Question

Solve for x,

Answer 1

\[\frac{ 5 }{ x^2-1 }-\frac{ 2 }{ x }=\frac{ 2 }{ x+1 }\]

Answer 2

I would solve this by multiplying every term on both sides of the equation by the LCD.

Answer 3

The LCD is \(x(x+1)(x-1)\).
\[x(x+1)(x-1)\left(\frac{ 5 }{ x^2-1 }\right)-x(x+1)(x-1)\left(\frac{ 2 }{ x }\right)=x(x+1)(x-1)\left(\frac{ 2 }{ x+1 }\right)\]
Simplifying:
\[5x -2(x+1)(x-1) = 2x(x-1)\]

Answer 4

I think its easier to do this:
\[\frac{5}{x^2-1}-\frac{2}{x}=\frac{2}{x+1}\]\[\frac{5}{x^2-1}-\frac{2(x-1)}{x^2-1}=\frac{2}{x}\]\[5x-2x^2-2x=2(x^2-1)\]\[-4x^2+3x+2=0\]

Answer 5

FOIL (x+1)(x-1)
\[5x - 2(x^2-1) = 2x^2 - 2x\]
Distribute to clear parenthesis
\[5x - 2x^2 + 2 = 2x^2- 2x\]
Gather all terms to one side (keep the \(x^2\) term positive)
\[0 = 4x^2-7x - 2\]

Solve the quadratic equation using factoring:
\[0 = (x-2)(4x+1)\]
\[x = 2\; \text{ and }\; x = \frac{1}{4}\]

Both solutions are ok since they won't make any of the original fractions undefined.

Answer 6

I messed up some +- there but the idea is the same.

Answer 7

Thanks to all of you!!