I'm new here, and I was wondering if any of you could help me out with this little problem that is already getting on my nerves since I've been trying to solve it for hours.

Studying for my next test on inequalities with absolute values, I found this one:

$$ |x-3|-|x-4|<x $$ (I precisely found the above inequality on this website, here to be precise, but, the problem is that when I try to solve it, my answer won't be $(-1,+\infty)$, but $(1,7)$. I took the same inequalities that the asker's teacher had given to him and then put them on a number line, and my result was definitely not $(-1,+\infty)$

Here are the inequalities: $$ x−3 < x−4 +x $$ $$ x−3 < −(x−4) +x $$ $$ −(x−3)<−(x−4)+x $$

And here are my answers respectively: $$ x>1, \quad x>-1, \quad x<7 $$

I will really appreciate if anyone could help me out, because I'm already stressed dealing with this problem, that by the way, it is not demanding that I solve it, but you know, why not?


6 Answers 6


How to solve $|x-3|-|x-4|<x$? Let $f(x)=|x-3|-|x-4|$

I would begin by noting $|x-3|=x-3$ for $x \geq 3$ whereas $|x-3|=3-x$ for $x \leq 3$. Likewise, $|x-4|=x-4$ for $x \geq 4$ whereas $|x-4|=4-x$ for $x \leq 4$

  1. if $x \geq 4$ then $x > 3$ hence $f(x)=x-3-(x-4)=1$ and we find $1<x$ which is true in this case. This puts $[4,\infty)$ in the solution set of the inequality.

  2. if $x \leq 3$ then $x < 4$ hence $f(x)=3-x-(4-x)=-1$ and we face $-1<x$ which is true for each $x$ with $-1<x\leq 3$. This shows $(-1,3]$ is also in the solution set of the inequality.

  3. if $3<x<4$ then $f(x) = x-3-(4-x)=2x-7$. Hence in this context we must solve $2x-7<x$ which gives $x<7$ which is true for each $x$ in the interval considered. This places $(3,4)$ in the solution set of the inequality.

Put all of this together, we obtain $(-1,3] \cup (3,4) \cup [4,\infty) = (-1, \infty)$ which is your answer. Usually I solve this sort of thing with a sign-chart method.


The standard semi-mechanical way to eliminate the absolute value signs is to divide the number line into segments. The critical point for $|x-4|$ is at $x=4$, and the critical point for $|x-3|$ is at $x=3$.

Suppose first that $x \ge 4$. Then $|x-4|=x-4$ and $|x-3|=x-3$. So we are looking at the inequality $(x-3)-(x-4)\lt x$, that is, at $1\lt x$, which is certainly true for $x\ge 4$.

Now suppose that $3\le x\lt 4$. Then $|x-4|=4-x$ and $|x=3|=x-3$, so we are looking at the inequality $(x-3)-(4-x)\lt x$, that is, $2x-7\lt x$. This simplifies to $x\lt 7$, which is certainly true in the interval $[3,4)$.

It is probably at this point that your calculation went astray. We were looking at the interval $[3,4)$ and asking which points in this interval satisfied our inequality. The manipulation told us that it was all points in this interval that satisfied $x\lt 7$. Well, they all do!

Finally, suppose that $x\lt 3$. Then $|x-4|=4-x$ and $|x-3|=3-x$. So we are looking at the inequality $(3-x)-(4-x)\lt x$. Calculate. The left side is $-1$, so for in the interval $(-\infty,3)$, the inequality holds precisely when $-1\lt x$.

Putting things together, we conclude that the original inequality holds (i) if $x\ge 4$; (ii) if $3\le x\lt 4$; and (ii) if $x\lt 3$ but $-1\lt x$. This complicated set of conditions can be summarized much more simply as $x\gt -1$.

There are other ways to describe the set of solutions. For example we could say that the set is $(-1,\infty)$.

  • $\begingroup$ You had me at "which simplifies to -1<x", I don't know how you concluded the original equality holds -1<x. Because I already found the answers to the intervals you suggested, but I don't know how those three answers give away that the one and only answer is -1<x $\endgroup$
    – ChairOTP
    Aug 27, 2012 at 5:33
  • $\begingroup$ @user38807: I have added a lot more explanation. Hope it will do the job. $\endgroup$ Aug 27, 2012 at 5:41
  • $\begingroup$ I already got another simpler answer that totally made me understand, but I surely appreciate the effort you did, and forgive me for being that dumb that I couldn't understand your answer. It's really late here, maybe I should get some sleep already. $\endgroup$
    – ChairOTP
    Aug 27, 2012 at 5:50
  • $\begingroup$ @user38807: While you are examining a particular interval, you are interested in what conditions your inequality imposes on numbers in that interval. So you proceed systematically, one interval at a time. I suggest you look at the wording used by James S. Cook (final paragraph). $\endgroup$ Aug 27, 2012 at 5:59

What you have is almost correct, the last step is to restrict your solution to the corresponding region.

