Question

Simplify$\frac{3d}{d+2}+\frac{4}{d}-\frac{d+8}{d^2+2d}$.

Short answer

The result is$\frac{3(d+1)}{d+2}$.

Step-by-step solution

Given Information     

The Given expression is $\frac{3d}{d+2}+\frac{4}{d}-\frac{d+8}{d^2+2d}$.

Find the LCD of denominators.      

Factor the denominators.

$d^2+2d=d(d+2)$

Rewrite the expression in factored form.

$=\frac{3d}{d+2}+\frac{4}{d}-\frac{d+8}{d(d+2)}$

Rewrite each fraction as an equivalent fraction with the LCD.    

$$\begin{gathered} =\frac{3d\left(d\right)}{d(d+2)}+\frac{4(d+2)}{d(d+2)}-\frac{d+8}{d(d+2)} \\ =\frac{3d^2}{d(d+2)}+\frac{4d+8}{d(d+2)}-\frac{d+8}{d(d+2)} \end{gathered}$$

Simplify.

Take common denominator.

$$\begin{gathered} =\frac{3d^2+4d+8-(d+8)}{d(d+2)} \\ =\frac{3d^2+4d+8-d-8}{d(d+2)} \\ =\frac{3d^2+3d}{d(d+2)} \end{gathered}$$

Cancel common factors.

$$\begin{gathered} =\frac{3d(d+1)}{d(d+2)} \\ =\frac{3(d+1)}{d+2} \end{gathered}$$