Sorry the title came off funny!
Given that "a" is a positive constant, solve the inequality:
l x - 3a l > l x - a l
"l l" is the modulus
Originally posted in BoAjjang, but I was stuck too... I know some of you are really good in Math. Please help, thank you!!!
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L X - 3a L > L X - A L Can anyone help solve?
#1
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Posted 26 April 2007 - 09:47 AM
credits: splotchydivinity
#2
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Posted 26 April 2007 - 02:32 PM
i thought those were shirt sizes. opps. 
substitute 1 for A. casue if A is positive, you can put 1 in, since 1 is positive, right?
so... would it be.... x-3 > x-1 ????
and then you solve from there...
no wait... is that the answer?
im not good in math.
that's how i would see it.
and then i would fail.
good luck.
substitute 1 for A. casue if A is positive, you can put 1 in, since 1 is positive, right?
so... would it be.... x-3 > x-1 ????
and then you solve from there...
no wait... is that the answer?
im not good in math.
that's how i would see it.
and then i would fail.
good luck.
#3
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Posted 26 April 2007 - 05:57 PM
there are no possible solutions.
there are 4 cases.
if both sides are positive, or both negative, then there is no value of x.
if the left side is positive, and the right side is negative. you get x - 3a > -x + a , so 2x > 4a, x>2a
if the left side is negative and the right is positive, you get -x + 3a > x -a, so then 4a > 2x, 2a > x
the intersection of x > 2a and x <2a is null.
there are 4 cases.
if both sides are positive, or both negative, then there is no value of x.
if the left side is positive, and the right side is negative. you get x - 3a > -x + a , so 2x > 4a, x>2a
if the left side is negative and the right is positive, you get -x + 3a > x -a, so then 4a > 2x, 2a > x
the intersection of x > 2a and x <2a is null.
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