Solve the inequality $|3x-2|-|x+2|>x$

When $|x+2|<0$:

$-(3x-2)+(x+2)>x\iff x <\frac{4}{3}$

When $|x+2|>0\land |3x-2|<0$:

$-(3x-2)-(x+2)>x\iff x < 0$

When $|3x-2|>0$:

$3x-2-(x+2) > x \iff x > 4$

Now, personally I would say that the solutions are:

$x < \frac{4}{3}\lor x> 4$

Since the first part $x <\frac{4}{3}$ includes the part with $x<0$. Now, apparently I'm wrong. However, I don't see why.

Why is the correct solution $x<0\lor x>4$ and not $x<\frac{4}{3}\lor x>4$?

  • $\begingroup$ Made sure the 'cases' were defined correctly this time around. $\endgroup$ – B. Lee Oct 28 '14 at 17:32

A few remarks:

1, $|x+2|<0$ and $|3x-2|<0$ never holds; what you probably mean is $x+2<0$ and $3x-2<0$.

2, You don't actually check $x+2=0$ as a case.

3, As for your question: you derived $x<\frac{4}{3}$ from the assumption that $x+2<0$. So what you've shown in the first case is that whenever $x+2<0$ and $x<\frac{4}{3}$, the inequality holds; this only gives you $x<-2$ as a solution. Similarly, the second case gives $-2<x<0$ and the third gives $x>4$, which taken together means that any $x$ which is less than $0$ or more than $4$ satisfies the inequality (with maybe the exception of $x=-2$; see the point above).

Hope that clears things up a bit! :)


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