Question

Add:$\frac{2n}{n^2-3n-10}+\frac{6}{n^2+5n+6}$

Short answer

The addition of the rational expression is$\frac{2\left(n^2+6n-15\right)}{\left(n+2\right)\left(n-5\right)\left(n+3\right)}$

Step-by-step solution

Given information

The given rational expression is$\frac{2n}{n^2-3n-10}+\frac{6}{n^2+5n+6}$

The simplification of the rational expression

$\frac{2n}{n^2-3n-10}+\frac{6}{n^2+5n+6}$

Factor the denominator

$$\begin{gathered} \frac{2n}{n^2-5n+2n-10}+\frac{6}{n^2+3n+2n+6} \\ \frac{2n}{n\left(n-5\right)+2\left(n-5\right)}+\frac{6}{n\left(n+3\right)+2\left(n+3\right)} \\ \frac{2n}{\left(n+2\right)\left(n-5\right)}+\frac{6}{\left(n+2\right)\left(n+3\right)} \end{gathered}$$

The LCD of the expression is $\left(n+2\right)\left(n-5\right)\left(n+3\right)$.

$$\begin{gathered} \frac{2n\left(n+3\right)}{\left(n+2\right)\left(n-5\right)\left(n+3\right)}+\frac{6\left(n-5\right)}{\left(n+2\right)\left(n+3\right)\left(n-5\right)} \\ \frac{2n^2+6n}{\left(n+2\right)\left(n-5\right)\left(n+3\right)}+\frac{6n-30}{\left(n+2\right)\left(n+3\right)\left(n-5\right)} \\ \frac{2n^2+12n-30}{\left(n+2\right)\left(n-5\right)\left(n+3\right)} \\ \frac{2\left(n^2+6n-15\right)}{\left(n+2\right)\left(n-5\right)\left(n+3\right)} \end{gathered}$$