Answer:
The values of p in the equation are 0 and 6
Explanation:
First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p
2
−7p−4=(2p+1)(p−4)
So then the equation looks like:
\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}
2p+1
p
−
(2p+1)(p−4)
2p
2
+5
=−
p−4
5
To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:
\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}
(2p+1)(p−4)
p
2
−4p
−
(2p+1)(p−4)
2p
2
+5
=−
(p−4)(2p+1)
10p+5
Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.
(p^2-4p)-(2p^2+5)=-(10p+5)(p
2
−4p)−(2p
2
+5)=−(10p+5)
Now, solve like a normal equation. Solve (p^2-4p)-(2p^2+5)(p
2
−4p)−(2p
2
+5) first:
(p^2-4p)-(2p^2+5)=-p^2-4p-5(p
2
−4p)−(2p
2
+5)=−p
2
−4p−5
-p^2-4p-5=-10p+5−p
2
−4p−5=−10p+5
Combine like terms:
-p^2-4p+0=-10p−p
2
−4p+0=−10p
-p^2+6p=0−p
2
+6p=0
Factor:
p=0, p=6p