Question

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

Answer 1

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