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: 36

Answer

{-5}

Work Step by Step

3(3y - 1) = 4(3 + 3y) Step 1 : Use distributive property 3.3y - 3.1 = 4.3 + 4.3y simplify 9y - 3 = 12 + 12y Step 2 : Collect variable terms on one side and constants on the other side. subtract 12y from both the sides 9y - 3 - 12y = 12 + 12y - 12y -3y -3 = 12 Add 3 on both the sides -3y -3 +3 = 12+3 -3y = 15 Divide both the sides by -3 $\frac{-3y}{-3}$ = $\frac{15}{-3}$ y = -5 Now we check the proposed solution, -5 , by replacing y with -5 in the original equation. Step 1: the original equation 3(3y - 1) = 4(3 + 3y) Step2: Substitute -5 for y 3(3(-5) - 1) = 4(3 + 3(-5)) Step 3: Multiply 3(-5) = -15, 3(-15 -1) = 4(3-15) Step 4: Solve 3(-16) = 4(-12) Multiply 3(-16) = -48 , 4(-12) = -48 -48 = -48 Since the check results in true statement, we conclude that the solution set of the given equation is {-5}
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