<s>[INST] Simple Absolute Value Inequality 
$$|\frac{x-1}{x+3}|\leq 2$$

I solve it as follow: 

$|\frac{x-1}{x+3}|\leq 2 \iff -2\leq \frac{x-1}{x+3} \leq 2 \iff -2\leq 1-\frac{4}{x+3} \leq 2 \iff -3\leq \frac{-4}{x+3} \leq 1 \iff \frac{3}{4}\geq \frac{1}{x+3} \geq -\frac{1}{4} \iff \frac{4}{3} \leq x+3 \leq -4 \iff -3+\frac{4}{3}=-\frac{5}{3}\leq x\leq -7$
I now see that I did not take into consideration $x\neq -3$ which can not be 
The book on the other side says that in answer is $x\leq -7$ or $x\geq -\frac{5}{3}$
Am I wrong?
 [/INST] $\left|\dfrac{x-1}{x+3}\right| \leq 2 \iff |x-1| \leq 2|x+3| \iff (x-1)^2 \leq 4(x+3)^2 \iff x^2-2x+1 \leq 4(x^2+6x+9) \iff 3x^2 + 26x+35 \geq 0 \iff (x+7)(3x+5) \geq 0 \iff x \leq -7$ or $x \geq -\dfrac{5}{3}$.
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