Solving inequalities with absolute values This is the question:
$$ 
\left| \frac{x+2}{3(x-1)} \right| \leq \frac{2}{3}
$$
And this is my working out, first I squared both the numerator and denominator, then solved it as if it was a normal inequality.
$$
\frac{(x+2)^2}{(3x-3)^2} \leq \frac{2}{3}
$$
$$
3(x+2)^2 \leq 2(3x-3)^2
$$
$$
9x^2 +36x+36 \leq 36x^2 -72x+36
$$
I took out a common factor of 9
$$
x^2 +5x+4 \leq 4x^2-8x+4
$$
$$
3x^2 -12x \geq 
\\
x(3x-4) \geq 0
\\
x \geq 0 \text{ and } x \geq \frac{4}{3}
$$
Usually, when dealing with inequalities like this, I end up with a quadratic equation, which I am then able to factorise and solve correctly,but I'm not sure what went wrong with this one. The correct answer is
$$
x \leq 0
\\
x \geq 4
$$
What have I done wrong, thank you in advance!

The procedure is sound. First a preliminary comment. The left-hand side is not defined at $x=1$, so whatever answer we get must exclude $1$.
We have
$$\left|\frac{x+2}{3(x-1)}\right|\le \frac{2}{3}\quad\text{if and only if}\quad \left(\frac{x+2}{3(x-1)}\right)^2\le \frac{4}{9}.$$
If $x\ne 1$, then the equality on the right above holds if and only if
$$9(x+2)^2 \le (4)(9)(x-1)^2.$$
The above inequality reduces quickly to $(9)(3)(x)(x-4)\ge 0$, which holds precisely if $x\le 0$ or $x\ge 4$. 
The point $x=1$ is not in the region just obtained, so we need not worry about it.
Remark: The procedure used in the post was sound. It was presented as a manipulation. The logic should have been made clearer (as in the "if and only if" of the answer above).
There was a minor arithmetical slip. It is not true that $3x^2-12x=x(3x-4)$. 
