Question

In the following exercises, perform the indicated operations

$\frac{n}{n+3}+\frac{2}{n-3}-\frac{n-9}{n^2-9}$

Short answer

The simplified expression is$\frac{n-5}{n-3}$

Step-by-step solution

Step 1. Given information

Expression:

$\frac{n}{n+3}+\frac{2}{n-3}-\frac{n-9}{n^2-9}$

Step 2. Find LCD     

Denominator of the first fraction is $n+3$

Denominator of the second fraction is $n-3$

Denominator of the third fraction is$n^2-9$

It can be factorize as

$$\begin{gathered} n^2-9=n^2-3^2 \\ =(n-3)(n+3) \end{gathered}$$

So LCD is$\left(n-3\right)\left(n+3\right)$

Step 3. Rewrite each rational expression as  an equivalent rational expression with the LCD      

$\frac{n(n-3)}{(n+3)(n-3)}+\frac{2(n+3)}{(n-3)(n+3)}-\frac{n-9}{(n-3)(n+3)}$

Step 4.Simplify the numerators and perform operations

$$\begin{gathered} \frac{(n^2-3n)}{(n+3)(n-3)}+\frac{(2n+6)}{(n+3)(n-3)}-\frac{(n-9)}{(n+3)(n-3)} \\ =\frac{(n^2-3n)+(2n+6)-(n-9)}{(n+3)(n-3)} \\ =\frac{n^2-3n+2n+6-n+9}{(n+3)(n-3)} \end{gathered}$$

Step 5. Simplify numerators

$\frac{n^2-2n+15}{(n+3)(n-3)}$

Step 6. factorize numerator

$$\begin{gathered} \frac{n^2-5n+3n+15}{(n+3)(n-3)} \\ =\frac{n(n-5)+3(n-5)}{(n+3)(n-3)} \\ =\frac{(n-5)(n+3)}{(n+3)(n-3)} \end{gathered}$$

Step 7. Cancel common factors

$\frac{n-5}{n-3}$