I solved the following problem from by book, but the answer of this problem at the end of book is $x \leq 3$. Please tell me how I can get this answer.
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    $\begingroup$ is the picture your solution or the solution from your book? supposing former: now 0=0 holds for any x; so you should include [in the answer] all x from the second case -- i.e. all x s.t. |x-3|=-(x-3) $\endgroup$ – Grigory M Aug 20 '10 at 22:19
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    $\begingroup$ Out of curiosity, did you create the image of the solution above? If so, was it done with TI-Nspire software or something else? $\endgroup$ – Isaac Aug 20 '10 at 22:41
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    $\begingroup$ On the right 0=0 is true, which means that all x such that x-3<0, or equivalently x<3, are valid solutions. This combined with the other solution on the left x=3, gives you x<=3 as a solution. $\endgroup$ – Américo Tavares Aug 20 '10 at 23:10

I'll answer by editing your solution slightly:

Depending on the sign of $x-3$: $$\begin{align} x-3=3-x&\text{ and }x-3\ge 0&\quad\text{ or }\quad&&-(x-3)=3-x\text{ and }&x-3<0 \\\\ x-3-3+x=0&\text{ and }x\ge 3&\quad\text{ or }\quad&& x-3=x-3\text{ and }&x<3 \\\\ 2x-6=0&\text{ and }x\ge 3&\quad\text{ or }\quad&& x-3-x+3=0\text{ and }&x<3 \\\\ 2x=6&\text{ and }x\ge 3&\quad\text{ or }\quad&& 0=0\text{ and }&x<3 \\\\ x=3&\text{ and }x\ge 3&\quad\text{ or }\quad&& \text{(true) and }&x<3 \\\\ &x=3&\quad\text{ or }\quad&& x<3& \\\\ &&x\le 3& \end{align}$$

edit: As a further explanation of the problem as a whole, consider the graph below, where $|x-3|$ is shown in blue and $3-x$ is shown in red.


The graphs coincide for $x\le 3$ and the blue graph is higher for $x>3$, so the original equation is true for $x\le 3$.


HINT It's obvious by a shift: put $\; z = x-3 \;$ in $\; |z| = -z \iff z \le 0 \; $ Making this substitution yields $|x-3| = 3-x \iff x-3 \le 0 \iff x \le 3$

  • $\begingroup$ I put $y=x-3$ instead, which is the same idea of yours. $\endgroup$ – Américo Tavares Aug 20 '10 at 23:15
  • $\begingroup$ I don't agree that your proof is the same. It doesn't appear to exploit any change of variable! I added a sentence to my proof to make it precisely clear how it finishes. I hadn't read the details of your proof before since a quick glance at its length led me to infer it was much different than what I envisioned. $\endgroup$ – Bill Dubuque Aug 21 '10 at 0:03
  • $\begingroup$ By no means I criticized you. $\endgroup$ – Américo Tavares Aug 21 '10 at 7:04
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    $\begingroup$ No problem. I merely wanted to emphasize the view of this solution as a transformation to a simpler problem since this is a ubiquitous problem solving technique - esp. when the transformation is some innate symmetry. $\endgroup$ – Bill Dubuque Aug 21 '10 at 17:26

By definition of absolute value:

$|x| = x$ if $x > 0$


$|x| = -x$ otherwise.

You are given $|x-3| = 3 - x$.

Now given a real $x$, either $x>3$ or $x \le 3$.

(The reason for splitting it this way is that we have $|x-3|$ and in order to get rid of the || we need to decide whether $x-3 > 0$ or not)

So we split into two cases.

Case 1) $x > 3$.

Then we have that $x-3 > 0$ and so by definition of absolute value, $|x-3| = x-3$.

Therefore you equation

$|x-3| = 3 - x$

is same as

$x-3 = 3 -x$

which is same as

$2x = 6$

which is same as

$x = 3$.

Since we assumed $x > 3$, there is no solution to your equation.

Case 2) $x \le 3$

Then we have that $x - 3 \le 0$ and so by definition of absolute value

$|x-3| = -(x-3) = 3-x$.

Therefore your equation is same as

$3-x = 3-x$ which is true for any $x$ (but keep in mind that we are only considering $x \le 3$).

Hence any $x \le 3$ satisfies this.

Combine the two solutions for both the cases and you get $x \le 3$.

The way you solved it, you get

$x = 3$ or $x < 3$.

If you combine the two, you can say $x \le 3$.

  • $\begingroup$ not helpful i could not understand this $\endgroup$ – Zia ur Rahman Aug 20 '10 at 22:03
  • $\begingroup$ @Zia: What part didn't you understand? $\endgroup$ – Aryabhata Aug 20 '10 at 22:09
  • $\begingroup$ how you assumed that x > 3 why you assumed that $\endgroup$ – Zia ur Rahman Aug 20 '10 at 22:11
  • $\begingroup$ @Zia: I edited the answer to make it clearer. Let me know if that helps. Please make sure to read the line: "Now given a real $x$, either $x>3$ or $x \le 3$." $\endgroup$ – Aryabhata Aug 20 '10 at 22:15

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