Algebra

what is the solution to this equation and what are the steps: (x^2-3x)/(x^2+1)=2

Multiply both sides by x^2 +1 to get rid of fractions, we get:

x^2 - 3x=2(x^2 +1)

x^2 - 3x=2x^2 +2

2x^2 + 2 -x^2 +3x=0x^

x^2 +3x +2=0, then we factor it into:

(x +2) (x +1)=0, so that,

x=-1 and x=-2

(x^2-3x)/(x^2+1)=2     multiply both sides by x^2 + 1

x^2 - 3x   = 2(x^2 + 1)   simplify

x^2 - 3x  = 2x^2 + 2    subtract x^2   and add  3x to both sides

0 = x^2 + 3x + 2       factor

(x + 1) ( x+ 2) = 0      and setting each factor to 0, we have that x =  - 1 or x = - 2