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

1 Answer
Feb 4, 2015

You can write the first denominator as:
#x^2+2x+1=(x+1)(x+1)=(x+1)^2#
So yo get:
#2/((x+1)^2)-3/(x+1)=4#

You find a common denominator for both sides and "rearrange" the numerators to adapt to this, so you get:

#2/((x+1)^2)-(3*(x+1))/((x+1)^2)=(4*(x+1)^2)/(x+1)^2#

Multiplying both sides by #(x+1)^2# you get:

#2-3(x+1)=4(x+1)^2#
#2-3(x+1)=4(x^2+2x+1)#
#2-3x-3=4x^2+8x+4#
#4x^2-11x+5=0#
Which is a second degree equation.

Hope it helps