# 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?

• Your answer is correct, the book just has the sign of the second answer flipped around. – Jeffrey L. Jun 5 '15 at 15:47
• @JeffreyL. my mistake, I fixed it – gbox Jun 5 '15 at 15:49
• Both solutions are equal. – callculus Jun 5 '15 at 15:52
• Yes, they are the same. Typically when dealing with less than signs, your inequality is partitioned because $x$ can't be in both ranges, just one or the other. For $4<x<7$, $x<7$ OR $x>4$, not both, but for $4>x>7$, $x>4$ AND $x<7$. – Jeffrey L. Jun 5 '15 at 15:54
• @gbox -3 is not part of the solution $x\leq -7 \cup x\geq -\frac{5}{3}$. Thus you do not have to considerate in your solution. – callculus Jun 5 '15 at 15:58

$\left|\dfrac{x-1}{x+3}\right| \leq 2 \iff |x-1| \leq 2|x+3| \iff (x-1)^2 \leq 4(x+3)^2 \iff x^2-2x+1 \leq 4(x^2+6x+9) \iff 3x^2 + 26x+35 \geq 0 \iff (x+7)(3x+5) \geq 0 \iff x \leq -7$ or $x \geq -\dfrac{5}{3}$.

• I did not understand what are the two inequalities you came to, thanks – gbox Jun 5 '15 at 15:56
• The $2$ numbers $-7,-\dfrac{5}{3}$ divide the real line into $3$ intervals $(-\infty,-7],(-7,-\dfrac{5}{3}),[-\dfrac{5}{3},\infty)$, and for the inequality to hold, $x$ needs to be in either the first or the last interval. – DeepSea Jun 5 '15 at 15:58
• 2nd equivalence holds because $|a|\le |b|\iff a^2\le b^2$. – user26486 Jun 5 '15 at 16:03
• There is a weird floating "or" in your solution. – Barry Jun 5 '15 at 16:41

Just break it into sections by sign:

$\left|\frac{x-1}{x+3}\right| = \left\{ \begin{array}{ll} \frac{x-1}{x+3} & \mbox{if } x \geq 1 \\ -\frac{x-1}{x+3} & \mbox{if } x \in (-3,1) \\ \frac{x-1}{x+3} & \mbox{if } x \lt -3 \end{array} \right.$

Then solve the three inequalities based on the sign of $x+3$:

$$x\geq1:\frac{x-1}{x+3} \leq 2 \iff x \geq -7$$ $$x\in(-3,1):-\frac{x-1}{x+3} \leq 2 \iff x \geq -\frac{5}{3}$$ $$x<-3:\frac{x-1}{x+3} \leq 2 \iff x \leq -7$$

That gives us the ranges $[1,\infty)$, $[-\frac53,1)$, and $(\infty,-7)$. So the final solution is $x \in (\infty,-7) \cup [-\frac53,\infty)$