I need to solve the following equation involving absolute value:

$$|x-1| = 1-x$$

Looking at the term $x-1$, I thought I'd divide the interval into parts: $x < 1$ and $x \geq 1$. Now, when $x<1$, $|x-1|$ is negative, which means $|x-1|$ needs to be written as $1-x$. But now I get:

$$1-x = 1-x$$

which is an identity and not a solution. Anyway, so now I moved on to the next part, that of $x \geq 1$. Here $|x-1|$ is positive, and so I have:

$$1-x = x-1$$

which gives $x = 1$.

When I checked the answer, it was $x \leq 1$. I'm not sure how it was reached, but looking back at the equation I can see that $x=0$ is also a solution, which got eliminated during my approach.

Please help me straighten out my thinking here!


You have $|x - 1| = 1 - x$. Yes, we break it in cases

  • if $x \geq 1$, the equation is $x - 1 = 1 - x$, and solving it wields us $x = 1$. Since $1 \geq 1$, our hypothesis, $1$ is a valid solution.

  • if $x < 1$, the equation is $1 - x = 1 - x$, which is true for every $x$, **such that $x < 1 $ **

Now we join all our solutions: every $x < 1 $ and $x = 1$, that is, $x \leq 1$. Ok?

  • $\begingroup$ Yes, okay! :-) I had overlooked the bit about $x<1$, it seems. $\endgroup$
    – ankush981
    May 11 '14 at 3:48

Note $|x-1|=|-(1-x)|=|1-x|$.

So, your question is finding $x$ such that $|1-x|=1-x$. If $1-x\geq 0$, then this is $1-x=1-x$, which is always true, so one set of solutions is $1-x \geq 0$ or $1 \geq x$.

If $1-x \leq 0$, then $|1-x| = -(1-x)$, so this is $-(1-x)=1-x$. This is true if and only if $1-x=0$ or $x=1$ ($a=-a$ is true if and only if $a=0$).


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