Answer
$\dfrac{3(x + 2)}{4(x - 3)}$
Restrictions: $x \ne -2, 2, 3$
Work Step by Step
First, factor all expressions in the original exercise:
$\dfrac{3(x - 2)}{12(x - 2)} \div \dfrac{(x - 3)(x - 2)}{3(x - 2)(x + 2)}$
To divide one fraction by another, multiply the first fraction by the reciprocal of the second fraction:
$\dfrac{3(x - 2)}{12(x - 2)} \cdot \dfrac{3(x - 2)(x + 2)}{(x - 3)(x - 2)}$
Multiply numerator with numerator and denominator with denominator:
$\dfrac{3(x - 2)\cdot 3(x - 2)(x + 2)}{12(x - 2) \cdot (x - 3)(x - 2)}$
Cancel common terms in the numerator and denominator:
$\require{cancel}
\dfrac{\cancel{3}\cancel{(x - 2)}\cdot 3\cancel{(x - 2)}(x + 2)}{^4\cancel{12}\cancel{(x - 2)} \cdot (x - 3)\cancel{(x - 2)}}$
$=\dfrac{3(x + 2)}{4(x - 3)}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set denominators equal to $0$ and solve for $x$ to find where the restrictions are:
First denominator:
$12x - 24 = 0$
$12x = 24$
$x = 2$
Second denominator:
$3x^2 - 12 = 0$
$3x^2 = 12$
$x^2 = 4$
Take the square root of each side of the equation:
$x = \pm 2$
Third denominator (from the reciprocal):
$x - 3 = 0$
$x = 3$
$x - 2 = 0$
$x = 2$
Restrictions: $x \ne -2, 2, 3$
