Absolute numbers question

Question

Here’s a question that took a run for my money.

Solve for $x$ for the following

$|x+3|+ |x - 4|\lt 0$

Answer 1

There are three cases to consider:

• CASE 1 ($x < -3$): We have $-x-3+4-x<0 \implies x > \frac{1}{2}$ but this is contradicting our first assumption $x < -3$ so there is no solution here.

• CASE 2 ($4>x>-3$): We have $x+3+4-x < 0 \implies 7 < 0$, which is a contradiction. Therefore, there is no solution here.

• CASE 3 ($x > 4$): We have $x+3+x-4 < 0 \implies x < \frac{1}{2}$, which is again contradicting our first assumption $x > 4$. So there is no solution here.

Therefore the solution set is $\emptyset$.

Alternatively, we have $$|x+3| < -|x-4|$$ which is always false since $|x-4| \ge 0$ for all $x$. So the result follows.

Answer 2

$|x+3| + |x -4| < 0\implies$

$|x+3| < -|x-4|$.

But absolute values are always non-negative so this is impossible. No solutions.

$|a| + |b| < 0$ is never possible.