## Algebra: A Combined Approach (4th Edition)

$r=3$
$\dfrac{4r-4}{r^{2}+5r-14}+\dfrac{2}{r+7}=\dfrac{1}{r-2}$ Factor the denominator of the first fraction: $\dfrac{4r-4}{(r+7)(r-2)}+\dfrac{2}{r+7}=\dfrac{1}{r-2}$ Multiply the whole equation by $(r+7)(r-2)$: $(r+7)(r-2)\Big[\dfrac{4r-4}{(r+7)(r-2)}+\dfrac{2}{r+7}=\dfrac{1}{r-2}\Big]$ $4r-4+2(r-2)=r+7$ $4r-4+2r-4=r+7$ Take all terms to the left side of the equation and simplify it by combining like terms: $4r-4+2r-4-r-7=0$ $5r-15=0$ Solve for $r$: $5r=15$ $r=\dfrac{15}{5}=3$ Since the equation is not undefined for this value of $r$, the solution to this equation is $r=3$