2.1 Properties and Operations Goal: Use properties of addition and multiplication. Commutative and Associative Properties Commutative Property of Addition Commutative Property of Multiplication Words In a sum, you can add the numbers in any order. Words In a product, you can multiply the numbers in any order. Numbers 4 (7) 7 4 Numbers 8(5) 5(8) Algebra a b b a Algebra ab ba Associative Property of Addition Associative Property of Multiplication Words Changing the grouping of the numbers in a sum does not change the sum. Words Changing the grouping of the numbers in a product does not change the product. Numbers (9 6) 2 9 (6 2) Numbers (3 p 10) p 4 3 p (10 p 4) Algebra (a b) c a (b c) Algebra (a p b) p c a p (b p c) Example 1 Using Properties of Addition Distance This week, you rode in a car for 42 miles, rode a bike for 5 miles, and rode in a bus for 23 miles. Find the total distance. Use properties of addition to group together distances that are easy to add mentally. Solution The total distance is the sum of the three distances. 42 5 23 (42 5) 23 ( ( ) 23 23) Answer: The total distance is Copyright © Holt McDougal. All rights reserved. Use order of operations. Commutative property of addition Associative property of addition Add Add and and . . miles. Chapter 2 • Pre-Algebra Notetaking Guide 23 2.1 Properties and Operations Goal: Use properties of addition and multiplication. Commutative and Associative Properties Commutative Property of Addition Commutative Property of Multiplication Words In a sum, you can add the numbers in any order. Words In a product, you can multiply the numbers in any order. Numbers 4 (7) 7 4 Numbers 8(5) 5(8) Algebra a b b a Algebra ab ba Associative Property of Addition Associative Property of Multiplication Words Changing the grouping of the numbers in a sum does not change the sum. Words Changing the grouping of the numbers in a product does not change the product. Numbers (9 6) 2 9 (6 2) Numbers (3 p 10) p 4 3 p (10 p 4) Algebra (a b) c a (b c) Algebra (a p b) p c a p (b p c) Example 1 Using Properties of Addition Distance This week, you rode in a car for 42 miles, rode a bike for 5 miles, and rode in a bus for 23 miles. Find the total distance. Use properties of addition to group together distances that are easy to add mentally. Solution The total distance is the sum of the three distances. 42 5 23 (42 5) 23 (5 5 ) 23 ( 42 23) 42 Use order of operations. Commutative property of addition Associative property of addition 5 65 Add 42 and 23 . 70 Add 5 and 65 . Answer: The total distance is 70 miles. Copyright © Holt McDougal. All rights reserved. Chapter 2 • Pre-Algebra Notetaking Guide 23 Example 2 Using Properties of Multiplication Evaluate 4xy when x 8 and y 15. ( )( ) [4( )]( ) [ ( )]( ) [( )( )] ( ) Substitute for x and for y. Multiply 4xy 4 Use order of operations. Commutative property of multiplication Associative property of multiplication Multiply and and . . Checkpoint Evaluate the expression when x 7 and y 25. 1. (2x y) 46 Example 3 2. 4x 2y Using Properties to Simplify Variable Expressions Simplify the expression. a. x 5 2 (x 5) 2 Use order of operations. x (5 2) x property of addition Add 5 and 2. b. 3(9y) (3 p 9)y property of multiplication Multiply 3 and 9. Checkpoint Simplify the expression. 3. n 6 7 24 Chapter 2 • Pre-Algebra Notetaking Guide 4. (4r)(3) Copyright © Holt McDougal. All rights reserved. Using Properties of Multiplication Example 2 Evaluate 4xy when x 8 and y 15. ( )( 15 ) [4( 8 )]( 15 ) [ 8 ( 4 )]( 15 ) 8 [( 4 )( 15 )] 8 ( 60 ) Substitute for x and for y. 480 Multiply 8 and 60 . 4xy 4 8 Use order of operations. Commutative property of multiplication Associative property of multiplication Multiply 4 and 15 . Checkpoint Evaluate the expression when x 7 and y 25. 1. (2x y) 46 2. 