1
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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?

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

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 $a=0$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.