Answer
$ \frac{3x-4}{(x-3)(x+2)}$.
Work Step by Step
In order to add the two rational expressions we must determine the Least Common Denominator (LCD). Therefore we need to factor the denominators.
Factor $x^2-4x+3$.
Rewrite the middle term $-4x$ as $-3x-1x$.
$\Rightarrow x^2-3x-1x+3$
Group the terms.
$\Rightarrow (x^2-3x)+(-1x+3)$
Factor each group.
$\Rightarrow x(x-3)-1(x-3)$
Factor out $(x-3)$.
$\Rightarrow (x-3)(x-1)$
Factor $x^2+x-2$.
Rewrite the middle term $x$ as $2x-1x$.
$\Rightarrow x^2+2x-1x-2$
Group the terms.
$\Rightarrow (x^2+2x)+(-1x-2)$
Factor each group.
$\Rightarrow x(x+2)-1(x+2)$
Factor out $(x+2)$.
$\Rightarrow (x+2)(x-1)$
Back substitute into the given expression.
$\Rightarrow \frac{2}{x^2-4x+3}+ \frac{3x}{x^2+x-2}=\frac{2}{(x-3)(x-1)}+ \frac{3x}{(x+2)(x-1)}$
The LCD is $(x-3)(x-1)(x+2)$.
Multiply the numerator and the denominator to form LCD at the denominators.
$\Rightarrow \frac{2(x+2)}{(x-3)(x-1)(x+2)}+ \frac{3x(x-3)}{(x-3)(x-1)(x+2)}$
Use the distributive property.
$\Rightarrow \frac{2x+4}{(x-3)(x-1)(x+2)}+ \frac{3x^2-9x}{(x-3)(x-1)(x+2)}$
Add the numerators because denominators are same.
$\Rightarrow \frac{2x+4+3x^2-9x}{(x-3)(x-1)(x+2)}$
Add like terms.
$\Rightarrow \frac{3x^2-7x+4}{(x-3)(x-1)(x+2)}$
Factor $3x^2-7x+4$.
Rewrite the middle term $-7x$ as $-4x-3x$.
$\Rightarrow 3x^2-4x-3x+4$
Group the terms.
$\Rightarrow (3x^2-4x)+(-3x+4)$
Factor each group.
$\Rightarrow x(3x-4)-1(3x-4)$
Factor out $(3x-4)$.
$\Rightarrow (3x-4)(x-1)$
Back substitute into the fraction.
$\Rightarrow \frac{(3x-4)(x-1)}{(x-3)(x-1)(x+2)}$
Cancel common factors.
$\Rightarrow \frac{3x-4}{(x-3)(x+2)}$.
