How do you solve x/(x+2)-(x+2)/(x-2)=(x+3)/(x-2)xx+2x+2x2=x+3x2?

1 Answer
Oct 5, 2015

x=-1x=1 and x=-10x=10

Explanation:

We have, at first, (x)/(x+2) - (x+2)/(x-2) = (x+3)/(x-2)xx+2x+2x2=x+3x2.
We pass -(x+2)/(x-2)x+2x2 to the other side adding.
We now have(x)/(x+2) = (x+3)/(x-2) + (x+2)/(x-2)xx+2=x+3x2+x+2x2.
As the denominator is common on both fractions of the right side, we add them.
We get (x)/(x+2) = (2x + 5)/(x-2)xx+2=2x+5x2.
We now cross multiply to get x(x-2) = (x+2)(2x+5)x(x2)=(x+2)(2x+5).
We multiply and get x^2 - 2x = 2x^2 + 9x + 10x22x=2x2+9x+10.
We pass the left hand side working to the right, getting x^2 + 11x + 10 = 0x2+11x+10=0 as an answer.
We factorise to get (x+1)(x+10) = 0(x+1)(x+10)=0.
Therefore, the answers are x = -1x=1 and x = -10x=10.
Hope it helps :D .