#### Answer

$x=-6$

#### Work Step by Step

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:
$$\dfrac{-2x}{-2} = \dfrac{12}{-2}
\\x=-6$$