Answer :
Starting with:
[tex] \frac{2p}{p+1} + \frac{3}{p-1} = \frac{15-p}{p^{2}-1} [/tex]
First, the difference of 2 squares shows that [tex]p^{2} - 1 = (p+1)(p-1)[/tex]
Sub that in and multiply both sides by [tex](p-1)(p+1)[/tex]:
[tex] \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)} [/tex]
This easily simplifies to [tex]2p(p-1) + 3(p+1) = 15 - p[/tex]
See if you can continue from here. If not, leave a comment and I or someone else can show you the rest.
[tex] \frac{2p}{p+1} + \frac{3}{p-1} = \frac{15-p}{p^{2}-1} [/tex]
First, the difference of 2 squares shows that [tex]p^{2} - 1 = (p+1)(p-1)[/tex]
Sub that in and multiply both sides by [tex](p-1)(p+1)[/tex]:
[tex] \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)} [/tex]
This easily simplifies to [tex]2p(p-1) + 3(p+1) = 15 - p[/tex]
See if you can continue from here. If not, leave a comment and I or someone else can show you the rest.