4x 2y 85 4900 Using Properties to Simplify Variable Expressions Example 3 Simplify the expression. a. x 5 2 (x 5) 2 Use order of operations. x (5 2) x 7 b. 3(9y) (3 p 9)y 27y Associative property of addition Add 5 and 2. Associative property of multiplication Multiply 3 and 9. Checkpoint Simplify the expression. 3. n 6 7 n 13 24 Chapter 2 • Pre-Algebra Notetaking Guide 4. (4r)(3) 12r Copyright © Holt McDougal. All rights reserved. Identity Properties Identity Property of Addition Identity Property of Multiplication Words The sum of a number and the additive identity, 0, is the number. Words The product of a number and the multiplicative identity, 1, is the number. Numbers 6 0 6 Numbers 4 p 1 4 Algebra a 0 a Algebra a p 1 a Example 4 Identifying Properties Statement Property Illustrated a. (3 2) 4 3 (2 4) b. 0 b b c. 1(7) 7 d. cd dc Checkpoint Identify the property that the statement illustrates. 5. (2 p 6) p 3 2 p (6 p 3) Copyright © Holt McDougal. All rights reserved. 6. q (r) r q Chapter 2 • Pre-Algebra Notetaking Guide 25 Identity Properties Identity Property of Addition Identity Property of Multiplication Words The sum of a number and the additive identity, 0, is the number. Words The product of a number and the multiplicative identity, 1, is the number. Numbers 6 0 6 Numbers 4 p 1 4 Algebra a 0 a Algebra a p 1 a Example 4 Identifying Properties Statement Property Illustrated a. (3 2) 4 3 (2 4) Associative property of addition b. 0 b b Identity property of addition c. 1(7) 7 Identity property of multiplication d. cd dc Commutative property of multiplication Checkpoint Identify the property that the statement illustrates. 5. (2 p 6) p 3 2 p (6 p 3) Associative property of multiplication Copyright © Holt McDougal. All rights reserved. 6. q (r) r q Commutative property of addition Chapter 2 • Pre-Algebra Notetaking Guide 25 Focus On Measurement Converting Between Systems of Measurement Use after Lesson 2.1 Goal: Convert between metric and U.S. customary units. Quantity Conversions Example 1 Length Capacity Weight/Mass 1 in. 2.54 cm 1 fl oz ≈ 29.573 mL 1 oz ≈ 28.35 g 1 ft 0.3048 m 1 qt ≈ 0.946 L 1 lb ≈ 0.454 kg 1 mi ≈ 1.609 km 1 gal ≈ 3.785 L 1 t ≈ 907.2 kg Converting Units of Measure Copy and complete the statement. Round to the nearest whole number. ?m a. 24 ft ≈ Solution ? qt b. 30 L ≈ 0.3048m ≈ a. 24 ft Answer: 24 ft ≈ b. 30 L ≈ Answer: 30 L ≈ m m 1 qt ≈ qt Checkpoint Copy and complete the statement. Round to the nearest whole number. ? kg 1. 3 t ≈ 26 Chapter 2 • Pre-Algebra Notetaking Guide ? L 2. 41 qt ≈ Copyright © Holt McDougal. All rights reserved. Focus On Measurement Converting Between Systems of Measurement Use after Lesson 2.1 Goal: Convert between metric and U.S. customary units. Quantity Conversions Example 1 Length Capacity Weight/Mass 1 in. 2.54 cm 1 fl oz ≈ 29.573 mL 1 oz ≈ 28.35 g 1 ft 0.3048 m 1 qt ≈ 0.946 L 1 lb ≈ 0.454 kg 1 mi ≈ 1.609 km 1 gal ≈ 3.785 L 1 t ≈ 907.2 kg Converting Units of Measure Copy and complete the statement. Round to the nearest whole number. ?m a. 24 ft ≈ Solution a. 24 ft 24 ft ? qt b. 30 L ≈ 0.3048m ≈ 7 m 1 ft Answer: 24 ft ≈ 7 m 1 qt b. 30 L ≈ 30 L ≈ 32 qt 0.946 L Answer: 30 L ≈ 32 qt Checkpoint Copy and complete the statement. Round to the nearest whole number. ? kg 1. 3 t ≈ 3 t ≈ 2722 kg 26 Chapter 2 • Pre-Algebra Notetaking Guide ? L 2. 41 qt ≈ 41 qt ≈ 39 L Copyright © Holt McDougal. All rights reserved. Comparing Units of Measure Example 2 ? 567 g using <, >, or . Copy and complete the statement 25 oz ? 567 g 25 oz ? 25 oz ? 25 oz 25 oz Answer: 20 oz 567 g Checkpoint Copy and complete the statement using <, >, or . ? 9 kg 4. 15 lb ? 65 cm 3. 28 in. Using Multiple Conversion Factors Example 3 ? L. Round to the Copy and complete the statement 180 fl oz ≈ nearest whole number. Solution 1. Convert fluid ounces to 180 fl oz ≈ 2. Convert ≈ to liters. ≈ Answer: . ≈ 180 fl oz ≈ Checkpoint Copy and complete the statement. Round to the nearest whole number. ? cm 5. 