For example, for $x>1$, the answer must be in the $x>4$ region, so your answer for this region is $x>4$.

For $x>-1$, your answer must be in $x<3$ region, so your answer for this region is $-1<x<3$.

And for the last one, $x<7$, your answer must be in the corresponding region we had first, namely $3<x<4$, so your answer for this region is $3<x<4$.

Now draw these three answers on the number line and you'll have $x>-1$, the desired final answer.

  • $\begingroup$ I totally appreciate this, so simple but effective. Now I do understand, and blame myself how could I not see this before. $\endgroup$
    – ChairOTP
    Aug 27, 2012 at 5:39
  • $\begingroup$ @user38807: Glad to hear that. $\endgroup$
    – Gigili
    Aug 27, 2012 at 5:41
  • $\begingroup$ Sorry to bother you again, my curiosity is driving me crazy. It is demanding to restrict the solution in every problem? Or in which cases, because I have one problem where, if I restrict the solution to the regions, it won't work. $\endgroup$
    – ChairOTP
    Aug 27, 2012 at 5:57
  • $\begingroup$ @user38807: Well, when you have multiple absolute values and you want to get rid of them, you'll have to assume where the term in the absolute sign is negative and where it is positive and you'll have to check your final answers with those primary assumption whether they are within that region or not. $\endgroup$
    – Gigili
    Aug 27, 2012 at 6:06
  • $\begingroup$ Feel free to share your other problem with me, so we can check if it works out to what you're told. $\endgroup$
    – Gigili
    Aug 27, 2012 at 6:08

Here's one way to do it.

If $x\ge4$, then $|x-3|=x-3$ and $|x-4|=x-4$, so you want $(x-3)-(x-4)\lt x$, which you should be able to solve (remembering to check your solution against the assumption $x\ge4$.

If $3\le x\le4$, then $|x-3|=x-3$ and $|x-4|=4-x$, so you get ....

If $x\le3$, then $|x-3|=\dots$


$|x-3|-|x-4|< x$, I write in $|x-3|< x+|x-4|$

but remember: $|r|< s \implies -s < r < s$.

So I write the equation in the form: $-x-|x-4| < x-3 < x+|x-4|$.

From this inequality I obtain 2 equations:

(a) $-x-|x-4| < x-3$

(b) $x-3 < x+|x-4|$.

Remember too : $|r| > s \implies r > s \text{ or } r < -s $.

So this concept I will apply to equation (a) and equation (b).

From equation (a):

$-x-|x-4| < x-3$ I write so absolute value is in one side: -|x-4| < 2x-3 then I multiply by -1 : |x-4| > -2x+3 . Now I write the equation in the form of: |r| > s----> r > s OR r < -s .As a result we obtain 2 additional equation: x-4 > -2x+3 OR x-4 < 2x-3 . From the first one: 3x > 7 --->x > 7/3 or (7/3,∞) . From the second inequality: -1 < x or (-1,∞) .The first and second inequality are OR function then: (7/3,∞) U (-1,∞) is: (-1,∞)

Now the equation (b): x-3 < x+|x-4| we write to put absolute value in one side: $-3 < |x-4|$ or $|x-4| > -3$ . To satisfy this inequality $x$ will take any value positive or negative. As a result we can write the result the value of $x$ for equation (b) as: $(-∞,+∞)$

The final result from the equation (a) AND (b) will be the intersection of their value: $(a)∩(b)$ or $(-1,∞)∩(-∞,∞)$ and find the final result for $x: (-1,∞)$ that satisfy the inequality $|x-3|-|x-4| < x$


for /x-3/-/x-4/

1st if both absolute values are positive. x=+1

2nd if both are negative, x=-1

3rd if the 1st one is negative and 2nd one positive x=7/3

4th if 1st one is posit. and 2nd one negat. x=7. so bcse all these numbers, give us a solution except -1 that is why all numbers above -1 are included in the solution. or S= x>-1. or from -1 to infinite.


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