Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 2 - Section 2.3 - Solving Linear Equations - Exercise Set - Page 141: 36

Answer

The solution is: $y = \frac{5}{3}$ We check to see if this solution is correct by substituting $\frac{5}{3}$ into the original equation to see if both sides equal one another. $\frac{(3)(\frac{5}{3})}{4} - \frac{2}{3} = \frac{7}{12}$ We multiply the numerator of the first term to simplify it: $\frac{5}{4} - \frac{2}{3} = \frac{7}{12}$ Find equivalent fractions for all three terms by finding the lowest common denominator: $\frac{15}{12} - \frac{8}{12} = \frac{7}{12}$ We can now subtract the fractions on the left-hand side: $\frac{7}{12} = \frac{7}{12}$ Both sides are equal; therefore, the solution is correct.

Work Step by Step

The first step to solving this equation is to rewrite the equation without fractions. We do this by finding the lowest common denominator for all three fractions in the equation. The lowest common denominator is $12$. By finding the lowest common denominator for each fraction, we can come up with equivalent fractions to better deal with adding and subtracting them. We rewrite the equation to get: $\frac{9y}{12} - \frac{8}{12} = \frac{7}{12}$ We can now multiply both sides of the equation by $12$ to get rid of the fraction: $9y - 8 = 7$ We can isolate the $y$ term on one side and the constant on the other side by adding $8$ to each side to get: $9y = 15$ To isolate $y$, divide each side by 9: $y = \frac{15}{9}$ We reduce the fraction by dividing both numerator and denominator by $3$ (the greatest common factor of both $15$ and $9$) to get: $y = \frac{5}{3}$ We check to see if this solution is correct by substituting $\frac{5}{3}$ into the original equation to see if both sides equal one another. $\frac{(3)(\frac{5}{3})}{4} - \frac{2}{3} = \frac{7}{12}$ We multiply the numerator of the first term to simplify it: $\frac{5}{4} - \frac{2}{3} = \frac{7}{12}$ Find equivalent fractions for all three terms by finding the lowest common denominator: $\frac{15}{12} - \frac{8}{12} = \frac{7}{12}$ We can now subtract the fractions on the left-hand side: $\frac{7}{12} = \frac{7}{12}$ Both sides are equal; therefore, the solution is correct.
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