Inequality absolute value help

[converting1]

$x - |x-|x|| > 2$

how would I go about solving something like this?

my initial thoughts was to consider if x >= 0

I get 2-x < 0 then x > 2 in that case

then consider if x < 0 which I get -|x+x| > 2-x then 2x > 2-x then x > 2/3 but I'm having troubles deciding which one is correct, and if there is another way to do it (I can't seem to sketch |x-|x||

edit: I know it's pretty obvious from looking at the equation that x > 2, but just wondering why I get the x >2/3 part from, and if I got given a question which is not obvious, then how would I know x > 2/3 would be wrong (for example), and which inequality would be the correct bit

 

[statdad]

You do need to consider cases, and your first bit is correct since, if $x \ge 0$, the absolute value term gives
$$| x - |x|| = |x - x| = 0$$

and your problem reduces to $x > 2$.

For the case $x < 0$, remember $|x| = -x$, so
$$|x - |x|= |x - (-x)| = |2x| = \text{ what?}$$

Simplify remembering $x < 0$: then take that simplification and plug into your original inequality statement and see what happens.

 

[converting1]

You do need to consider cases, and your first bit is correct since, if $x \ge 0$, the absolute value term gives
$$| x - |x|| = |x - x| = 0$$

and your problem reduces to $x > 2$.

For the case $x < 0$, remember $|x| = -x$, so
$$|x - |x|= |x - (-x)| = |2x| = \text{ what?}$$

Simplify remembering $x < 0$: then take that simplification and plug into your original inequality statement and see what happens.

$|x - |x|= |x - (-x)| = |2x| = -2x$ if x < 0

plugging this in I get x - (-2x) > 2 so x > 2/3

 

[statdad]

Ok, so for the inequality to be satisfied you need to have both $x > \frac 2 3$ and $x > 2$ true. Where on the number line do those two inequalities intersect (where do their graphs overlap)? The answer to that gives the solution set to the problem.

 

[converting1]

x>2 obviously! So stupid

thanks, lol.

 

[statdad]

"So stupid"

Not at all. At some time we've all benefited from a little push to look at a problem with a different eye.