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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 ?

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    $\begingroup$ how did you find your solutions? $\endgroup$ – Jonas Lenz Dec 2 '17 at 11:36
  • $\begingroup$ How did you find the solutions $0,2,3,4,5,6$ it would be helpful for us to help you identify the mistake you found. Other than that for $x>6$ you have that $2x-3-x>6$ for $x<0$ we have $3-3x>3$ so checking integers between and inclusive with $0$ and $6$ is enough. $\endgroup$ – kingW3 Dec 2 '17 at 11:36
  • $\begingroup$ For $0\lt x\lt 3/2$, the inequality is equivalent to $-(2x-3)-x\le 3,$ i.e. $-3x\le 0,$ i.e. $x\color{red}{\ge}0$. $\endgroup$ – mathlove Dec 2 '17 at 11:48
  • $\begingroup$ Oh, I forgot changing the sign of dividing -3. Damn my mistake hits me in my heart. $\endgroup$ – Harshit Dec 2 '17 at 11:51
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Expression $$-3x<0$$ simplifies to $$x>0$$ not $$x<0$$ so $$3/2>x>0$$ and there is your $$x=1$$ answer.

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

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  • $\begingroup$ In the 2nd case, since the condition given is $f(x) \le 3$, this would come up as $-3x + 3 \le 3$ or $-3x \le 0$, which gives us no solutions with respect to the case taken $\endgroup$ – Harshit Dec 2 '17 at 11:46
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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 $$-2x+3+x\le 3$$

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

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 $x\in[0,6]$

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