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: 50

Answer

{15}

Work Step by Step

$\frac{z}{5}$ - $\frac{1}{2}$ = $\frac{z}{6}$ Step 1 : Collect variable terms on one side and constants on the other side. Add $\frac{1}{2}$ on both the sides $\frac{z}{5}$ - $\frac{1}{2}$ + $\frac{1}{2}$ = $\frac{z}{6}$ + $\frac{1}{2}$ $\frac{z}{5}$ = $\frac{z}{6}$ + $\frac{1}{2}$ Subtract $\frac{z}{6}$ from both the sides $\frac{z}{5}$ - $\frac{z}{6}$ = $\frac{z}{6}$ + $\frac{1}{2}$ - $\frac{z}{6}$ $\frac{z}{5}$ - $\frac{z}{6}$ = $\frac{1}{2}$ Step 2 : Multiply both the sides by 30 ($\frac{z}{5}$ - $\frac{z}{6}$)*30 = $\frac{1}{2}$*30 $\frac{z}{5}$ * 30 - $\frac{z}{6}$ * 30= $\frac{1}{2}$*30 6z - 5z = 15 Step 3: Add 6z - 5z = z z = 15 Now we check the proposed solution, 15 , by replacing z with 15 in the original equation. Step 1: the original equation $\frac{z}{5}$ - $\frac{1}{2}$ = $\frac{z}{6}$ Step2: Substitute 15 for z $\frac{15}{5}$ - $\frac{1}{2}$ = $\frac{15}{6}$ Step 3: Divide $\frac{15 }{5}$ = 3 3 - $\frac{1}{2}$ = $\frac{15}{6}$ Step 4 : Multiply both the sides by 6 (3 - $\frac{1}{2}$)*6 = $\frac{15}{6}$*6 18 - 3 = 15 Step 5 : Subtract 18 - 3 = 15 15 = 15 Since the check results in true statement, we conclude that the solution set of the given equation is {15}.
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