Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 3 - Solving Inequalities - Get Ready - Page 161: 27

Answer

$y=\frac{1}{2}$

Work Step by Step

$\frac{1}{3}+\frac{4y}{6}=\frac{2}{3}\longrightarrow$ convert all fractions to a common denominator $\frac{1}{3}+\frac{4y\div2}{6\div2}=\frac{2}{3}\longrightarrow$ simplify $\frac{1}{3}+\frac{2y}{3}=\frac{2}{3}\longrightarrow$ subtract $\frac{1}{3}$ from each side $\frac{1}{3}+\frac{2y}{3}-\frac{1}{3}=\frac{2}{3}-\frac{1}{3}\longrightarrow$ subtract $\frac{2y}{3}=\frac{1}{3}\longrightarrow$ multiply both sides by $\frac{3}{2}$ $\frac{2y}{3}\times\frac{3}{2}=\frac{1}{3}\times\frac{3}{2}\longrightarrow$ multiply $y=\frac{3}{6}\longrightarrow$ reduce $y=\frac{3\div3}{6\div3}\longrightarrow$ simplify $y=\frac{1}{2}$ When we plug our solution back into the equation, we find that each side of the equation is equal, making our answer correct.
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