## Precalculus (6th Edition)

$x=-6$
Multiply $3(x-3)$ to both side of the equation to obtain: $$3(x-3)\left(\dfrac{3x+4}{3}-\dfrac{2x}{x-3}\right)=x(3)(x-3) \\3(x-3)\left(\dfrac{3x+4}{3}\right)-3(x-3)\left(\dfrac{2x}{x-3}\right)=3x(x-3) \\(x-3)(3x+4)-3(2x)=3x(x)-3x(3) \\x(3x)+x(4)-3(3x)-3(4)-6x=3x^2-9x \\3x^2+4x-9x-12-6x=3x^2-9x \\3x^2-11x-12=3x^2-9x$$ Subtract $3x^2$ and add $9x$ to both sides of the equation, then combine like terms to obtain: $$3x^2-11x-12-3x^2+9x=3x^2-9x-3x^2+9x \\(3x^2-3x^2)+(-11x+9x)-12=(3x^2-3x^2)+(-9x+9x) \\-2x-12=0$$ Add $12$ to both sides of the equation to obtain: $$-2x-12+12=0+12 \\-2x=12$$ Divide $-2$ to both sides of the equation to obtain: $$\dfrac{-2x}{-2} = \dfrac{12}{-2} \\x=-6$$