Solve $|3-x|=x-3$.

Question

Solve: $|3 - x| = x - 3$.

Answer: $|u| = -u$ when and only when $u \le 0$. So, $|3 - x| = x - 3$ when and only when $3 - x \le 0$; that is, $3 \le x$.

Hi! I'm new here. I'm working out of this book called Schaum's Outlines 3,000 Solved Problems in Calculus. I understand that when solving for absolute values that you have to draw two equations, for example when solving for $|x+3|<5$ you must solve for both $x+3<5$ AND $x+3>-5$. But can someone please explain the process above to me? I don't understand it. Thanks!

Answer 1

$|-u|=|-1 \cdot u|=|-1| \cdot |u|=1 \cdot |u|=|u| \\ |u|=u \text{ if } u \ge0 \\ |3-x|=|(-1)(x-3)|=|x-3| \\ |x-3|=x-3 \text{ if } x-3 \ge 0$

Answer 2

the graph of $y = |3-x|$ is $v$-shaped with the vertex at $(3,0)$ and it lies in the upper half plane. on the left of $x = 3$ is the line $y = 3 - x$ and to the right is the line $y = x - 3$ the equation $|3-x| = x - 3$ is satisfied for all $x \ge 3.$

Answer 3

This is the definition of the absolute value of a real number that we are using:

$$|x| = \begin{cases}x; & x > 0 \\ 0; & x = 0 \\ -x; & x < 0\end{cases}$$

The equation to be considered is:

$$|3 - x| = x - 3 = \begin{cases}3 - x; & 3 - x > 0 \\ 0; & 3 - x = 0 \\ -(3 - x); & 3 - x < 0\end{cases}$$

We see that we don't need to consider $ 3 - x > 0$ because that case doesn't make sense. Only $3 - x = 0$ and $3 - x < 0$ will give you sensible answers. Conclude $ 3 \le x$.