Question 286145
{{{x/(x^2-1)+2/(x-1)=1+1/(2x-2)}}}
.
{{{x^2-1 = (x+1)(x-1)}}}
{{{2x-2 = 2(x-1)}}}
.
{{{(2(x+1)(x-1)*x)/((x+1)(x-1)) + 2(x+1)(x-1)*2/(x-1) = 2*(x+1)(x-1)*1 + 2(x+1)(x-1)*1/(2(x-1))}}}
.
The idea of the LCD is to get rid of the denominators...so cancel the terms you can:
.
{{{(2*x) + 2(x+1)*2 = 2*(x+1)(x-1)*1 + 2(x+1)*1/(2)}}}
.
Of course we can eliminate any '1' that is a multiplier, and cancel the 2/2.  And 2*2 is 4.
.
{{{ 2x + 4(x+1) = 2(x+1)(x-1) + (x+1) }}}
.
{{{ 2x + 4x + 4 = 2(x^2 -1) + x + 1 }}}
.
{{{ 6x + 4 = 2x^2 -2 + x + 1 }}}
.
Subtract 6x from both sides.
.
{{{4 = 2x^2 -2 + x +1 -6x }}}
.
{{{4 = 2x^2 -5x -1 }}}
.
Subtract 4 from both sides.
.
{{{0 = 2x^2 -5x -5 }}}
.
This cannot be factored easily, so the quadratic equation can be used.
.
*[invoke quadratic "x", 2, -5, -5 ]