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It common in the literature to solve the modulus equation like $|x+5|+|x-1|=8$ by dividing into cases when $x<-5$, $-5\leq x<1$ and $x\geq1$.

My question is whether dividing into cases is really necessary, since what we care is just whether the thing inside the absolute value bars is positive or negative, so why can't we just solve each of the four possible combinations of $x+5+x-1$, $x+5-x+1$, $-x-5+x-1$, $-x-5-x+1$. It will also give the correct solutions anyway.

Is there any cases where just by solving the four possible combinations will not give the correct solutions? Or are they just the same method in disguise? If they are the same, why do I never see anyone mention it? I hope my question is understandable. Many thanks!

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    $\begingroup$ I'm not certain as to what happens when you have higher-order terms (e.g. $x^3$), but I believe your method works fine for all linear equalities. $\endgroup$ – apnorton May 6 '14 at 2:35
  • $\begingroup$ @anorton. Thanks. And that is basically my doubt, will it always work? for higher order terms as well? $\endgroup$ – user71346 May 6 '14 at 2:39
  • $\begingroup$ Note, if points $A$ and $B$ on the number line have coordinates $a$ and $b$ respectively, then an equation of the segment $\overline{AB}$ is $$|a-x| + |x-b| = |a-b|$$ Note also that $|x-y|=|y-x|$. $\endgroup$ – steven gregory Jun 2 '18 at 21:43
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No. Taking the max of the four possible combinations will always give the right answer, but as you suspect in the question, they're essentially the same method in disguise.

When $x \le -5$, "$-x-5-x+1$" gives the maximum value.

When $-5 \le x \le 1$, "$x+5-x+1$" gives the maximum value.

When $x \ge 1$, "$x+5+x-1$" gives the maximum value.

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Let us take the the related equation $|x| +|x-2|+|x-6|=10$, and the choice of signs $-,+,+$.

So we arrive at $-x+x-2+x-6=10$, giving $x=18$. Of course $x=18$ is not a solution of the original equation.

The problem is that the choices $-,+,+$ are incompatible. The issue of compatibilitity does not arise when we look at equations that involve the absolute values of $2$ linear expressions.

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  • $\begingroup$ Thanks for the answer. But I have checked the solutions are also x=2 and x=-6 with either method (cases and listing all expressions). I don't quite understand the reason why solving all possible expressions will give more solutions, would you mind explaining the reason? Thanks. $\endgroup$ – user71346 May 6 '14 at 9:00

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