Squaring across an inequality with an unknown number This should be something relatively simple. I know there's a trick to this, I just can't remember it.
I have an equation
$$\frac{3x}{x-3}\geq 4.$$
I remember being shown at some point in my life that you could could multiply the entire equation by $(x-3)^2$ 
in order to get rid of the divisor. However, 
$$3x(x-3)\geq 4(x-3)^2.$$ 
Really doesn't seem to go anywhere.  
Anyone have any clue how to do this, that could perhaps show me/ point me in the right direction?

$$\begin{align}
3x(x-3)&\geq 4(x-3)^2 \\
3x^2-9x &\geq 4(x^2-6x+9) \\
3x^2-9x &\geq 4x^2-24x+36
\end{align}$$
By rearranging we get
$$\begin{align}
x^2-15x+36 &\leq 0 \\
(x-3)(x-12) &\leq 0
\end{align}
$$
So $(x-3)> 0 \land (x-12)\leq0$ or $(x-3)< 0 \land (x-12)\geq 0$. ($x-3\neq0$)
