1
$\begingroup$

$$|\frac{x-1}{x+3}|\leq 2$$

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?

$\endgroup$
  • $\begingroup$ Your answer is correct, the book just has the sign of the second answer flipped around. $\endgroup$ – Jeffrey L. Jun 5 '15 at 15:47
  • $\begingroup$ @JeffreyL. my mistake, I fixed it $\endgroup$ – gbox Jun 5 '15 at 15:49
  • $\begingroup$ Both solutions are equal. $\endgroup$ – callculus Jun 5 '15 at 15:52
  • $\begingroup$ 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 $x<7$. $\endgroup$ – Jeffrey L. Jun 5 '15 at 15:54
  • $\begingroup$ @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. $\endgroup$ – callculus Jun 5 '15 at 15:58
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I did not understand what are the two inequalities you came to, thanks $\endgroup$ – gbox Jun 5 '15 at 15:56
  • $\begingroup$ 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. $\endgroup$ – DeepSea Jun 5 '15 at 15:58
  • $\begingroup$ 2nd equivalence holds because $|a|\le |b|\iff a^2\le b^2$. $\endgroup$ – user26486 Jun 5 '15 at 16:03
  • $\begingroup$ There is a weird floating "or" in your solution. $\endgroup$ – Barry Jun 5 '15 at 16:41
2
$\begingroup$

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 $x \in (\infty,-7) \cup [-\frac53,\infty)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.