I have finals in a couple of days. I need help with inequalities. I have spent around 15 hours trying my hand at inequalities and I am still trying to figure things out. In this case I need help with an inequality with two absolute values with greater than symbol.

The following is a picture of the inequality ($|4x-1|>|3x+2|$) and the procedure I tried, which is something I applied from an answer by Isaac to another question in this site, although I think I did something wrong because it looks like I got the answer wrong or incomplete.

Please tell me what I did wrong and how to do it the right way.

Thanks in advance.

PS: Does this method also apply if there is a less than sign?

possible wrong procedure of inequality with two absolute values

  • $\begingroup$ Alternatively (and I'd say it's faster) you could note that $|a|>|b|$ is equivalent to $a^2>b^2$. $\endgroup$
    – dfnu
    Nov 15, 2019 at 8:32
  • $\begingroup$ Two years later I have completely forgotten how to do this. :D $\endgroup$ Nov 19, 2021 at 23:50

3 Answers 3


Here’s another way to look at it: $$|4x-1|>|3x+2| \iff (4x-1)^2-(3x+2)^2>0 \iff (7x+1)(x-3)>0$$

Now LHS has only two obvious zeros, and outside the interval with those end points, it has to be positive.


P.S. When you have something like $f(x)>0$, where $f$ is continuous, it can change sign only when there are roots. So its roots effectively partition the real line into intervals which are either solutions or not. In the above case, two roots are $-\frac17, 3$, so we need to consider only the three intervals $(-\infty, -\frac17), (-\frac17, 3)$ and $(3, \infty)$. Obviously the LHS is positive for $x\to \pm\infty$ and negative for $x=0$, so the first and last intervals are solutions, the middle one isn’t.

BTW, whether the intervals to consider are open / closed depends on whether the inequality is strict or $\geqslant$.

  • $\begingroup$ 7x+1>0 gives me (-1/7,+) which seems to be invalid. What is wrong? $\endgroup$ Nov 15, 2019 at 16:18
  • $\begingroup$ @dfnu +macavity x-3 gives me (3,+), with a combined solution of (-1/7,+). $\endgroup$ Nov 15, 2019 at 16:56
  • $\begingroup$ @freethinker36 If you need $A\cdot B>0$, you need $A, B$ to have the same sign. Not just $A$ or just $B$. $\endgroup$
    – Macavity
    Nov 15, 2019 at 16:58
  • $\begingroup$ It would be great if you can elaborate. I'm learning, thus what may look evident, for me may not be evident. And tomorrow is my test and have to study two classes; already taking time of the second class to try to cement my understanding of inequalities. But I appreciate whatever guidance has been provided! $\endgroup$ Nov 15, 2019 at 17:05
  • 1
    $\begingroup$ OK, I have added some explanations, do let me know if youre having more doubts after studying it. $\endgroup$
    – Macavity
    Nov 15, 2019 at 17:11

You're doing almost everything right; you're just misinterpreting the results at the end.

In particular, the idea is to combine intervals where the inequality holds. In the first region, $\left(-\infty, -\frac23\right)$, you arrived at $x<3$. It seems that you rejected this result, judging by the "$\times$" beside your work. This is where you went wrong.

Instead, think of it like this:

Within the region $\left(-\infty, -\frac23\right)$, which $x$ satisfy $x<3$?

All $x$ do within this region. Another way to see this is that you are effectively taking the intersection of the two: $$\left(-\infty, -\frac23\right) \cap (-\infty,3) = \left(-\infty, -\frac23\right)$$

The other two regions are handled the same way.

  • $\begingroup$ My rationale for rejecting it was that 3>-2/3, so x<3 is not within the region of x<-2/3. $\endgroup$ Nov 15, 2019 at 6:33

I tried again, following the answer of Théophile, and I was able to reach the correct solution. Below is the pic of the procedure I used. Thanks!

solved inequality


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