SOLUTION: I don't know how to solve this inequalitie.: It gave me this information. y=x^2 , squareroot y= squareroot x^2, squareroot y=|x| where therefore since x The question is |x-3|>|2x+

Algebra ->  Absolute-value -> SOLUTION: I don't know how to solve this inequalitie.: It gave me this information. y=x^2 , squareroot y= squareroot x^2, squareroot y=|x| where therefore since x The question is |x-3|>|2x+      Log On


   



Question 125946: I don't know how to solve this inequalitie.: It gave me this information. y=x^2 , squareroot y= squareroot x^2, squareroot y=|x|
where therefore since x The question is |x-3|>|2x+1|
I asked this question before but is there another way to figure this out without graphing like can we actually solve it by doing the math?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
|x-3|>|2x+1|
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if x = 0, you get |-3|>|1|, which is true: so x=0 is part of the solution.
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If both x-3 and 2x+1 are positive you get:
x-3 > 2x+1
x<-4
Testing x=-10 you get:
|-13| > |-19| which is false.
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If both x-3 and 2x+1 are negative you get:
-x+3 > -2x-1
x > -4
Testing x=0 you know already this is true so x>-4 is part of the solution.
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If x-3 is positive and 2x+1 is negative you get:
x-3 > -2x-1
3x > 2
x>2/3
Testing x=1 you get:
|1-3| > |2+1| which is false
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If x-3 is negative and 2x+1 is positive you get:
-x+3 > 2x+1
3x<2
x <2/3
Testing 0 you know already this is true so x < 2/3
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Putting that all together you get -4 < x < 2/3
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PS: Graphing it is easier.
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Cheers,
Stan H.