## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{x(5x+1)}{(x-4)(x+3)} \quad \text{ or } \quad \dfrac{5x^2+x}{(x-4)(x+3)}, \quad x\ne-3,4$
Since the LCM of the denominators is $(x-4)(x+3)$, then we have $$\frac{3x}{x-4}+\frac{2x}{x+3}=\frac{3x(x+3)}{(x-4)(x+3)}+\frac{2x(x-4)}{(x-4)(x+3)} .$$ Now, since the denominators are the same then we have \begin{align*} \frac{3x(x+3)}{(x-4)(x+3)}+\frac{2x(x-4)}{(x-4)(x+3)} &=\frac{3x(x+3)+2x(x-4)}{(x-4)(x+3)}\\ \\&=\frac{3x^2+9x+2x^2-8x}{(x-4)(x+3)}\\ \\&=\frac{5x^2+x}{(x-4)(x+3)}\\ \\&=\frac{x(5x+1)}{(x-4)(x+3)} \end{align*} Note that $x\ne-3,4$ because both of them will make the denominator equal to $0$.