Question about solving absolute values.

Question

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.

Answer 1

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 $x \le 3$.

Answer 2

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 $x\le 3$.

Answer 3

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