Modulus inqeualities
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Ok, heres the question:
(3x - 2)/ l x + 1 l > 1 , x =/= -1.
I think the first step is to multiply both sides by (l x + 1 l)^2, but how to continue on after that? Do I make 2 seperate equations and changing the inequality/sign for 1 equation like the way we do linear modulus inequalities?
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(3x - 2) / l x + 1 l > 1
3x - 2 > |x + 1| ==> no need to square because |x+1| is always positiveso,
Sketch the graph, and find that they intersect at x = 1.5
So answer to the above inequality: x > 1.5 -
don't we have to split up into 2 equations? because the modulus function causes a value to be positive? and we cannot assume that x +1 is always positive?(and i remembered, squaring both sides happens only when both sides are mod-ded
)so i had 3x - 2 < -x -1 = x<1/2..
paiseh for the late reply
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Fwah so chim.
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Originally posted by donkhead333:
don't we have to split up into 2 equations? because the modulus function causes a value to be positive? and we cannot assume that x +1 is always positive?(and i remembered, squaring both sides happens only when both sides are mod-ded
)so i had 3x - 2 < -x -1 = x<1/2..
paiseh for the late reply
you have to draw out the graph to see
removing the mod in this case may not be correct because -x-1 can be negative.
Note in your case where x < 1/2, -x-1 is negative already, but the mod makes it positive.Maybe you wouldn't understand the above paragraph. Believe me, just sketch out the graph, and you will see it. Or you could plot it with your graphic calculator.... Sketching is faster though... :D
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Originally posted by eagle:
you have to draw out the graph to see
removing the mod in this case may not be correct because -x-1 can be negative.
Note in your case where x < 1/2, -x-1 is negative already, but the mod makes it positive.Maybe you wouldn't understand the above paragraph. Believe me, just sketch out the graph, and you will see it. Or you could plot it with your graphic calculator.... Sketching is faster though... :D
ppl don't believe in sketching anymore what with the gc :(
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Originally posted by tinuviel07:
ppl don't believe in sketching anymore what with the gc :(
sketching in this case takes only 0.5 minutes or less. Or if you have lots of experience, you can even visualise the graph in your head, as what I have done by typing the solution right out without sketching at all.
Not sure how long a GC will take. -
Originally posted by donkhead333:
don't we have to split up into 2 equations? because the modulus function causes a value to be positive? and we cannot assume that x +1 is always positive?(and i remembered, squaring both sides happens only when both sides are mod-ded
)so i had 3x - 2 < -x -1 = x<1/2..
paiseh for the late reply
3x - 2 > |x + 1|
one look at this equation, and you realise that the modulus part, smallest can go is 0.
so x must be bigger than 1.5.
smaller than 1.5 is rejected...
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Originally posted by skythewood:
3x - 2 > |x + 1|
one look at this equation, and you realise that the modulus part, smallest can go is 0.
so x must be bigger than 1.5.
smaller than 1.5 is rejected...
Another very smart logical deduction
great for quick instant checks.
thanks for pointing out!
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er, so the x< 1/2 one is invalid in the first place? and we work with only the equation of 3x-2> l x+1 l?
and the graph...think you're referring to the one with the curve and the x-axis only, right?
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3x-2 is a straight line
|x+1| is a V-shaped graph
then you plot on the same axes. There will be one point where 3x-2 intersects |x+1|
you can see where 3x-2 > |x+1| from the graph
or you can use your GC