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

$\displaystyle \frac{4(x+y)}{3(x-y)}$
Factor what we can: $x^{2}+2xy+y^{2}$ = (square of a sum) = $(x+y)^{2}$ $x^{2}-2xy+y^{2}$ = (square of a difference) = $(x-y)^{2}$ $4x-4y$ = $4(x-y)$ $3x+3y=3(x+y)$ Rewrite the problem: $\displaystyle \frac{(x+y)(x+y)}{(x-y)(x-y)}\cdot\frac{4(x-y)}{3(x+y)}=\qquad$ ... reduce common factors =$\displaystyle \frac{(1)(x+y)}{(1)(x-y)}\cdot\frac{4(1)}{3(1)}=$ =$\displaystyle \frac{4(x+y)}{3(x-y)}$