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

1 Answer
Oct 5, 2015

Answer:

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

Explanation:

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