Question

Perform the indicated operation and simplify.

$\frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}$

Short answer

The solution is$\frac{3n-2}{n-1}$.

Step-by-step solution

Step 1. Given information.

The given expression is$\frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}$.

Step 2. Simplify the expression.

$$\begin{gathered} \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{2n^2+8n-1}{n^2-1}+\frac{n^2-7n-1}{n^2-1} \\ \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{2n^2+8n-1+n^2-7n-1}{n^2-1} \\ \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{2n^2+n^2+8n-7n-1-1}{n^2-1} \\ \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{3n^2+n-2}{\left(n^2-1\right)} \\ \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{3n^2+3n-2n-2}{n^2-1} \\ \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{3n\left(n+1\right)-2\left(n+1\right)}{\left(n-1\right)\left(n+1\right)} \\ \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{\left(n+1\right)\left(3n-2\right)}{\left(n-1\right)\left(n+1\right)}(Removethecommonfactor) \\ \frac{2n^2+8n-1}{n^2-1}-\frac{n^2-7n-1}{1-n^2}=\frac{3n-2}{n-1} \end{gathered}$$

Step 3. Simplified answer.

Hence, the simplified expression is$\frac{3n-2}{n-1}$.