Answer
$\{-\frac{3}{4},2\}$
Work Step by Step
Multiply both sides by the LCD $x(x-3)$ to get rid f the denominators:
$x(x-3) \cdot \left [\dfrac{4(x-2)}{x-3}+\dfrac{3}{x}\right ]=x(x-3) \cdot \left [\dfrac{-3}{x(x-3)}\right ]$
Use distributive property.
$x(x-3) \cdot \left (\frac{4(x-2)}{x-3}\right )+x(x-3) \cdot \left (\frac{3}{x}\right )=x(x-3) \cdot \left (\frac{-3}{x(x-3)}\right )$
Cancel common factors.
$\begin{align*}
\require{cancel}
x\cancel{(x-3)} \cdot \left (\frac{4(x-2)}{\cancel{x-3}}\right )+\cancel{x}(x-3) \cdot \left (\frac{3}{\cancel{x}}\right )&=\cancel{x(x-3)} \cdot \left (\frac{-3}{\cancel{x(x-3)}}\right )\\
\\4x(x-2)+3(x-3) &=-3
\end{align*}$
Use distributive property.
$4x^2-8x+3x-9 =-3$
Simplify.
$4x^2-5x-9 =-3$
Add $3$ to both sides.
$4x^2-5x-9+3 =-3+3$
Simplify.
$4x^2-5x-6 =0$
Rewrite $-5x$ as $-8x+3x$.
$4x^2-8x+3x-6 =0$
Group the terms.
$(4x^2-8x)+(3x-6) =0$
Factor out the GCF in each group.
$4x(x-2)+3(x-2) =0$
Factor out $(x-2)$.
$(x-2)(4x+3) =0$
Use Zero-Product Property.
$x-2=0 \quad or $\quad 4x+3 =0$
Solve for $x$.
$x=2 \quad $ or $\quadx =-\frac{3}{4}$
Hence, the solution set of the given equation is $\{-\frac{3}{4},2\}$.