5 ft ≈ Copyright © Holt McDougal. All rights reserved. ? m 6. 34 mi ≈ Chapter 2 • Pre-Algebra Notetaking Guide 27 Comparing Units of Measure Example 2 ? 567 g using <, >, or . Copy and complete the statement 25 oz ? 567 g 25 oz Original statement 1 oz ? 567 g 25 oz Convert grams to ounces. 28.35 g ? 20 oz 25 oz Simplify. 25 oz > Compare. 20 oz 20 oz > 567 g Answer: Checkpoint Copy and complete the statement using <, >, or . ? 9 kg 4. 15 lb ? 65 cm 3. 28 in. 28 in. > 65 cm 15 lb < 9 kg Using Multiple Conversion Factors Example 3 ? L. Round to the Copy and complete the statement 180 fl oz ≈ nearest whole number. Solution 1. Convert fluid ounces to milliliters . 180 fl oz ≈ 180 fl oz 29.573 mL ≈ 5323.1 mL 1 fl oz 2. Convert 5323.1 milliliters to liters. 5323.1 mL ≈ 5323.1 mL 1L ≈ 5L 1000 mL Answer: 180 fl oz ≈ 5 L Checkpoint Copy and complete the statement. Round to the nearest whole number. ? cm 5. 5 ft ≈ 5 ft ≈ 152 cm Copyright © Holt McDougal. All rights reserved. ? m 6. 34 mi ≈ 34 mi ≈ 54,706 m Chapter 2 • Pre-Algebra Notetaking Guide 27 2.2 The Distributive Property Goal: Use the distributive property. Vocabulary Equivalent numerical expressions: Equivalent variable expressions: The Distributive Property Algebra a(b c) ab ac Numbers 4(6 3) (b c)a ba ca (6 3)4 a(b c) ab ac 5(7 2) (b c)a ba ca (7 2)5 Example 1 Using the Distributive Property Crafts You are buying beads for a craft project. You need gold, silver, and white beads. A bag of each type of bead costs $3.99. Use the distributive property and mental math to find the total cost of the beads. Solution Total cost 3(3.99) Write expression for total cost. ( ) 3( ) 3( ) Rewrite 3.99 as Multiply using mental math. 3 Chapter 2 • Pre-Algebra Notetaking Guide . Distributive property Subtract using mental math. Answer: The total cost of the beads is $ 28 . Copyright © Holt McDougal. All rights reserved. 2.2 The Distributive Property Goal: Use the distributive property. Vocabulary Equivalent numerical Numerical expressions that have the same value expressions: Equivalent variable Equivalent variable expressions have the same value for all values of the variable(s). expressions: The Distributive Property Algebra a(b c) ab ac Numbers 4(6 3) 4(6) 4(3) (b c)a ba ca (6 3)4 6(4) 3(4) a(b c) ab ac 5(7 2) 5(7) 5(2) (b c)a ba ca (7 2)5 7(5) 2(5) Example 1 Using the Distributive Property Crafts You are buying beads for a craft project. You need gold, silver, and white beads. A bag of each type of bead costs $3.99. Use the distributive property and mental math to find the total cost of the beads. Solution Total cost 3(3.99) Write expression for total cost. ( ) 3( 4 ) 3( 0.01 ) Rewrite 3.99 as 4 0.01 . 12 0.03 Multiply using mental math. 11.97 Subtract using mental math. 3 4 0.01 Distributive property Answer: The total cost of the beads is $ 11.97 . 28 Chapter 2 • Pre-Algebra Notetaking Guide Copyright © Holt McDougal. All rights reserved. Checkpoint Use the distributive property to evaluate the expression. 1. 2(9 4) 2. (12 3)3 3. (4 11)(4) Evaluate the expression using the distributive property and mental math. 4. 5(103) Example 2 5. 4(3.8) 6. 3(6.03) Writing Equivalent Variable Expressions Use the distributive property to write an equivalent variable expression. a. 2(x 10) Distributive property Multiply. b. (m 3)(4) Multiply. Definition of subtraction c. 3(2y 6) Copyright © Holt McDougal. All rights reserved. Distributive property Distributive property Multiply. Definition of subtraction Chapter 2 • Pre-Algebra Notetaking Guide 29 Checkpoint Use the distributive property to evaluate the expression. 1. 2(9 4) 2. (12 3)3 26 3. (4 11)(4) 27 28 Evaluate the expression using the distributive property and mental math. 4. 5(103) 5. 