Solving an inequality

  • Thread starter SengNee
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  • #1
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1) [tex]|2x+1|<4x-2[/tex]

2) [tex]|2x-1|>x+2[/tex]

3) [tex]|\frac {x-2}{x+1}|<3[/tex]

4) [tex]|2x-1|>\frac {1}{x}[/tex]


(Show me as many methods as possible. Thanks)
 

Answers and Replies

  • #2
symbolipoint
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Each of those shows only a single absolute value in the left member. Ask yourself what happens when the expression inside the absolute value function is positive or zero; and what happens when the expression inside is less than zero. Continue with each condition to find a solution set for each exercise.
 
  • #3
arildno
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Take 1:

Split up in the two cases:
A) 2x+1<4x-2 AND 2x+1>0
B) -(2x+1)<4x-2 AND 2x+1<0

Take A:
The second inequality requires x>-1/2
The first requires 3<2x, that is x>3/2

Thus, to fulfill both of these inequalities, we must have x>3/2 as the solution to A.

Now, let us tackle B:
The second inequality requires x<-1/2

The first requires:
1<6x, implying x>1/6

But these two inequalities cannot be fulfilled simultaneously, i.e, there are no solutions to case B

Thus, the entire solution to 1) is x>3/2
 
  • #4
epenguin
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Since you ask as many methods as possible, sketch or drwa a graph of both functions. Eg the | | part for the first two has a V-shape. It will help give you a sense of what is happening and, also as a habit, pick out mistakes sometimes.
 
  • #5
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1) [tex]|2x+1|<4x-2[/tex]

[tex]2x+1<4x-2[/tex]
[tex]3<2x[/tex]
[tex]3/2<x[/tex] ...... i


[tex]2x+1>2-4x[/tex]
[tex]6x>1[/tex]
[tex]x>1/6[/tex] ...... ii

To fulfill both i and ii, therefore [tex]3/2<x[/tex].

Can I do like that?
 
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  • #6
symbolipoint
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You could always take the three different regions around the two critical points in case you feel confused.
 

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