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: 39

Answer

{6}

Work Step by Step

5(2x - 8) - 2 = 5(x - 3) + 3 Step 1 : Use distributive property 5.2x -5.8 -2 = 5.x -5.3 + 3 Multiply 10x -40 - 2 = 5x -15 +3 Solve 10x -42 = 5x -12 Step 2 : Collect variable terms on one side and constants on the other side. subtract 5x from both the sides 10x - 42 -5x = 5x -12 - 5x 5x - 42 = -12 Add 42 on both the sides 5x - 42 + 42 = -12 + 42 5x = 30 Divide both the sides by 5 $\frac{5x}{5}$ = $\frac{30}{5}$ x = 6 Now we check the proposed solution, 6 , by replacing x with 6 in the original equation. Step 1: the original equation 5(2x - 8) - 2 = 5(x - 3) + 3 Step2: Substitute 6 for x 5(2.6 - 8) - 2 = 5(6 - 3) + 3 Step 3: Simplify 5(12 - 8) - 2 = 5(3) + 3 5(4) -2 = 5(3) + 3 Multiply 5(4) = 5.4 = 20, 5(3) = 5.3 = 15 20 - 2 = 15 + 3 Solve 18 = 18 Since the check results in true statement, we conclude that the solution set of the given equation is {6}
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