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: 49

Answer

{24}

Work Step by Step

20 - $\frac{z}{3}$ = $\frac{z}{2}$ Step 1 : Collect variable terms on one side and constants on the other side. Add $\frac{z}{3}$ on both the sides 20 - $\frac{z}{3}$ + $\frac{z}{3}$= $\frac{z}{2}$ + $\frac{z}{3}$ 20 = $\frac{z}{2}$ + $\frac{z}{3}$ or $\frac{z}{2}$ + $\frac{z}{3}$ = 20 Step 2 : Multiply both the sides by 6 ($\frac{z}{2}$ + $\frac{z}{3}$ )* 6= 20* 6 $\frac{z}{2}$ * 6+ $\frac{z}{3}$ * 6= 20* 6 3z + 2z = 120 Step 3: Add 3z + 2z = (3+2 )z = 5z 5z = 120 Step 4: Divide both the sides by 5 $\frac{5z}{5}$ = $\frac{120}{5}$ z = 24 Now we check the proposed solution, 24 , by replacing z with 24 in the original equation. Step 1: the original equation 20 - $\frac{z}{3}$ = $\frac{z}{2}$ Step2: Substitute 24 for z 20 - $\frac{24}{3}$ = $\frac{24}{2}$ Step 3: Divide $\frac{24}{3}$ = 8, $\frac{24}{2}$ = 12 20 - 8 = 12 Step 4 : Subtract 20 -8 =12 12 = 12 Since the check results in true statement, we conclude that the solution set of the given equation is {24}.
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