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!

Answer:  
The inequality reduces to two separate inequalities:
They are $$ 
\frac{x+2}{3(x-1)} \leq \frac{2}{3}
$$
and 
$$ 
-\frac{x+2}{3(x-1)} \leq \frac{2}{3}
$$
Solving the first one $$(x+2) \leq 2(x-1)$$
$$x\geq 4$$
Solving the second one $$x+2\geq 2-2x$$
$$3x\leq 0$$
$$x \leq 0$$
