 2 for the given values of a, b, and c. Guided Practice In Exercises 19–24, find the

value of each algebraic expression when the given substitutions are made: 19. 2x(x + 5y) – 3y(y + 3) 20. 6y(yx – 4) – 5(yx – 4) if x = 2, y = 4 if x = 1, y = –1 if a = 0, b = 3, n = 1 7 if x = 4, y = –8 if a = 4, b = 0.2 21. 7n(30 + b) – 4a(7n + 2b) 22. –2y(3yx + 2) 23. –4a(b – 4) 24. 2x(k – a) – k(x – a) + xk(a – 3x) if a = –4.2, x = 0.1, k = 1 3 Independent Practice In Exercises 1–3, simplify the algebraic expressions: 1. –7a(8bc – 3a) 2. 6n(yn + 7) – 7n(n – yn) 3. 2ab(c – d) + 4cb(c – d) – 3ac(2b + d) 4. Find the value of 2a(4a – 3) + (8 – a), if a = 3. 5. Find the value of 2ab(c – d) + 4cb(c – d) – 3ac(2b + d), if a = 8, b = 7, c = 6, d = 0. 6. The formula P = 6.5h + 0.10x is used to find the weekly pay of a salesperson at a local electronic store, where P is the pay in dollars, h is the number of hours worked, and x is the total value of merchandise sold (in dollars) by the salesperson. If the salesperson worked 40 hours and sold $4,250 worth of merchandise, how much pay did she earn? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The material on equations in this Topic leads neatly on to the next Section. The rest of the Sections in this Chapter are all about forming and manipulating equations. Section 2.1 Section 2.1 Section 2.1 — Algebra Basics Section 2.1 Section 2.1 7575757575 TTTT

Topicopicopicopicopic 2.2.12.2.1 2.2.12.2.1 2.2.1 Section 2.2 Equality Equality ties of ties of oper oper PrPrPrPrProper Equality ties of Equality operties of Equality ties of oper PrPrPrPrProper Equality Equality ties of ties of oper oper ties of Equality operties of Equality ties of oper Equality California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as sucsucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll solve one-step equations using properties of equality. Key words: expression equation variable solve isolate equality Now it’s time to use the material on expressions you learned in Section 2.1. An equation contains two expressions, with an equals sign in the middle to show that they’re equal. In this Topic you’ll solve equations that involve addition, subtraction, multiplication, and division. e Equal e Equal essions ar essions ar o Expr tion Shows ws ws ws ws TTTTThahahahahat t t t t TTTTTwwwwwo Expr o Expr tion Sho tion Sho An Equa An Equa e Equal essions are Equal o Expressions ar An Equation Sho e Equal essions ar o Expr tion Sho An Equa An Equa An equation is a way of stating that two expressions have the same value. This equation contains only numbers — there are no unknowns: The expression on the left

-hand side... 24 – 9 = 15...has the same value as the expression on the right-hand side Some equations contain unknown quantities, or variables. The left-hand side... 2x – 3 = 5...equals the right-hand side The value of x that satisfies the equation is called the solution (or root) of the equation. tions tions action in Equa action in Equa dition and Subtr AdAdAdAdAddition and Subtr dition and Subtr tions action in Equations dition and Subtraction in Equa tions action in Equa dition and Subtr Addition Property of Equality For any real numbers a, b, and c, if a = b, then a + c = b + c. Subtraction Property of Equality For any real numbers a, b, and c, if a = b, then a – c = b – c. These properties mean that adding or subtracting the same number on both sides of an equation will give you an equivalent equation. This may allow you to isolate the variable on one side of the equals sign. Finding the possible values of the variables in an equation is called solving the equation. 7676767676 Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 Example Example Example Example Example 11111 Solve x + 9 = 16. Solution Solution Solution Solution Solution You want x on its own, but here x has 9 added to it. So subtract 9 from both sides to get x on its own. (x + 9) – 9 = 16 – 9 x + (9 – 9) = 16 – 9 x + 0 = 16 – 9 x = 16 – 9 x = 7 In Example 1, x = 7 is the root of the equation. If x takes the value 7, then the equation is satisfied. If x takes any other value, then the equation is not satisfied. For example, if x = 6, then the left-hand side has the value 6 + 9 = 15, which does not equal the right-hand side, 16. When you’re actually solving equations, you won’t need to go through all the stages each time — but it’s really important that you understand the theory of the properties of equality. If you have a “+ 9” that you don’t want, you can get rid of it by just subtracting 9 from

