Inequality absolute value help

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To solve the inequality x - |x| > 2, consider two cases based on the value of x. For x ≥ 0, the equation simplifies to x > 2. For x < 0, the absolute value leads to the condition x > 2/3. The correct solution requires both conditions to be satisfied, which ultimately means x must be greater than 2. The discussion highlights the importance of case analysis in solving absolute value inequalities.
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## 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
 
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You do need to consider cases, and your first bit is correct since, if x \ge 0, the absolute value term gives
<br /> | x - |x|| = |x - x| = 0<br />

and your problem reduces to x &gt; 2.

For the case x &lt; 0, remember |x| = -x, so
<br /> |x - |x|= |x - (-x)| = |2x| = \text{ what?}<br />

Simplify remembering x &lt; 0: then take that simplification and plug into your original inequality statement and see what happens.
 
statdad said:
You do need to consider cases, and your first bit is correct since, if x \ge 0, the absolute value term gives
<br /> | x - |x|| = |x - x| = 0<br />

and your problem reduces to x &gt; 2.

For the case x &lt; 0, remember |x| = -x, so
<br /> |x - |x|= |x - (-x)| = |2x| = \text{ what?}<br />

Simplify remembering x &lt; 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
 
Ok, so for the inequality to be satisfied you need to have both x &gt; \frac 2 3 and x &gt; 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.
 
x>2 obviously! So stupid

thanks, lol.
 
"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.
 

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