## Algebra 2 (1st Edition)

No solution as $x \ne 4$
We have $\dfrac{3x}{x-4}=\dfrac{2(x-4)}{x-4}+\dfrac{12}{x-4}$ Re-write as: $3x=2(x-4)+12$ or, $3x=2x-8+12$ or, $x =4$ and $x -4 \ne 0 \implies x \ne 4$ No solution since there is the domain restriction $x \ne 4$