both sides. If you have a “– 9” that you want to get rid of, you can just add 9 to both sides. In other words, you just need to use the inverse operations. Example Example Example Example Example 22222 Solve x + 10 = 12. Solution Solution Solution Solution Solution x + 10 = 12 x = 12 – 10 Subtr Subtr Subtr act 10 fr act 10 fr om both sides om both sides Subtract 10 fr act 10 from both sides om both sides Subtr act 10 fr om both sides x = 2 Example Example Example Example Example 33333 Solve x – 7 = 8. Solution Solution Solution Solution Solution = 15 d 7 to both sides d 7 to both sides AdAdAdAdAdd 7 to both sides d 7 to both sides d 7 to both sides Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 7777777777 Guided Practice In Exercises 1–8, solve the equation for the unknown variable. 1. x + 7 = 15 2. x + 2 = –8 3. 4 3 + x = 2 3 5. –9 + x = 10 7. 4 9 = x – 1 3 4. x – (–9) = –17 6. x – 0.9 = 3.7 8. –0.5 = x – 0.125 tions tions vision in Equa vision in Equa tion and Di tion and Di Multiplica Multiplica tions vision in Equations tion and Division in Equa Multiplication and Di tions vision in Equa tion and Di Multiplica Multiplica Multiplication Property of Equality For any real numbers a, b, and c, if a = b, then a × c = b × c. Division Property of Equality For any real numbers a, b, and c, such that c π 0, if a = b, then a c = b c. These properties mean that multiplying or dividing by the same number on both sides of an equation will give you an equivalent equation. That can help you to isolate the variable and solve the equation. vide to Get the VVVVVariaariaariaariaariabbbbble on Its Own le on Its Own le on Its Own vide to Get the vide to Get the y or Di y or Di Multipl Multipl le on Its Own y or Divide to Get the Multiply or Di le on Its

Own vide to Get the y or Di Multipl Multipl As with addition and subtraction, you can get the variable on its own by simply performing the inverse operation. If you have “× 3” on one side of the equation, you can get rid of that value by dividing both sides by 3. If you have a “÷ 3” that you want to get rid of, you can just multiply both sides by 3. Once again, you just need to use the inverse operations. 7878787878 Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 Example Example Example Example Example 44444 Solve 2x = 18. Solution Solution Solution Solution Solution You want x on its own... but here you’ve got 2x. on its ownwnwnwnwn on its o on its o y 2 to get et et et et x on its o y 2 to g y 2 to g vide both sides b vide both sides b DiDiDiDiDivide both sides b vide both sides by 2 to g on its o y 2 to g vide both sides b x = 2 2 18 2 x = 9 2 2 1x = 9 x = 9 Example Example Example Example Example 55555 Solve m 3 = 7. Solution Solution Solution Solution Solution × 3 = 7 × 3 Multiply both sides by 3 to get Multiply both sides by 3 to get Multiply both sides by 3 to get on its own on its own Multiply both sides by 3 to get m on its own on its own Multiply both sides by 3 to get on its own m 3 m = 21 Some equations are a bit more complicated. Take them step by step. Example Example Example Example Example 66666 Solve 4x – 2(2x – 1) = 2 – x + 3(x – 4). Solution Solution Solution Solution Solution 4x – 2(2x – 1) = 2 – x + 3(x – 4) 4x – 4x + 2 = 2 – x + 3x – 12 2 = –10 + 2x 2x = 12 x = 6 Clear out an Clear out an Clear out any gy gy gy gy grrrrrouping symbols ouping symbols ouping symbols ouping symbols Clear out an Clear out an ouping symbols TTTTThen combine lik hen combine lik hen combine lik e ter e ter e termsmsmsmsms hen