Answer
$\dfrac{4x}{3y}$
Restrictions: $x \ne -1, 0, \text{ and } y \ne 0$
Work Step by Step
Rewrite the exercise using the division ($\div$) symbol:
$=\dfrac{8x^{2}y}{x + 1} \div \dfrac{6xy^{2}}{x + 1}$
Invert the divisor and change the operation to multiplication:
$=\dfrac{8x^{2}y}{x + 1} \cdot \dfrac{x + 1}{6xy^{2}}$
Cancel common factors in the numerator and denominator:
$\require{cancel}
=\dfrac{\cancel{8}^4x^{\cancel{2}}\cancel{y}}{\cancel{x + 1}} \cdot \dfrac{\cancel{x + 1}}{\cancel{6}^3\cancel{x}y^{\cancel{2}}}$
$=\dfrac{4x}{3y}$
Restrictions occur when the denominator is $0$, meaning the expression becomes undefined. To find the restrictions, set each expression in the denominators of the original rational expressions and in the reciprocal equal to zero and solve:
First restriction:
$x + 1 = 0$
$x = -1$
Second restriction:
$6xy^{2} = 0$
For this expression, neither $x$ nor $y$ can be $0$.
Thus, the restrictions are:
$x \ne -1, 0$
$y \ne 0$
