Question about solving absolute values. 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.


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