Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2 - Page 362: 35

Answer

{-4}

Work Step by Step

8(y + 2) = 2(3y + 4) Step 1 : Use distributive property 8.y + 8.2 = 2.3y + 2.4 simplify 8y + 16 = 6y + 8 Step 2 : Collect variable terms on one side and constants on the other side. subtract 6y from both the sides 8y + 16 - 6y = 6y + 8 -6y 2y + 16 = 8 subtract 16 from both the sides 2y+ 16 - 16 = 8 -16 2y = -8 Divide both the sides by 2 $\frac{2y}{2}$ = $\frac{-8}{2}$ y = -4 Now we check the proposed solution, -4 , by replacing y with -4 in the original equation. Step 1: the original equation 8(y + 2) = 2(3y + 4) Step2: Substitute -4 for for 8(-4 + 2) = 2(3(-4) + 4) Step 3: Multiply 3(-4) = -12 8(-4 + 2) = 2(-12 + 4) Step 4: Solve 8(-2) = 2(-8) Multiply 8(-2) = -16, 2(-8) = -16 -16 =-16 Since the check results in true statement, we conclude that the solution set of the given equation is {-4}
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