4(3.8) 515 Example 2 6. 3(6.03) 15.2 18.09 Writing Equivalent Variable Expressions Use the distributive property to write an equivalent variable expression. a. 2(x 10) 2(x) 2(10) Distributive property 2x 20 Multiply. b. (m 3)(4) m(4) 3(4) Distributive property 4m (12) Multiply. 4m 12 Definition of subtraction c. 3(2y 6) 3(2y) (3)(6) Copyright © Holt McDougal. All rights reserved. Distributive property 6y (18) Multiply. 6y 18 Definition of subtraction Chapter 2 • Pre-Algebra Notetaking Guide 29 Checkpoint Use the distributive property to write an equivalent variable expression. 7. (x 7)4 8. 3(4m 7) Finding Areas of Geometric Figures Example 3 Find the area of the rectangle or triangle. a. b. 12 3y 5 3x 2 14 Solution a. Use the formula for the area of a rectangle. 1 2 ( )( ) ( ) ( ) 1 2 A bh A lw b. Use the formula for the area of a triangle. ( ( ( ) )( ) ) ( ) Answer: The area is square units. Answer: The area is square units. Checkpoint Find the area of the rectangle or triangle. 9. 10. 9 2y 7 5x 3 30 Chapter 2 • Pre-Algebra Notetaking Guide 10 Copyright © Holt McDougal. All rights reserved. Checkpoint Use the distributive property to write an equivalent variable expression. 7. (x 7)4 8. 3(4m 7) 4x 28 12m 21 Finding Areas of Geometric Figures Example 3 Find the area of the rectangle or triangle. a. b. 12 3y 5 3x 2 14 Solution a. Use the formula for the area of a rectangle. b. Use the formula for the area of a triangle. 1 1 A bh 14 A lw 2 )( 12x 3y ) ( 12 3y ) ( 12 ) 7 ( 3y ) 2 ( ( 3x 2 )( 5 ) 3x ( 5 ) 2 ( 5 ) 7 15x 10 84 21y Answer: The area is (15x 10) square units. 7 Answer: The area is (84 21y) square units. Checkpoint Find the area of the rectangle or triangle. 9. 10. 9 2y 7 5x 3 35x 21 30 Chapter 2 • Pre-Algebra Notetaking Guide 10 45 10y Copyright © Holt McDougal. All rights reserved. 2.3 Simplifying Variable Expressions Goal: Simplify variable expressions. Vocabulary Terms of an expression: Coefficient of a term: Constant term: Like terms: Example 1 Identifying Parts of an Expression Identify the terms, like terms, coefficients, and constant terms of the expression 5 2x 3 x. Solution 1. Write the expression as a sum: . 2. Identify the parts of the expression. Note that because x x, the coefficient of x is . Terms: Like terms: Coefficients: Constant terms: Checkpoint Identify the terms, like terms, coefficients, and constant terms of the expression. 1. 4y 6 3y Copyright © Holt McDougal. All rights reserved. 2. 9 w 5 8w Chapter 2 • Pre-Algebra Notetaking Guide 31 2.3 Simplifying Variable Expressions Goal: Simplify variable expressions. Vocabulary Terms of an The parts of an expression that are added together expression: are called terms. Coefficient of The coefficient of a term with a variable is the number part of the term. a term: Constant term: A constant term has a number but no variable. Like terms: Example 1 Like terms are terms that have identical variable parts. Identifying Parts of an Expression Identify the terms, like terms, coefficients, and constant terms of the expression 5 2x 3 x. Solution 1. Write the expression as a sum: 5 (2x) (3) x . 2. Identify the parts of the expression. Note that because x 1 x, the coefficient of x is 1 . Terms: 5, 2x, 3, x Like terms: 5 and 3; 2x and x Coefficients: 2, 1 Constant terms: 5, 3 Checkpoint Identify the terms, like terms, coefficients, and constant terms of the expression. 1. 4y 6 3y Terms: 4y, 6, 3y Terms: 9, w, 5, 8w Like terms: 4y and 3y Like terms: 9 and 5; w and 8w Coefficients: 4, 3 Constant term: 6 Copyright © Holt McDougal. All rights reserved. 2. 