5/x^2 - 1 - 2/x = 2/x+1

Question

hero · Answer

Remember to use parentheses

5/(x^2 - 1) - 2/x = 2/(x+1)

hero · Answer

Otherwise it can be interpreted differently

hero · Answer

Anyway....

$\large\frac{5}{x^2-1} - \frac{2}{x} = \frac{2}{x+1}$

hero · Answer

Make sure the denominators are factored: 

$\large\frac{5}{(x+1)(x-1)} - \frac{2}{x} = \frac{2}{x+1}$

hero · Answer

Multiply both sides by (x-1)

hero · Answer

$\frac{5}{x+1} - \frac{2(x-1)}{x} = \frac{2(x-1)}{x+1}$

hero · Answer

Add the second fraction to both sides; Subtract the third fraction from both sides: 

$\frac{5}{x+1} - \frac{2x -2}{x+1} = \frac{2x-2}{x}$

hero · Answer

Combine the fractions on the left hand side: 

$\frac{5 - 2x + 2}{x+1} = \frac{2x-2}{x}$

hero · Answer

Then simplify to get

$\frac{7 - 2x}{x+1} = \frac{2x-2}{x}$

hero · Answer

Then cross multiply:

x(7-2x) = 2(x-1)(x+1)

hero · Answer

$7x-2x^2 = 2(x^2 - 1)$

hero · Answer

$7x - 2x^2 = 2x^2 - 2$

hero · Answer

$4x^2 - 7x - 2 = 0$

hero · Answer

$4x^2 -8x + x - 2 = 0 \ 4x(x-2) + 1(x-2) = 0 \ (x-2)(4x+1) = 0$

Continue Solving for x

hero · Answer

You should get: 
x = -1/4
x = 2

hero · Answer

Any questions?