Question

Select correct answer

Answer 1

Answer:

The values of p in the equation are 0 and 6

Step-by-step 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

Answer 2

Answer:

The values of p in the equation are 0 and 6

Step-by-step explanation:

First, you have to make the denominators the same. to do that, first factor $2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)$

So then the equation looks like:

$\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}$

To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:

$\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}$

Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.

$(p^2-4p)-(2p^2+5)=-(10p+5)$

Now, solve like a normal equation. Solve $(p^2-4p)-(2p^2+5)$ first:

$(p^2-4p)-(2p^2+5)=-p^2-4p-5$

$-p^2-4p-5=-10p+5$

Combine like terms:

$-p^2-4p+0=-10p$

$-p^2+6p=0$

Factor:

$p=0, p=6$