9 w 5 8w Coefficients: 1, 8 Constant terms: 9, 5 Chapter 2 • Pre-Algebra Notetaking Guide 31 Example 2 Simplifying an Expression 5m 8 3m 7 5m 8 5m [ ( )( ( ) ( )]m Write as a sum. ) ( ) ( Example 3 Commutative property ) Distributive property Simplify. Simplifying Expressions with Parentheses a. 3(x 2) x 9 x9 Distributive property Group like terms. Combine like terms. b. 2k 5(k 4) 2k Distributive property c. 5a (5a 7) 5a Combine like terms. (5a 7) Identity property 5a Distributive property Combine like terms. Simplify. Checkpoint Simplify the expression. 32 3. 4y 6 3y 4. 9 w 5 8w 5. 4(x 1) 2 x 7 6. 6(k 3) 5k Chapter 2 • Pre-Algebra Notetaking Guide Copyright © Holt McDougal. All rights reserved. Example 2 Simplifying an Expression 5m 8 3m 7 5m 8 5m [5 ( 3m ) ( 7 ) ( 3m ) 8 ( 3 )]m ( 7 ) 8 Write as a sum. Commutative property ( 7 ) 2m 1 Example 3 Distributive property Simplify. Simplifying Expressions with Parentheses a. 3(x 2) x 9 3x 6 x 9 Distributive property 3x x 6 9 Group like terms. 2x 15 Combine like terms. b. 2k 5(k 4) 2k 5k 20 3k 20 Distributive property Combine like terms. c. 5a (5a 7) 5a 1 (5a 7) Identity property 5a 5a 7 Distributive property 07 Combine like terms. 7 Simplify. Checkpoint Simplify the expression. 3. 4y 6 3y 7y 6 5. 4(x 1) 2 x 7 2x 11 32 Chapter 2 • Pre-Algebra Notetaking Guide 4. 9 w 5 8w 7w 4 6. 6(k 3) 5k k 18 Copyright © Holt McDougal. All rights reserved. 2.4 Variables and Equations Goal: Solve equations with variables. Vocabulary Equation: Solution of an equation: Solving an equation: Example 1 Writing Verbal Sentences as Equations Verbal Sentence Equation a. The sum of x and 4 is 8. b. The difference of 7 and y is 13. c. The product of 2 and p is 24. d. The quotient of n and 3 is 5. Example 2 Checking Possible Solutions Tell whether 7 or 8 is a solution of x 3 5. a. Substitute 7 for x. b. Substitute 8 for x. x35 x35 35 35 5 Answer: 7 solution. Copyright © Holt McDougal. All rights reserved. 5 a Answer: 8 solution. a Chapter 2 • Pre-Algebra Notetaking Guide 33 2.4 Variables and Equations Goal: Solve equations with variables. Vocabulary Equation: An equation is a mathematical sentence formed by placing an equal sign, , between two expressions. A solution of an equation with a variable is a Solution of an number that produces a true statement when equation: it is substituted for the variable. Solving an Finding all solutions of an equation is called solving equation: the equation. Example 1 Writing Verbal Sentences as Equations Verbal Sentence Equation a. The sum of x and 4 is 8. x48 b. The difference of 7 and y is 13. 7 y 13 c. The product of 2 and p is 24. 2p 24 d. The quotient of n and 3 is 5. n 5 3 Example 2 Checking Possible Solutions Tell whether 7 or 8 is a solution of x 3 5. a. Substitute 7 for x. b. Substitute 8 for x. x35 x35 7 35 8 35 5 5 Answer: 7 is not a solution. Answer: 8 solution. 4 Copyright © Holt McDougal. All rights reserved. 5 is a Chapter 2 • Pre-Algebra Notetaking Guide 33 Checkpoint Write the verbal sentence as an equation. 1. The sum of x and 7 is 12. 2. The quotient of n and 4 is 16. 3. Tell whether 8 or 10 is a solution of x 4 6. Example 3 Solving Equations Using Mental Math Equation Question Solution Check a. x 4 7 47 b. 12 n 5 12 c. 18 3t 18 3 y 4 d. 5 5 ( ) 4 5 Checkpoint Solve the equation using mental math. 4. x 8 10 34 Chapter 2 • Pre-Algebra Notetaking Guide 5. 24 4m c 3 6. 9 Copyright © Holt McDougal. All rights reserved. Checkpoint Write the verbal sentence as an equation. 1. The sum of x and 7 is 12. x 7 12 2. The quotient of n and 4 is 16. n 16 4 3. Tell whether 8 or 10 is a solution of x 4 6. 8 is not a solution; 10 is a solution Example 3 Solving Equations Using Mental Math Equation Question Solution a. x 4 7 What number plus 4 equals 7? 3 3 47 b. 12 n 5 12 minus what number equals 5? 7 12 7 5 c. 18 3t 18 equals 3 times what number? 6 18 3 6 y 4 d. 5 What number divided by 4 equals 5? Check ( ) 20 4 5 20 Checkpoint Solve the equation using mental math. 4. x 8 10 18 34 Chapter 2 • Pre-Algebra Notetaking Guide 5. 24 4m 6 c 3 6. 9 27 Copyright © Holt McDougal. All rights reserved. 