Solutions of Absolute Value inequality

Question

Find the integral solutions of the inequality $$|2x-3|-|x| \le 3$$ The solutions i found out were $0,2,3,4,5,6.$But how do i find 1 as a solution to it ? I'm now fully confused of the cases. The solution 2,3,4,5,6 came out from the condition when $x \gt 3/2$. The solution x=0 came up in the case $x \le 0$. But in the case $0\lt x \lt 3/2 $, i get the following expression, $$3-2x-x \lt 3$$which simplifies to $$-3x \lt 0$$or $x \lt 0$. But the given condition inequality does not satisfy in the case we took. So how do i find x=1 as an answer ?

Answer 1

Expression $$-3x<0$$ simplifies to $$x>0$$ not $$x<0$$ so $$3/2>x>0$$ and there is your $$x=1$$ answer.

Answer 2

Analyse the function $f(x) = |2x-3|-|x|$ in different intervals, namely:

1. $x\in (-\infty, 0]$, $f(x) = -x+3$.

2. $x\in (0, 3/2]$, $f(x) = -3x+3$.

3. $x\in (3/2, \infty)$, $f(x) = x-3$.

Answer 3

in the first case we have $$x\geq \frac{3}{2}$$ then we have $$2x-3-x\le 3$$ $$0\le x <\frac{3}{2}$$ and we get $$-(2x-3)-x\le 3$$ last case $$x<0$$ then we have $$-2x+3+x\le 3$$

Answer 4

$\begin{array}{ll]} |2x-3|-|x|-3 &=\max(2x-3,3-2x)-\max(x,-x)-3\\ &=\max(2x-3,3-2x)+\min(-x-3,x-3)\\ &=\max(\min(x-6,3x-6),\min(-3x,-x))\\ &\le 0 \end{array}$

So $\min(-3x,-x)\le 0\iff x\ge 0$

Then $\min(x-6,3x-6)=x-6\le 0\iff x\le 6$

Thus $x\in[0,6]$