# Math Help - Help me...

1. ## Help me...

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

2) $|2x-1|>x+2$

3) $|\frac {x-2}{x+1}|<3$

4) $|2x-1|>\frac {1}{x}$

(Show me as many methods as possible. Thanks)

2. 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
first do the case $2x+1 \geq 0$

We want the intersection of $2x+1 \geq 0$ and 2x+1<4x-2
$2x \geq -1$
$x \geq -1/2$

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.

3. 1) $|2x+1|<4x-2$

$2x+1<4x-2$
$3<2x$
$3/2 ...... i

$2x+1>2-4x$
$6x>1$
$x>1/6$ ...... ii

To fulfill both i and ii, therefore $3/2.

Can I do like that?

4. For
$|2x+1|<4x-2$

It would be
$2x+1<4x-2$

and
$2x+1>2-4x$

Note the change from < to > in the second expression.

-Dan

5. 2) $|2x-1|>x+2$

$2x-1>x+2$
$x>3$ ... i

$2x-1<-x-2$
$3x<-1$
$x<-1/3$ ... ii

6. Originally Posted by SengNee
2) $|2x-1|>x+2$

$2x-1>x+2$
$x>3$ ... i

$2x-1<-x-2$
$3x<-1$
$x<-1/3$ ... ii

Looks good to me. Can you plug in some numbers from your solution set to see if it looks good to you?

-Dan

7. ## linear inequalility solution

Originally Posted by SengNee
1) $|2x+1|<4x-2$

2) $|2x-1|>x+2$

3) $|\frac {x-2}{x+1}|<3$

4) $|2x-1|>\frac {1}{x}$

(Show me as many methods as possible. Thanks)
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

8. 1.

set the postive mod to equal the straightline graph to get point of intersection

cant go wrong

9. ## 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.