Answer
$\dfrac{6r-18}{9r^{2}+6r-24}\cdot\dfrac{12r-16}{4r-12}=\dfrac{2}{r+2}$
Work Step by Step
$\dfrac{6r-18}{9r^{2}+6r-24}\cdot\dfrac{12r-16}{4r-12}$
Factor both rational expressions completely:
$\dfrac{6r-18}{9r^{2}+6r-24}\cdot\dfrac{12r-16}{4r-12}=\dfrac{6(r-3)}{3(r+2)(3r-4)}\cdot\dfrac{4(3r-4)}{4(r-3)}=...$
Evaluate the product of the two fractions:
$...=\dfrac{(6)(4)(r-3)(3r-4)}{(3)(4)(r+2)(3r-4)(r-3)}=...$
Simplify by removing the factors that appear both in the numerator and in the denominator of the resulting expression:
$...=\dfrac{6}{3(r+2)}=\dfrac{2}{r+2}$