Question

Solve: 8/x^2-1=(3/x+1)+(4/x-1)
a. no solution
b. 7
c. 1
d. 1/7

Answer 1

Make sure the denominators are factored first

\(\large\frac{8}{(x+1)(x-1)} = \frac{3}{x+1} + \frac{4}{x-1}\)

Answer 2

Now multiply both sides by x + 1:

\(\large\frac{8}{x-1} = 3 + \frac{4(x+1)}{x-1}\)

Answer 3

Subtract \(\frac{4(x+1)}{x-1}\) from both sides

Answer 4

\(\large\frac{8}{x-1} - \frac{4x + 4}{x-1} = 3\)

Answer 5

Combine fractions on the left side:

\(\large\frac{4 - 4x}{x-1} = 3\)

Answer 6

factor the numerator to get:

\(\large\frac{-4(x-1)}{x-1} = 3\)

Answer 7

Cancel the factor of 1 to get

\(-4 \ne 3\)

Answer 8

No solution

Answer 9

I hope I didn't lose you

Answer 10

Nope, I got it. thank you very much! that was helpful :)

Answer 11

@waterineyes is a fan of my methods now.

Answer 12

Need not to do factorization there..

Answer 13

It was necessary to continue with my methods

Answer 14

\[\frac{8}{x^2-1} = \frac{7x + 1}{(x+1)(x-1)} = \frac{7x+1}{x^2 - 1}\]

\[x = 1\]

But when plugging in 1 back in the equation we get undefined..
So no solution...

Answer 15

I told you my methods avoid having to solve for x where there are extraneous roots.

Answer 16

Remember, half of the class will solve for x and get it wrong.

Answer 17

Half of the class would pick c as the answer.

Answer 18

Stop showing off please..

Answer 19

I'm not. I'm stating a fact.