Question

I need help solving this for pre calc

Answer 1

Starting with:

$\frac{2p}{p+1} + \frac{3}{p-1} = \frac{15-p}{p^{2}-1}$

First, the difference of 2 squares shows that $p^{2} - 1 = (p+1)(p-1)$

Sub that in and multiply both sides by $(p-1)(p+1)$:

$\frac{2p(p+1)(p-1)}{p+1} + \frac{3(p-1)(p+1)}{p-1} = \frac{(15-p)(p-1)(p+1)}{(p-1)(p+1)}$

This easily simplifies to $2p(p-1) + 3(p+1) = 15 - p$

See if you can continue from here. If not, leave a comment and I or someone else can show you the rest.