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Find x in Z from this inequality:

$$\frac32\left|x-\frac32\right|=\frac53|2x|-\frac16$$

I tried to solve it,but i don't know how to continue:

$$\frac32\left|2x-\frac32\right|=\frac103|x|-\frac16$$ $$\frac34|2x-3|=\frac{20|x|-1}6$$ $$9|2x-3|=40|x|-2$$

Here i stoped because i don't know what's the next step.

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  • $\begingroup$ It's very hard to understand what you wrote. Please do use LaTeX to write mathematics. $\endgroup$
    – DonAntonio
    Aug 4, 2013 at 15:02
  • $\begingroup$ BTW, what you did is completely correct... $\endgroup$
    – DonAntonio
    Aug 4, 2013 at 15:49

2 Answers 2

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Ideas:

$$\frac32\left|\;x-\frac32\;\right|=\frac53|2x|-\frac16\implies 9\left|\;x-\frac32\;\right|=20|x|-1\implies9\left|\;2x-3\;\right|=40|x|-2$$

In this case perhaps it's easier to check different cases:

$$\begin{align*}\bullet& x<0 :&\;\;27-18x=-40x-2\implies& 22x=-29\ldots \\{}\\ \bullet& 0\le x<\frac32:&\;\;27-18x=40x-2\implies&58x=29\ldots\\{}\\ \bullet&x\ge\frac32:&\;\;18x-27=40x-2\implies&22x=-25\ldots \end{align*}$$

Note: the third case above has no solution (why?)

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  • $\begingroup$ If i take 2 cases: 1.when x<0(negative) and when x>0(positive).It would be right? $\endgroup$
    – marinaaaa
    Aug 4, 2013 at 20:42
  • $\begingroup$ I don't understand what you mean, @marinaaaa: I already wrote above all three possible cases and even said the last one has no solution...! $\endgroup$
    – DonAntonio
    Aug 4, 2013 at 20:46
  • $\begingroup$ Ok,now i got the idea!Thank you very much! $\endgroup$
    – marinaaaa
    Aug 4, 2013 at 20:53
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HINT:

We know for real $x,|x|=\begin{cases} x &\mbox{if } x\ge0 \\ -x & \mbox{if } x<0 \end{cases} $

So,$|2x|=\begin{cases} 2x &\mbox{if } 2x\ge0\iff x\ge0 \\ -2x & \mbox{if } x<0 \end{cases} $

and $|x-\frac32|=\begin{cases} x-\frac32 &\mbox{if } x-\frac32\ge0\iff x\ge\frac32 \\ -(x-\frac32)=\frac32-x & \mbox{if } x<\frac32 \end{cases} $

Now consider the ranges, $-\infty<x<0,$ $0\le x\le\frac32,$ $\frac32<x\le2,$ and $2<x<\infty$

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