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(Show me as many methods as possible. Thanks)

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- Jan 8th 2008, 01:19 AM #1

- Jan 8th 2008, 02:23 AM #2

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When you have inequalities involving moduluses you need to separate it into the greater than and less than 0 cases as follows

1) |2x+1|<4x-2

We want the intersection of and 2x+1<4x-2

2x+1<4x-2

3<2x

x>3/2

So the intersection is x > 3/2

Next we look at 2x+1 < 0

So we want the intersection of 2x+1<0 and -(2x+1)<4x-2

solving this gives x<-1/2 and x>1/6. This has no intersection so the final answer is simply x > 3/2.

If the intersection did exist, we would take the union of it and the first solution to get the final result.

- Jan 8th 2008, 08:58 PM #3

- Jan 8th 2008, 09:12 PM #4

- Jan 8th 2008, 10:18 PM #5

- Jan 9th 2008, 09:49 AM #6

- Mar 3rd 2008, 06:33 PM #7

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## linear inequalility solution

1) 2x +1< 4x -2

Collect like terms

2x-4x < -2-1

-2x < -3

Divide both sides by -2

X > 1½

2) 2x – 1 > x +2

2x- x > 2+1

x> 3

from clement

cokhale@yahoo.com

- Mar 6th 2008, 01:49 AM #8

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- Mar 6th 2008, 05:44 AM #9

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## inequality solution

[QUOTE=cokhale;112884]1) 2x +1< 4x -2

Collect like terms

2x-4x < -2-1

-2x < -3

Divide both sides by -2

X > 1½

2) 2x – 1 > x +2

2x- x > 2+1

x> 3

3) X -2\ x +1 < 3

Solution

-----------

Multiple both sides by (x +1)²

X -2\ x +1 (x +1)² < 3(x +1)² = (x -2)(x +1)< 3(x +1)(x +1) = (x -2)(x +1) < 3( x² +2x +1) = (x -2)(x +1)<3x² +6x +3 = x² -x-2< 3x² +6x +3

= 2x² +7x +5<0

2x² +7x +5 =0

2x² + 5x +2x +5

X(2x +5) +1(2x +5)

(2x +5)(x +1)<0

2x +5 =0

2x = -5

Divide both sides by 2

X = -5\2 = -2½

Or

X +1 =0

X= -1

_+____-_______+__

-2.5 0 1 2

Since,< is a negative solution.

-2½ < x < 2.