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.



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 $x\le 3$.
