#### Answer

$r=3$

#### Work Step by Step

$\dfrac{4r-4}{r^{2}+5r-14}+\dfrac{2}{r+7}=\dfrac{1}{r-2}$
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 $r=3$