Section 2 3 Using the Addition and Multiplication

Section 2. 3 Using the Addition and Multiplication Principles Together Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc.

Example Solve for w. 3 w 9 = 24 3 w 9 + 9 = 24 + 9 3 w = 15 Use the division principle to divide both sides of the equation by 3. w = 5 Check your answer in the original equation. 3( 5) 9 = 24 15 9 = 24 24 = 24 Use the addition principle to add 9 to both sides of the equation. Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 2

Example Solve for x. 9 x = 6 x + 15 9 x + ( 6 x) = 6 x + ( 6 x) + 15 3 x = 15 Add – 6 x to both sides. Notice 6 x + (– 6 x) eliminates the variable on the right side. Use the division principle to divide both sides of the equation by 3. x=5 Check your answer in the original equation. 9(5) = 6(5) + 15 45 = 30 + 15 45 = 45 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 3

Example Solve for c. 9 c + 5 = 3 c 13 9 c + 5 + ( 5) = 3 c 13 + ( 5) 9 c = 3 c 18 9 c + ( 3 c) = 3 c 18 + ( 3 c) 6 c = 18 c = 3 Check your answer in the original equation. Add 5 to both sides of the equation. Add ( 3 c) to both sides of the equation. Divide both sides of the equation by 6. 9( 3) + 5 = 3( 3) 13 27 + 5 = 9 13 22 = 22 Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 4

Example Solve for x. 5 + 7 x 19 = 8 x 6 + 3 x Add like terms on both sides of the equation. 7 x 14 = 11 x 6 7 x 14 + 14 = 11 x 6 + 14 7 x = 11 x + 8 7 x 11 x = 11 x + 8 11 x 4 x = 8 Add 14 to both sides of the equation. Add ( 11 x) to both sides of the equation. Divide both sides of the equation by 4. Be sure to check your answer! Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 5

Solving Equations with Parentheses The equations that you just solved are simpler versions of the equations that will now discuss. These equations contain parentheses. If the parentheses are first removed, the problem then becomes just like those encountered previously. We can use the distributive property to remove the parentheses. Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 6

Example Solve for d. 5 d 6(d + 1) = 2 d 6 5 d 6 d + ( 6) = 2 d 6 Distribute to remove the parentheses. Add like terms. d + ( 6) + 6 = 2 d 6 + 6 Add 6 to both sides of the equation. d = 2 d d 2 d = 2 d 3 d = 0 d=0 Simplify. Add ( 2 d) to both sides of the equation. Simplify. Be sure to check your answer! Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 7

Example Solve for x. 0. 3(1. 2 x – 3. 6) = 4. 2 x – 16. 44 0. 36 x – 1. 08 = 4. 2 x – 16. 44 Remove parentheses. 0. 36 x – 1. 08 = 4. 2 x – 0. 36 x – 16. 44 Subtract 0. 36 x from both sides. – 1. 08 = 3. 84 x – 16. 44 Combine like terms. – 1. 08 + 16. 44 = 3. 84 x – 16. 44 + 16. 44 Add 16. 44 to both sides. 15. 36 = 3. 84 x Simplify. Divide both sides by 3. 84. 4=x Be sure to check your answer! Copyright © 2012, 2009, 2005, 2002 Pearson Education, Inc. 8