# Thread: Solve 2/[(x^2)-x] = (1/x) + 4/(x^2-1)

1. ## Solve 2/[(x^2)-x] = (1/x) + 4/(x^2-1)

Can somebody help me solve this equation? I have gotten to a certain point then get stuck. Thank you!

2/x^2-x = (1/x) + (4/x^2-1)

2. Originally Posted by connroy17
Can somebody help me solve this equation? I have gotten to a certain point then get stuck. Thank you!

2/x^2-x = (1/x) + (4/x^2-1)
Are the terms in the red part of exponent or not?

3. 2/((x^2)-x) = (1/x) + (4/((x^2)-1)

Hopefully that makes it easier to understand. The exponents are in red.

4. $\displaystyle \frac{2}{x^2}-x=\frac{1}{x}+\frac{4}{x^2}-1$

Move to one side and simplify:

$\displaystyle \frac{2}{x^2}+x+\frac{1}{x}-1=0$

Obtain same common denominator.

You can do that one.

5. Originally Posted by connroy17
2/((x^2)-x) = (1/x) + (4/((x^2)-1)

Hopefully that makes it easier to understand. The exponents are in red.
$\displaystyle \frac{2}{x^2-x} = \frac{1}{x} + \frac{4}{x^2-1}$

$\displaystyle \frac{2}{x(x-1)} = \frac{1}{x} + \frac{4}{(x+1)(x-1)}$

common denominator is $\displaystyle x(x+1)(x-1)$ ...

$\displaystyle \frac{2(x+1)}{x(x-1)(x+1)} = \frac{1(x+1)(x-1)}{x(x+1)(x-1)} + \frac{4x}{x(x+1)(x-1)}$

note that x cannot equal 0 , 1 , or -1 (why?)

with a common denominator, the numerators form the equation ...

$\displaystyle 2(x+1) = (x+1)(x-1) + 4x$