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Recently, I came across the following math problem:

Find the range of values of $x$ that satisfy $|x-1|+|6-2x|=|5-x|$.

To solve this problem, I considered the 3 cases:

  1. $x\le 1$, which gives $7-3x=5-x$, giving $x=1$
  2. $1\lt x\lt 3$, which gives $5-x=5-x$, suggesting that $1<x<3$ are all solutions.
  3. $x\ge 3$, which gives $3x-7=5-x$ or $3x-7=x-5$, giving $x=3$ or $1$ (which I am slightly confused as to why I get $x=1$ when I assumed $x\ge3$).

So my first question is, is my method above correct? Why did I arrive at the contradiction as pointed out?

Anyhow, I arrived at the solution that $1\le x \le3$ is the solution range, which is the correct answer.

Upon closer inspection, it is clear than this solution set is that to $(x-1)(6-2x)\ge0$ as well, which is all that the suggested solutions said.

This led me to think that there is some correlation between the 3 linear equations, and it is observed that $x-1+6-2x=5-x$. Further investigation suggests that for other linear equations $a,b$, where $|a|+|b|=|a+b|$, $ab\ge0$ appears to give the desired solution set. The logic here does not seem trivial to me.

My second question: Is there a theorem describing this observation? Or is it something trivial that I am missing out? Thanks.

Alessio K
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QuIcKmAtHs
  • 1,451

4 Answers4

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For the first question, you have three quantities to decide the sign of: $x-1$, $6-2x$ and $5-x$. Specifically:

  • $x-1\ge0\Leftrightarrow x\ge1$;
  • $6-2x\ge0\Leftrightarrow x\le 3$;
  • $5-x\ge 0\Leftrightarrow x\le 5$.

Therefore your four cases are:

  1. $x<1$, where $\lvert x-1\rvert=1-x$, $\lvert 6-2x\rvert=6-2x$ and $\lvert 5-x\rvert=5-x$;

  2. $1\le x\le 3$ where some absolute values must be in another algebraic form and some must stay the same;

  3. $3<x\le 5$, similar;

  4. $x>5$, and you should know the drill.

For the second question: since both sides are non-negative, $$\lvert a+b\rvert=\lvert a\rvert+\lvert b\rvert\Leftrightarrow \lvert a+b\rvert^2=(\lvert a\rvert+\lvert b\rvert)^2\Leftrightarrow (a+b)^2=a^2+b^2+2\lvert ab\rvert\Leftrightarrow ab=\lvert ab\rvert$$

Which is the case if and only if $ab\ge0$.

  • Thanks for the solution, appreciate it very much!

    Another question, why does my "trial-and-error" method present a contradiction? ie, when i assume x > 5 and solve the equation x-1 + 2x -6 = x-5, I get x = 1?

     – QuIcKmAtHs Sep 04 '20 at 08:28
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If it is not true that $ab \geq 0$ then either $a>0$ and $b <0$ or $a<0$ and $b >0$. In the first case we get $a-b=|a+b|$ which means either $a-b =a+b$ or $a-b=-(a+b)$. But this means $b=0$ or $a=0$ which a contradiction. Similarly the second case leads to a contradiction.

Kavi Rama Murthy
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We know that for all $a,b$,

$$|a+b|\le|a|+|b|$$

and equality holds when $a,b$ have the same sign (because the magnitudes add up, otherwise they subtract).

Hence the solution

$$(x-1\ge0\land6-2x\ge0)\lor(x-1\le0\land6-2x\le0)$$

which simplifies as

$$1\le x\le3.$$

0

Alternative solution:

An expression of the form $|a|$ can be written as $-a$ or $a$ and the switch occurs at the root(s). In this particular problem, we have three "switch points", $x=1,3,5$, and we can rewrite the equation in the four subintervals so defined.

  • $x\le1\to1-x+6-2x=5-x$, or $x=1$;

  • $1\le x\le3\to x-1+6-2x=5-x$, or $5=5$;

  • $3\le x\le5\to x-1+2x-6=5-x$, or $x=3$;

  • $5\le x\to x-1+2x-6=x-5$, or $x=1$, which is not possible.

By collecting the results,

$$3\le x\le5.$$