## Thinking Mathematically (6th Edition)

Published by Pearson

# Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2 - Page 362: 51

y = -6

#### Work Step by Step

We start with the original equation: $\frac{y}{3}$ + $\frac{2}{5}$ = $\frac{y}{5}$ - $\frac{2}{5}$ We can start by finding the least common denominator for all of the fractions. The least common denominator is 15. We can now write all of the fractions as equivalents with a denominator of 15. This gives us: $\frac{5y}{15}$ + $\frac{6}{15}$ = $\frac{3y}{15}$ - $\frac{6}{15}$ We can now subtract $\frac{3y}{15}$ from both sides of the equation. $\frac{5y}{15}$ - $\frac{3y}{15}$ + $\frac{6}{15}$ = $\frac{3y}{15}$ - $\frac{3y}{15}$ - $\frac{6}{15}$ Completing the arithmetic, we get: $\frac{2y}{15}$ + $\frac{6}{15}$ = -$\frac{6}{15}$ Next, subtract $\frac{6}{15}$ from both sides of the equation. $\frac{2y}{15}$ + $\frac{6}{15}$ - $\frac{6}{15}$ = -$\frac{6}{15}$ - $\frac{6}{15}$ Complete the arithmetic: $\frac{2y}{15}$ = $\frac{-12}{15}$ Now, cross multiply so that: 2y(15) = (-12)(15) And, 30y = -180 Last, divide both sides by 30. $\frac{30y}{30}$ = $\frac{-180}{30}$ Complete the division and get y = -6. We can check by substituting -6 back into the original equations for y. $\frac{-6}{3}$ + $\frac{2}{5}$ = $\frac{-6}{5}$ - $\frac{2}{5}$ Now, complete the arithmetic on each side of the equation. -2 + $\frac{2}{5}$ = $\frac{-8}{5}$ $\frac{-10}{5}$ + $\frac{2}{5}$ = $\frac{-8}{5}$ In the step above, we rewrote -2 as a fraction using a denominator of 5. We get $\frac{-8}{5}$ = $\frac{-8}{5}$ Showing that our solution of y = -6 is correct.

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