How do you solve #(x+3)/(x^2-1)-(2x)/(x-1)=1#?

1 Answer
Dec 3, 2016

#(x + 3)/((x + 1)(x- 1)) - (2x(x + 1))/((x + 1)(x - 1)) = (x^2 -1)/((x +1)(x- 1))#

#x + 3 - (2x^2 + 2x) = x^2 - 1#

#x + 3 - 2x^2 - 2x = x^2 - 1#

#0 = 3x^2 + x - 4#

#0 = 3x^2 - 3x + 4x - 4#

#0 = 3x(x -1) + 4(x - 1)#

#0 = (3x + 4)(x - 1)#

#x = -4/3 and 1#

However, our restrictions in the original equation are #x!= +-1#, so #x= -4/3# is the only valid solution.

Hopefully this helps!