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


$\frac{2}{3}y+\frac{1}{2}(y-3)=\frac{y+1}{4}$ Solution:$y=\frac{21}{11}$

Work Step by Step

1. $\frac{2}{3}y+\frac{1}{2}(y-3)=\frac{y+1}{4}$ 2. Distribute $\frac{1}{2}$ to simply the equation to:$\frac{2}{3}y+\frac{1}{2}y-\frac{3}{2}=\frac{y+1}{4}$ 3. Multiply both sides by 4: $4\frac{2}{3}y-6=y+1$ 4. Subtract y from both sides:$3\frac{2}{3}y-6=1$ 5. Add 6 to both sides: $3\frac{2}{3}y=7$ 6. Change $3\frac{2}{3}$ to an improper fraction: $\frac{11}{3}y = 7$ 7. Divide both sides by $\frac{11}{3}$ (equivalent to multiplying both sides by $\frac{3}{11}$) :$y=\frac{7}{\frac{11}{3}} = \frac{21}{11}$
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