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?

I'll give a more mechanical way to solve the inequality that avoids the (neat!) trick of multiplying by $(x-3)^2$. But note that this is essentially the same as what Bill and Eivind showed. From
$$
\frac{3x}{x-3} \geq 4 ,
$$
we get
$$
\frac{3x}{x-3} - 4 \geq 0 \hspace{.2in} \implies \frac{3x-4x+12}{x-3} \geq 0.
$$
Simplifying this, we get
$$
\frac{12-x}{x-3} \geq 0.
$$
From this point, the solution is similar to Eivind's. This inequality implies one of the following two cases:


*

*$12 - x \geq 0$ and $x-3 > 0$. That is, $3 < x \leq 12$.

*$12 - x \leq 0$ and $x - 3 < 0$. That is, $x < 3$ and $x \geq 12$; clearly this is impossible. 
Hence the valid solutions are $3 < x \leq 12$. 