2.5 Solving Equations Using Addition or Subtraction Goal: Solve equations using addition or subtraction. Vocabulary Inverse operations: Equivalent equations: Subtraction Property of Equality Words Subtracting the same number from each side of an equation produces an equivalent equation. Numbers If x 3 5, then x 3 Algebra If x a b, then x a x 5 b , or x . , or . Example 1 Solving an Equation Using Subtraction Solve x 5 2. Solution When you solve an equation, your goal is to write an equivalent equation that has the variable by itself on one side. This process is called solving for the variable. Use the subtraction property of equality to solve for x. x 5 2 x5 Write original equation. 2 x x 5 2 5 2 2 Copyright © Holt McDougal. All rights reserved. from each side. Simplify. Answer: The solution is Check: Subtract . Write original equation. Substitute for x. . Chapter 2 • Pre-Algebra Notetaking Guide 35 2.5 Solving Equations Using Addition or Subtraction Goal: Solve equations using addition or subtraction. Vocabulary Inverse operations: Inverse operations are two operations that undo each other, such as addition and subtraction. Equivalent equations: Equivalent equations are equations that have the same solution(s). Subtraction Property of Equality Words Subtracting the same number from each side of an equation produces an equivalent equation. Numbers If x 3 5, then x 3 3 5 3 , or x 2 . Algebra If x a b, then x a a b a , or x ba . Example 1 Solving an Equation Using Subtraction Solve x 5 2. Solution When you solve an equation, your goal is to write an equivalent equation that has the variable by itself on one side. This process is called solving for the variable. Use the subtraction property of equality to solve for x. x 5 2 x 5 5 2 5 x 7 Write original equation. Subtract 5 from each side. Simplify. Answer: The solution is 7 . Check: x 5 2 7 5 2 2 Copyright © Holt McDougal. All rights reserved. 2 Write original equation. Substitute for x. Solution checks . Chapter 2 • Pre-Algebra Notetaking Guide 35 Addition Property of Equality Words Adding the same number to each side of an equation produces an equivalent equation. Numbers If x 3 5, then x 3 5 Algebra If x a b, then x a x b , or x . , or . Solving an Equation Using Addition Example 2 Solve 12 y 7. Solution Use the addition property of equality to solve for y. 12 y 7 12 Write original equation. y7 Add y Simplify. Answer: The solution is to each side. . Checkpoint Solve the equation. Check your solution. 1. x 6 19 36 Chapter 2 • Pre-Algebra Notetaking Guide 2. 5 y 12 3. m 3 11 Copyright © Holt McDougal. All rights reserved. Addition Property of Equality Words Adding the same number to each side of an equation produces an equivalent equation. Numbers If x 3 5, then x 3 3 5 3 , or x 8 . Algebra If x a b, then x a a b a , or x ba . Example 2 Solving an Equation Using Addition Solve 12 y 7. Solution Use the addition property of equality to solve for y. 12 y 7 Write original equation. 12 7 y 7 7 19 y Add 7 to each side. Simplify. Answer: The solution is 19 . Checkpoint Solve the equation. Check your solution. 1. x 6 19 13 36 Chapter 2 • Pre-Algebra Notetaking Guide 2. 5 y 12 17 3. m 3 11 8 Copyright © Holt McDougal. All rights reserved. 2.6 Solving Equations Using Multiplication or Division Goal: Solve equations using multiplication or division. Division Property of Equality Words Dividing each side of an equation by the same nonzero number produces an equivalent equation. 3x 12 Numbers If 3x 12, then , or x Remember that you cannot divide a number or an expression by 0. ax . b Algebra If ax b and a 0, then , or x . Solving an Equation Using Division Example 1 Solve 7x 42. Solution 7x 42 Write original equation. 7x 42 Divide each side by x Simplify. Answer: The solution is 7x 42 Check: 7 . ( ) 42 42 . Write original equation. Substitute for x. . Checkpoint Solve the equation. Check your solution. 