## Introductory Algebra for College Students (7th Edition)

The solution is: $$y = -6$$ To check if this solution is correct, we plug $-6$ into the original equation to see if both sides equal one another: $$\frac{-6}{3} + \frac{2}{5} = \frac{-6}{5} - \frac{2}{5}$$ We now multiply by the least common denominator ($15$) to get rid of the fractions: $$15(\frac{-6}{3} + \frac{2}{5}) = 15(\frac{-6}{5} - \frac{2}{5})$$ Dividing out common factors, we get: $$-30 + 6 = -18 - 6$$ Simplifying, we get: $$-24 = -24$$ We see that both sides are equal, so we know that the solution is correct.
To solve this equation, we want to transform it so that we don't have to work with fractions. We find the least common denominator for all fractions, which, in this case, is $15$. We now multiply both sides by $15$ to get: $$15(\frac{y}{3} + \frac{2}{5}) = 15(\frac{y}{5} - \frac{2}{5})$$ We now use the distributive property to simplify the equation: $$15\frac{y}{3} + 15\frac{2}{5} = 15\frac{y}{5} - 15\frac{2}{5}$$ Divide out the common factors to get rid of the fractions: $$5y + 6 = 3y - 6$$ Subtract $3y$ from each side to isolate the variable on one side: $$2y + 6 = -6$$ Subtract $6$ from each side to keep variables on one side and constants on the other: $$2y = -12$$ We can now solve for $y$ by dividing each side by $2$: $$y = -6$$ To check if this solution is correct, we plug $-6$ into the original equation to see if both sides equal one another: $$\frac{-6}{3} + \frac{2}{5} = \frac{-6}{5} - \frac{2}{5}$$ We now multiply by the least common denominator ($15$) to get rid of the fractions: $$15(\frac{-6}{3} + \frac{2}{5}) = 15(\frac{-6}{5} - \frac{2}{5})$$ Dividing out common factors, we get: $$-30 + 6 = -18 - 6$$ Simplifying, we get: $$-24 = -24$$ We see that both sides are equal, so we know that the solution is correct.