Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 22

Answer

$y=\frac{21}{11}$

Work Step by Step

Given: $\frac{2}{3}y+\frac{1}{2}(y-3)=\frac{y+1}{4}$ Distribute $\frac{1}{2}$ and $\frac{1}{4}$ to simply the equation to: $\frac{2}{3}y+\frac{1}{2}y-\frac{3}{2}=\frac{1}{4}y+\frac{1}{4}$ Subtract $\frac{1}{4}y$ from both sides: $\frac{2}{3}y+\frac{1}{2}y-\frac{3}{2}-\frac{1}{4}y=\frac{1}{4}$ Add $\frac{3}{2}$ to both sides: $\frac{2}{3}y+\frac{1}{2}y-\frac{1}{4}y=\frac{1}{4}+\frac{3}{2}$ Group like terms: $\frac{8}{12}y+\frac{6}{12}y-\frac{3}{12}y=\frac{1}{4}+\frac{6}{4}$ $\frac{11}{12}y=\frac{7}{4}$ Multiply both sides by $\frac{12}{11}$: $y=\frac{7}{4}\cdot\frac{12}{11}=\frac{21}{11}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.