1. 5x 45 Copyright © Holt McDougal. All rights reserved. 2. 56 8y Chapter 2 • Pre-Algebra Notetaking Guide 37 2.6 Solving Equations Using Multiplication or Division Goal: Solve equations using multiplication or division. Division Property of Equality Words Dividing each side of an equation by the same nonzero number produces an equivalent equation. 3x 12 3 3 Numbers If 3x 12, then , or x 4 . Remember that you cannot divide a number or an expression by 0. ax b b Algebra If ax b and a 0, then , or x a . a a Solving an Equation Using Division Example 1 Solve 7x 42. Solution 7x 42 Write original equation. 7x 42 7 7 Divide each side by 7 . x 6 Simplify. Answer: The solution is 6 . 7x 42 Check: ( 7 6 42 ) 42 42 Write original equation. Substitute for x. Solution checks . Checkpoint Solve the equation. Check your solution. 1. 5x 45 2. 56 8y 9 Copyright © Holt McDougal. All rights reserved. 7 Chapter 2 • Pre-Algebra Notetaking Guide 37 Multiplication Property of Equality Words Multiplying each side of an equation by the same nonzero number produces an equivalent equation. x 3 x 3 Numbers If 12, then p x a Algebra If b and a 0, then p 12, or x x a p . p b, or x . Solving an Equation Using Multiplication Example 2 w 11 Solve 5 . Solution w 11 5 p5 Write original equation. w 11 p w Answer: The solution is Multiply each side by . Simplify. . Checkpoint Solve the equation. Check your solution. m 4 3. 11 38 Chapter 2 • Pre-Algebra Notetaking Guide c 6 4. 9 Copyright © Holt McDougal. All rights reserved. Multiplication Property of Equality Words Multiplying each side of an equation by the same nonzero number produces an equivalent equation. x 3 x 3 Numbers If 12, then 3 p 3 p 12, or x 36 . x a x a Algebra If b and a 0, then a p a p b, or x ab . Solving an Equation Using Multiplication Example 2 w 11 Solve 5 . Solution w 11 5 w 11 p 5 11 p 11 55 w Write original equation. Multiply each side by 11 . Simplify. Answer: The solution is 55 . Checkpoint Solve the equation. Check your solution. c 6 m 4 3. 11 4. 9 44 38 Chapter 2 • Pre-Algebra Notetaking Guide 54 Copyright © Holt McDougal. All rights reserved. 2.7 Decimal Operations and Equations with Decimals Goal: Solve equations involving decimals. Example 1 Adding and Subtracting Decimals a. Find the sum 1.7 (3.4). Use the rule for adding numbers with the same sign. Add and . Both decimals are , so the sum is . 1.7 (3.4) b. Find the difference 21.29 (34.62). First rewrite the difference as a sum: 21.29 34.62. Then use the rule for adding numbers with different signs. Subtract from . > , so the sum has the same sign as . 21.29 (34.62) Checkpoint Find the sum or difference. 1. 2.8 (5.9) Copyright © Holt McDougal. All rights reserved. 2. 7.12 (3.46) Chapter 2 • Pre-Algebra Notetaking Guide 39 2.7 Decimal Operations and Equations with Decimals Goal: Solve equations involving decimals. Example 1 Adding and Subtracting Decimals a. Find the sum 1.7 (3.4). Use the rule for adding numbers with the same sign. Add 1.7 and 3.4 . Both decimals are negative , so the sum is negative . 1.7 (3.4) 5.1 b. Find the difference 21.29 (34.62). First rewrite the difference as a sum: 21.29 34.62. Then use the rule for adding numbers with different signs. Subtract 21.29 from 34.62 . 34.62 > 21.29 , so the sum has the same sign as 34.62 . 21.29 (34.62) 13.33 Checkpoint Find the sum or difference. 1. 2.8 (5.9) 2. 7.12 (3.46) 8.7 10.58 Copyright © Holt McDougal. All rights reserved. Chapter 2 • Pre-Algebra Notetaking Guide 39 You can use estimation to check the results of operations with decimals. For instance, notice that 29.07 (1.9) 15.3 is about 30 (2), or 15. So, an answer of 15.3 is reasonable. Multiplying and Dividing Decimals Example 2 a. 0.4(13.7) Different signs: Product is . b. 2.5(6.75) Same signs: Product is . c. 23.49 (2.9) Same signs: Quotient is . d. 18.05 (1.9) Different signs: Quotient is . Checkpoint Find the product or quotient. 