How do you solve #2/(x+3)-4/(x^2+2x-3)=1/(1-x)#?

1 Answer
Apr 9, 2016

Answer:

The given equation is invalid.
There are no possible solutions.

Explanation:

Given
#color(white)("XXX")2/(x+3)-4/(x^2+2x-3)=1/(1-x)#

Noting that
#color(white)("XXX")x^2+2x-3=(x+3)(x-1)#
the given equation can be re-written as
#color(white)("XXX")(2(x-1))/(x^2+2x-3)-4/(x^2+2x-3)= ((-1)(x+3))/(x^2+2x+3)#

Provided #x!=-3# and #color(red)(x!=1)#
then #x^2+2x-3 != 0#

and we have
#color(white)("XXX")2(x-1)-4=-(x+3)#

#color(white)("XXX")2x-6=-x-3#

#color(white)("XXX")3x = 3#

#color(white)("XXX")x=1# ...but this is one of the "forbidden values".

Both the terms #4/(x^2+2x-3)# and #1/(1-x)# are undefined when #x=1#