Solve modulus inequality $|\frac{x+9}{x-9}|≤2$ In a problem, I've been asked to  solve the inequality
$$|\frac{x+9}{x-9}|≤2$$
So I've done $\frac{x+9}{x-9}≤2$ and $\frac{x+9}{x-9}≥-2$, which gave $x≥27$ and $x≥3$, which doesn't seem quite correct because why would it give that $x$ is greater or equal to both 27 and 3? 
What did I do wrong?

$$|\frac{x+9}{x-9}|≤2$$
Is the same as
$$(\frac{x+9}{x-9})^2≤4$$
$$(x+9)^2≤4(x-9)^2$$
$$0≤(2x-18)^2-(x+9)^2$$
Using $a^2-b^2=(a-b)(a+b)$
$$0≤(2x-18-x-9)(2x-18+x+9)$$
$$0≤(x-27)(3x-9)$$
$$0≤(x-27)(x-3)$$
The inequality holds when $x-27$ and $x-3$ have the same sign
