Harder inequalities... (1 Viewer)

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Ive a couple problems that would be great if someone could show the working out for me...

Code:
|5/2x-1| < 1
Code:
(2-x)(2x-1)(x+3) less than/equal to 0
Code:
2x+3/x-4 > 1
Code:
-3x^2+10x+8 less than/equal to 0
Cheers guys!
 

kami

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I'm a bit rusty, but here you go...

Code:
|5/2x-1| < 1
2x - 1 <0 and 2x - 1 >5 [because dividing by 0 gives infinity, dividing by 5 gives 1, and anything in between will push it above 1]
∴ x< 1/2 and x >3

Code:
(2-x)(2x-1)(x+3) less than/equal to 0
to be equal to 0, x must be either 2, 1/2 or -3
a quick way to then find where it is less than 0 would then be to graph it but I'll just use some substitution here:
  • when x = 1, y = 4 so between 2 and 1/2 is not less than 0
  • when x = -1, y = -18 so when x is between 1/2 and -3, y is less than 0
  • when x = 3, y = -30 so when x is greater than 2, y is less than 0
  • when x = -5, y= 154 so when x is less than -3 it is not less than 0
∴ your answer is 1/2 ≥ x ≥ -3 and x ≥ 2

I might give the others a go later after I've done my uni hw
 
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Mountain.Dew

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just a general tip...when you do have a factorised form, draw the actual curve of it on a number plane. then, the solution is for x-values where the curve is below the x-axis (for <= 0) or above x-axis (for >= 0). i find this much better than testing points.
 
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pLuvia

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But isn't this the same thing? if you just test the point in between the 2 numbers let's say and it doesn't fit the condition then the x/y values are outside i.e -a<x x>b and vice versa.

Your way is for the more visual person I guess :p
 

Mountain.Dew

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pLuvia said:
But isn't this the same thing? if you just test the point in between the 2 numbers let's say and it doesn't fit the condition then the x/y values are outside i.e -a<x x="">b and vice versa.

Your way is for the more visual person I guess :p
yeah, i suppose so. either way is equally valid and effective. its all a matter of personal preference. :)
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Riviet

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Mountain.Dew said:
just a general tip...when you do have a factorised form, draw the actual curve of it on a number plane. then, the solution is for x-values where the curve is below the x-axis (for <= 0) or above x-axis (for >= 0). i find this much better than testing points.
I like to draw the curve in my head. :p
 

Mountain.Dew

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Riviet said:
I like to draw the curve in my head. :p
although it would be nice if u did draw the curve on the test paper so the marker actually knows how you got your final solution.:rolleyes:
 

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