How to solve $|x-5| + |x-4| ≥ 3$

Question

I was given the following question to solve for homework. The solution $S$ I got was $$S = \{x : x ≥ 6 \text{ or } x\leq3\}$$ I checked my solution with the answers provided and it was correct. My working out can be seen below:

$|x-5| + |x-4| - 3 = \begin{cases} 2x-12 , & \text{if } x>5\\ -2, & \text{if }4<x≤5\\ -2x+6,&\text{if }x\leq4\end{cases}$

I then solved $2x-12 ≥3, -2 ≥ 3,$ and $-2x+6 ≥3$.

This is how I was instructed to do it but it does not make sense to me because why do we just dismiss the regions stated above, namely $x > 5,4<x≤5$ and $x≤4$.

Can someone please provide an explicit explanation?

Answer 1

I) $U_1=\{x: 5\leq x\}$. Then, $x-5+x-4\geq3\implies x\geq6$. So, $V_1=\{x: 6\leq x\}$ and the first solution set is $S_1=U_1\cap V_1=\{x: 6\leq x\}.$

II) $U_2=\{x: 4\leq x\leq 5\}$. Then, $-x+5+x-4\geq3\implies 1\geq3$, false. So , $V_2=\{\}$ and the second solution set is $S_2=U_2\cap V_2=\emptyset$.

III) $U_3=\{x: x\leq 4\}$. Then, $-x+5-x+4\geq3\implies x\leq3$. So, $V_3=\{x: x\leq3\}$ and the third solution set is $S_3=U_3\cap V_3=\{x: x\leq3\}.$

The solution set is then $S=S_1\cup S_2\cup S_3=\{x: x\leq3 \lor 6\leq x\}$

Answer 2

In short, you do not disregard those conditions. A more complete solution might proceed as follows:

If $x > 5$, then $|x - 5| + |x - 4| - 3 = 2x - 12$, which we require to be greater than or equal to $0$. So $2x - 12 \ge 0$ and $x > 5$, hence $x \ge 6$. This single condition $x \ge 6$ meets both criteria.

Similarly, if $4 < x \le 5$, then $|x - 5| + |x - 4| - 3 = -2$, which is always negative, so no solutions exist in the interval $4 < x \le 5$.

Finally, if $x \le 4$, then $|x - 5| + |x - 4| - 3 = -2x + 6$, and if we want this to be nonnegative, then $x \le 3$, so the combined conditions $x \le 3$ and $x < 4$ means we must have $x \le 3$.

All together, this yields the solution $x \ge 6$ or $x \le 3$.

Answer 3

Deal in Cases...

Case(i) $x\ge 5$

$|x-5|+|x-4|\geq 3$ is same as $$x-5+x-4\geq3$$ $$2x-9 \geq3 \implies x\geq6$$

Case(ii) $4 \le x <5$

$|x-5|+|x-4|\geq 3$ is same as $$5-x+x-4 \geq3\implies 1\geq3$$ which cannot happen and hence we can discard this case.

Case(iii) $x<4$

$|x-5|+|x-4|\geq 3$ is same as $$4-x+5-x \geq 3 \implies x \leq 3$$

for case(i), what it means is if I need a solution greater than 5, then it should be greater than 6, which means $ x \in [5,6)$ can't be a solution

similarly on case(iii), $x \in (3,4]$ can't be a solution

Answer 4

$|x-5|+|x-4|\ge 3$

The algebra is simplified with a symmetric substitution.

$u=x-9/2$

$|u-1/2|+|u+1/2|\ge 3$

Case $u>1/2$:

$2u\ge 3\implies u\ge 3/2$

Case $-1/2\le u\le 1/2 :$

$1/2-u+u+1/2\ge 3$

Case $u \le -1/2$

$1/2-u-u-1/2\ge 3\implies -2u\ge 3 \implies u \le -3/2$

$u\le 3$.

Or

$u\ge 6.$

Answer 5

We may consider the problem in two dimensions. The condition $$\|(x,y)-(4,0)\|+\|(x,y)-(5,0)\|\ge 3$$ describes the exterior of the ellipse with foci at $(4,0),\ (5,0)$ and the distance $3.$ The left most point of the ellipse is located at $(3,0)$ while the right most one at $(6,0).$ Therefore the solution is $x\le 3$ or $x\ge 6.$