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.
