$$|x-3|+|6-x| ≤x $$ I tried to solve it with parts for $x<3$, then for $3< x <6$ and for $x>6$ I found the set of each and i did the intersection and it resulted $[3,9]$ but I don't know if this is the correct answer. Can anyone help me?
2026-03-25 01:14:12.1774401252
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Solve the absolute value inequality
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write your inequality in the form $$|x-3|+|x-6|\le x$$ then you have the cases to solve: $$x\geq 6$$ $$3\le x<6$$ $$x<3$$ the solution is given by $$3\le x\le 9$$
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"I tried to solve it with parts for x<3, then for 36"
Perhaps not the slickest way, but it's the way I'd be most comfortable with.
If $x \le 3$ then
$|x-3|+|6-x| = 3-x + 6-x = 9-2x \le x \implies x \ge 3 \implies x = 3$
If $3 \le x \le 6$ then
$|x-3|+|6-x| = x-3 + 6-x = 3 \le x \implies 3 \le x \le 6$.
If $x \ge 6$ then
$|x-3| + |6-x| = x-3 + x -6 = 2x -9 \le x \implies x \le 9\implies 6\le x \le 9$.
So solutions are precisely $3\le x \le 9$.
As I said, not the slickest but very direct and clear.
By triangle inequality we have $$x\geq |x-3|+|6-x| \geq |x-3+6-x|=3$$ so $x\geq 3$. Now we have $$x-3+|6-x|\leq x $$ so we have $$|6-x|\leq 3$$and thus $$-3\leq 6-x\leq 3$$ which give us ...