# A Question About Solving An Equation Involving The Addition Of Absolute Value Functions

I am trying to solve the following problem: $$|4-x| \leq |x|-2$$. I am trying to do it algebraically, but I'm getting a solution to the problem that makes no sense. I fail to see the error in my reasoning though. I hope to get an explanation where I went wrong.

$$|4-x| \leq |x|-2$$

$$|4-x|-|x| \leq -2$$

If $$4-x$$ and $$x$$ are both negative, then for them to be equal to $$2$$, we need to multiply both expressions by $$-1$$.

$$-4+x+x \leq -2$$

$$-4+2x \leq -2$$

$$2x \leq 2$$

$$x \leq 1$$

But if you sub in any $$x$$ less than or equal to $$1$$, the inequality doesn't work! Can you please explain where in my logic, where in the steps, have I gone wrong? Thank you!

• If you type \leq and \geq when you are in math mode, you will obtain $\leq$ and $\geq$, respectively. – N. F. Taussig Sep 22 '18 at 9:11

For both $x$ and $4-x$ to be negative, $x<0$ and $x>4$, which is impossible.

As for the solution, the right side has to be positive, hence $| x | \geq 2$. Now, we solve it in $3$ parts.

In $( -\infty,-2 ]$, we have, $4+|x| \leq |x|-2$, which is not true.

In $(4,\infty )$, we have, $x-4 \leq x-2$, which is true.

In $[2,4]$, we have, $4-x \leq x-2 \implies x \geq 3$.

Hence, the solution is, $x \in [3, \infty )$.

Subtract all terms to the left side of the inequality to obtain

$$|4-x|-|x|+2\le0$$

Then define $$f(x)=|4-x|-|x|+2$$. We want to find all values of $$x$$ for which $$f(x)\le0$$.

Important changes occur in the function at $$x=0$$ and at $$x=4$$

$$|x|=\begin{cases} x&\text{ for }x\ge0\\-x&\text{ for }x<0\end{cases}$$ $$|4-x|=\begin{cases} 4-x&\text{ for }x\le4\\x-4&\text{ for }x>4\end{cases}$$

This allows us to re-write $$f(x)$$ as a piecewise defined function: $$f(x)=\begin{cases} 6&\text{ for }x<0\\6-2x&\text{ for }0\le x<4\\-2&\text{ for }x\ge4\end{cases}$$

In the middle interval we see that $$f(x)$$ is decreasing and reaches a value of $$0$$ at $$x=3$$ and we get $$f(x)\le0$$ for $$x\ge3$$. So the solution set is $$[3,\infty)$$.

hint

observe that there is no solution if the RHS is negative which gives $$|x|<2$$

so, solve in $[2,4], [4,+\infty)$ and $(-\infty,-2]$.