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.
Question about solving absolute values.
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2is 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) – 2010-08-20
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1Out of curiosity, did you create the image of the solution above? If so, was it done with TI-Nspire software or something else? – 2010-08-20
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1On 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. – 2010-08-20
3 Answers
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 $|x-3| = 3-x \iff x-3 \le 0 \iff x \le 3$
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0I put $y=x-3$ instead, which is the same idea of yours. – 2010-08-20
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0I 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. – 2010-08-21
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0By no means I criticized you. – 2010-08-21
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1No 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**. – 2010-08-21
By definition of absolute value:
$|x| = x$ if $x > 0$
and
$|x| = -x$ 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 $x \le 3$.
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0not helpful i could not understand this – 2010-08-20
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0@Zia: What part didn't you understand? – 2010-08-20
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0how you assumed that x > 3 why you assumed that – 2010-08-20
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0@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 $x \le 3$." – 2010-08-20