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

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Answer 1 · 2015-02-04T11:58:12

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

How do you solve 2/(x^2+2x+1)-3/(x+1)=42/(x^2+2x+1)-3/(x+1)=4?