Question

Simplify$\frac{6d}{d-5}+\frac{1}{d+4}-\frac{7d-5}{d^2-d-20}$.

Short answer

The result is$\frac{6d^2+18d}{(d-5)(d+4)}$.

Step-by-step solution

Given Information 

The Given expression is$\frac{6d}{d-5}+\frac{1}{d+4}-\frac{7d-5}{d^2-d-20}$.

Find the LCD of denominators.    

Factor the denominators.

$d^2-d-20=(d-5)(d+4)$

LCD$=(d-5)(d+4)$

Rewrite the expression in factored form.

$=\frac{6d}{d-5}+\frac{1}{d+4}-\frac{7d-5}{(d-5)(d+4)}$

Rewrite each fraction as an equivalent fraction with the LCD.    

$$\begin{gathered} =\frac{6d(d+4)}{(d-5)(d+4)}+\frac{d-5}{(d-5)(d+4)}-\frac{7d-5}{(d-5)(d+4)} \\ =\frac{6d^2+24d}{(d-5)(d+4)}+\frac{d-5}{(d-5)(d+4)}-\frac{7d-5}{(d-5)(d+4)} \end{gathered}$$

Simplify. 

Take common denominator.

$=\frac{6d^2+24d+d-5-(7d-5)}{(d-5)(d+4)}$

Simplify numerator.

$$\begin{gathered} =\frac{6d^2+25d-5-7d+5}{(d-5)(d+4)} \\ =\frac{6d^2+18d}{(d-5)(d+4)} \end{gathered}$$