How do you simplify $(3x-2)/(x+3)+7/(x^2-x-12)$ ?

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Answer 1 · 2017-07-02T11:31:02

$((x-3)(3x-5))/((x+3)(x-4))$

First, factor out the denominator of the right hand term

$(3x-2)/(x+3)+7/((x+3)(x-4))$

Next multiply the numerator and denominator of the first term by $(x-4)$ in order to get a common denominator.

$(x-4)/(x-4)(3x-2)/(x+3)+7/((x+3)(x-4))$

Combine the two fractions.

$((x-4)(3x-2)+7)/((x+3)(x-4))$

Expand the numerator

$(3x^2-12x-2x+8+7)/((x+3)(x-4))$

Simplify the numerator

$(3x^2-14x+15)/((x+3)(x-4))$

Factor the numerator

$((x-3)(3x-5))/((x+3)(x-4))$

How do you simplify (3x-2)/(x+3)+7/(x^2-x-12)(3x-2)/(x+3)+7/(x^2-x-12)?