# How do you simplify \frac { x ^ { 2} - 16y ^ { 2} } { x ^ { 2} + 3x y - 4y ^ { 2} } \div \frac { x ^ { 2} - 8x y + 16y ^ { 2} } { x - y }?

May 3, 2018

$\frac{1}{x - 4 y}$

#### Explanation:

Recall that dividing something by $x$ is the same as multiplying it by the inverse $\frac{1}{x}$. That is, $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$. We use this algebraic fact to help us simplify.

$\frac{{x}^{2} - 16 {y}^{2}}{{x}^{2} + 3 x y - 4 {y}^{2}} \div \frac{{x}^{2} - 8 x y + 16 {y}^{2}}{x - y}$
$= \frac{\left({x}^{2} - 16 {y}^{2}\right) \left(x - y\right)}{\left({x}^{2} + 3 x y - 4 {y}^{2}\right) \left({x}^{2} - 8 x y + 16 {y}^{2}\right)}$

Now we note that most of these terms can be factored. See that the following are true:

${x}^{2} - 16 {y}^{2} = \left(x - 4 y\right) \left(x + 4 y\right)$
${x}^{2} - 8 x y + 16 {y}^{2} = \left(x - 4 y\right) \left(x - 4 y\right)$
${x}^{2} + 3 x y - 4 {y}^{2} = \left(x + 4 y\right) \left(x - y\right)$

Making the replacements as needed, this gives

$\frac{\left(x - 4 y\right) \left(x + 4 y\right) \left(x - y\right)}{\left(x + 4 y\right) \left(x - y\right) {\left(x - 4 y\right)}^{2}}$
$= \frac{1}{x - 4 y} \cdot \frac{x - 4 y}{x - 4 y} \cdot \frac{x + 4 y}{x + 4 y} \cdot \frac{x - y}{x - y}$
$= \frac{1}{x - 4 y}$

Thus, our final answer is $\frac{1}{x - 4 y}$.