How do you combine $(3x-2)*[x/(3x^2+x-2)+2/(x+1)]$ ?

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Answer 1 · 2016-10-01T15:41:23

$(3x-2)*[x/(3x^2+x-2)+2/(x+1)]=(7x-4)/(x+1)$

Let us first factorize $3x^2+x-2$

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

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

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

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

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

= $(3x-2)xx x/((3x-2)(x+1))+(3x-2)xx2/(x+1)$

= $cancel(3x-2)xx x/(cancel((3x-2))(x+1))+(2(3x-2))/(x+1)$

= $x/(x+1)+(2(3x-2))/(x+1)$

= $(x+2(3x-2))/(x+1)$

= $(x+6x-4)/(x+1)$

= $(7x-4)/(x+1)$

How do you combine (3x-2)*[x/(3x^2+x-2)+2/(x+1)](3x-2)*[x/(3x^2+x-2)+2/(x+1)]?