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If you have an inequality that has two absolute value bars like $|4x+1|<|3x|$, how do you go about doing this? I know that if $4x+1<3x$, then those $x$'s will work but what else do I do? I think you do $4x+1<-3x$. Is this correct?

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    $\begingroup$ What don't you look first at a plot of the problem ? This will give you some ideas. Cheers. $\endgroup$ – Claude Leibovici Aug 14 '14 at 5:13
  • $\begingroup$ I dont't know why no one just said set both possibilities equal to zero and test a point in between each zero. $\endgroup$ – King Squirrel Aug 24 '14 at 6:17
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You could also square everything $$ |f(x)| < |g(x)| \Leftrightarrow |f(x)|^2 < |g(x)|^2 \\ \Leftrightarrow f(x)^2 < g(x)^2 \\ \Leftrightarrow 0< g(x)^2-f(x)^2 \\ \Leftrightarrow 0< (g(x)-f(x))(g(x)+f(x)), \\ $$ which means that $g(x)-f(x)$ and $g(x)+f(x)$ have the same sign.

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Just take the different cases. For example:

You know that $$ |3x|=\left\{ \begin{align} 3x & \text{ , if }x\geq 0 \\ -3x & \text{ , if }x <0 \end{align} \right\} $$ $$ |4x+1|=\left\{ \begin{align} 4x+1 & \text{ , if }x\geq \frac{-1}{4} \\ -(4x+1) & \text{ , if }x <\frac{-1}{4} \end{align} \right\} $$ This gives you a few different cases to check: $x<\frac{-1}{4}$, $\frac{-1}{4}\leq x <0$, and $x\geq 0$.

So for instance, take the first cases: $x<\frac{-1}{4}$ so that $|4x+1|=-(4x+1)$ and $|3x|=-3x$. Then we have $$ \begin{align} |4x+1|&<|3x| \\ -(4x+1)&<-3x \\ -4x-1&<-3x \\ -x-1&<0 \\ -x&<+1\\ x&>-1 \end{align} $$

Just be sure to check the $\leq$ or $\geq$ cases and be sure the answers from the different regions agree! (Meaning if you found in one case $x>1$ and found in another case $x<1$ there would be no solutions)

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  • $\begingroup$ I have a triplet of absolute values inequality, thus I have 4 different regions of x which need to be checked. What happens if one region yields a false clause (e.g. $-2 \ge 0$) and another region yields a true clause (e.g. $0 \ge 0$) ? The other regions do relate to x. $\endgroup$ – Dor Nov 26 '14 at 11:18
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There's three parts here that you need to consider: the area where both 4x+1 and 3x are positive (0 < x); the area where one's positive and the other's negative (-1/4 < x < 0); the area where both are negative (x < -1/4). Union the results of those together.

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Need to break up into 2 cases:

(3x) >= 0 or x >= 0: |3x| = 3x -3x < (4x+1) < 3x 7x > -1 and x < -1 and x>=0 (no solution)

(3x) <= 0 or x<=0 |3x| = -3x 3x < (4x+1) < -3x x > -1 and 7x < -1 Solution: -1 < x < -1/7

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I just solve the equalities without absolute values: LHS = RHS and LHS = -RHS. Then I use a number line test to check values in each interval to determine the intervals of the solution set. This led me to all values between -1 and -1/7, not including engpoints, since the original inequality was strict.

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    $\begingroup$ Why do you use $=$ sign?It is a question on inequations. $\endgroup$ – tatan Oct 14 '15 at 17:31
  • $\begingroup$ @tatan Scott uses equations to find the endpoints for the intervals where the inequality holds. $\endgroup$ – user147263 Oct 14 '15 at 17:34
  • $\begingroup$ @NormalHuman-OK...I understand...but is it efficint enough? $\endgroup$ – tatan Oct 15 '15 at 4:36
  • $\begingroup$ What do you mean "efficient enough?" The method solves the problem just fine. $\endgroup$ – Scott Eisner Oct 15 '15 at 15:00

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