Question

add and simplify the result.
\frac{8x}{x - 8}+\frac{x + 2}{x + 5}

\frac{8x}{x - 8}+\frac{x + 2}{x + 5}=square
(type your answer in factored form.)

Explanation:

Step1: Find a common denominator

The common denominator of $x - 8$ and $x + 5$ is $(x - 8)(x + 5)$.

Step2: Rewrite each fraction with the common denominator

$\frac{8x}{x - 8}\times\frac{x + 5}{x + 5}+\frac{x + 2}{x + 5}\times\frac{x - 8}{x - 8}=\frac{8x(x + 5)}{(x - 8)(x + 5)}+\frac{(x + 2)(x - 8)}{(x - 8)(x + 5)}$

Step3: Expand the numerators

$\frac{8x^2+40x}{(x - 8)(x + 5)}+\frac{x^2-8x+2x - 16}{(x - 8)(x + 5)}=\frac{8x^2+40x+x^2-6x - 16}{(x - 8)(x + 5)}$

Step4: Combine like - terms in the numerator

$\frac{(8x^2+x^2)+(40x-6x)-16}{(x - 8)(x + 5)}=\frac{9x^2+34x - 16}{(x - 8)(x + 5)}$

Step5: Factor the numerator

$9x^2+34x - 16=9x^2+36x - 2x - 16=9x(x + 4)-2(x + 8)=(9x - 2)(x + 8)$
So the result is $\frac{(9x - 2)(x + 8)}{(x - 8)(x + 5)}$

Answer:

$\frac{(9x - 2)(x + 8)}{(x - 8)(x + 5)}$