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

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Answer 1 · 2018-06-04T11:34:39

x= $1/3$ or x=-2

Multiply $(2x)/(x-2)$ by x+1 so that the fractions are all over a common denominator

$[2x(x+1)+x^2+3x]/[(x+1)(x-2)]=2/[(x+1)(x-2)]$

$[2x^2+2x+x^2+3x]/[(x+1)(x-2)]=2/[(x+1)(x-2)]$

$[3x^2+5x]/[(x+1)(x-2)]=2/[(x+1)(x-2)]$

mutiply throughout by (x+1)(x-2) to remove the fractions

$3x^2+5x=2$

$3x^2+5x-2=0$

(3x-1)(x+2)=0

3x-1=0 or x+2=0

x= $1/3$ or x=-2

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