Question

how do you solve this fast ?

Answer 1

(1/((p^2)+7p-8))-(1/(p+8))=((p-3)/((p^2)+7p-8))

Answer 2

Or you can refer to this:
http://www.wolframalpha.com/input/?i=%281%2F%28%28p%5E2%29%2B7p-8%29%29-%281%2F%28p%2B8%29%29%3D%28%28p-3%29%2F%28%28p%5E2%29%2B7p-8%29%29

Answer 3

I tried solving it but I think it is long and I got 2 values for p
unlike in wolfram it has only 1 value

Answer 4

\[\frac{1}{p^2+7p-8}-\frac{1}{p+8}=\frac{p-3}{p^2+7p-8}\]
?

Answer 5

@UnkleRhaukus Yup :)
I got the values
p=5/2
p=-8

Answer 6

Do you know some way to solve it fast?

Answer 7

\[\frac{1}{p^2+7 p-8}-\frac{1}{p+8}-\frac{p-3}{p^2+7 p-8}=0\]
Combine terms and simplify.\[\frac{2 p-5}{(p-1) (p+8)}=0 \]

Answer 8

My last equation was
-2p^2 -11p +40= 0

Answer 9

after that I removed the negative sign
-(2p^2 +11p -40)= 0
(2p-5)(p+8)=0
p=5/2
p=-8
is this correct?

Answer 10

\[\frac{1}{p^2+7p-8}-\frac{1}{p+8}=\frac{p-3}{p^2+7p-8}\]
\[\frac{1}{(p+8)(p-1)}-\frac{1}{p+8}=\frac{p-3}{(p+8)(p-1)}\]
\[\frac{1}{(p+8)(p-1)}-\frac{(p-1)}{(p+8)(p-1)}=\frac{p-3}{(p+8)(p-1)}\]
\[{1-(p-1)}={p-3}\]
\[{2-p}={p-3}\]
\[5=2p\]\[p=\frac52\]

Answer 11

rob gives a good way. notice that the approach that leads to a quadratic introduces a spurious solution

Answer 12

I think what you did @UnkleRhaukus is better than what I did.
I'll just try the solution of rob :)

Answer 13

watch out for solutions that make the denominator of the original expressions go zero tho

Answer 14

Do you think that the method of unkle is the fastest ?
because it is stated on our sample questionnaire to watch out for extraneous solutions :)

Answer 15

I like your method :)

Answer 16

\[(p^2+7p-8)\left(\frac{1}{p^2+7p-8}-\frac{1}{p+8}=\frac{p-3}{p^2+7p-8}\right)\]
\[1-\frac{1}{p-1}=p-3\]
\[(p-1)\left(1-\frac{1}{p-1}=p-3\right)\]
\[p-1-1=(p-3)(p-1)\]

dunno what is considered a "fast" way

Answer 17

I tried solving a similar problem could you check it?

Answer 18

sure