3. 2.8(5.9) 4. 7.093 (3.46) Solving Addition and Subtraction Equations Example 3 Solve the equation. a. x 6.3 4.8 b. y 5.74 3.51 Solution x 6.3 4.8 a. x 6.3 4.8 x y 5.74 3.51 y 40 Chapter 2 • Pre-Algebra Notetaking Guide Subtract from each side. Simplify. y 5.74 3.51 b. Write original equation. Write original equation. Add to each side. Simplify. Copyright © Holt McDougal. All rights reserved. You can use estimation to check the results of operations with decimals. For instance, notice that 29.07 (1.9) 15.3 is about 30 (2), or 15. So, an answer of 15.3 is reasonable. Example 2 Multiplying and Dividing Decimals a. 0.4(13.7) 5.48 Different signs: Product is negative . b. 2.5(6.75) 16.875 Same signs: Product is positive . c. 23.49 (2.9) 8.1 Same signs: Quotient is positive . d. 18.05 (1.9) 9.5 Different signs: Quotient is negative . Checkpoint Find the product or quotient. 3. 2.8(5.9) 4. 7.093 (3.46) 2.05 16.52 Example 3 Solving Addition and Subtraction Equations Solve the equation. a. x 6.3 4.8 b. y 5.74 3.51 Solution x 6.3 4.8 a. x 6.3 6.3 4.8 6.3 x 1.5 Subtract 6.3 from each side. Simplify. y 5.74 3.51 Write original equation. y 5.74 5.74 3.51 5.74 Add 5.74 to each side. b. y 2.23 40 Write original equation. Chapter 2 • Pre-Algebra Notetaking Guide Simplify. Copyright © Holt McDougal. All rights reserved. Checkpoint Solve the equation. Check your solution. 5. x 5.6 9.4 Example 4 6. 3.5 y 1.2 7. m 5.3 7.2 Solving Multiplication and Division Equations Solve the equation. a. 0.8m 4.8 n 5 b. 2.15 Solution a. 0.8m 4.8 0.8m 4.8 m b. n 2.15 5 n (2.15) 5 n Write original equation. Divide each side by . Simplify. Write original equation. Multiply each side by . Simplify. Checkpoint Solve the equation. Check your solution. 8. 6x 43.2 Copyright © Holt McDougal. All rights reserved. y 3.1 9. 8.4 Chapter 2 • Pre-Algebra Notetaking Guide 41 Checkpoint Solve the equation. Check your solution. 5. x 5.6 9.4 6. 3.5 y 1.2 4.7 3.8 Example 4 7. m 5.3 7.2 1.9 Solving Multiplication and Division Equations Solve the equation. a. 0.8m 4.8 n 5 b. 2.15 Solution a. 0.8m 4.8 0.8m 4.8 0.8 0.8 m 6 b. 5 n 2.15 5 n 5 (2.15) 5 n 10.75 Write original equation. Divide each side by 0.8 . Simplify. Write original equation. Multiply each side by 5 . Simplify. Checkpoint Solve the equation. Check your solution. 8. 6x 43.2 7.2 Copyright © Holt McDougal. All rights reserved. y 3.1 9. 8.4 26.04 Chapter 2 • Pre-Algebra Notetaking Guide 41 2 Words to Review Give an example of the vocabulary word. Additive identity Multiplicative identity Equivalent numerical expressions Equivalent variable expressions Term, coefficient, constant term, like terms Equation Solution of an equation Solving an equation Inverse operations Equivalent equations Review your notes and Chapter 2 by using the Chapter Review on pages 108–111 of your textbook. 42 Chapter 2 • Pre-Algebra Notetaking Guide Copyright © Holt McDougal. All rights reserved. 2 Words to Review Give an example of the vocabulary word. Additive identity 0 Equivalent numerical expressions 3(3.99) 3(4 0.01) Term, coefficient, constant term, like terms 5 2x 3 x Terms: 5, 2x, 3, x Like terms: 5 and 3; 2x and x Coefficients: 2, 1 Constant terms: 5, 3 Solution of an equation 5 is a solution of 2x 10. Inverse operations Addition and subtraction; Multiplication and division Multiplicative identity 1 Equivalent variable expressions 2(x 10) 2x 20 Equation 2x 10 Solving an equation x 5 2 x 5 5 2 5 x 7 Equivalent equations 2x 10 and x 5 Review your notes and Chapter 2 by using the Chapter Review on pages 108–111 of your textbook. 42 Chapter 2 • Pre-Algebra Notetaking Guide Copyright © Holt McDougal. All rights reserved.