- #1

- 2

- 0

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)

- Thread starter SengNee
- Start date

- #1

- 2

- 0

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)

- #2

symbolipoint

Homework Helper

Education Advisor

Gold Member

- 6,054

- 1,128

- #3

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 132

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

Homework Helper

Gold Member

- 3,766

- 807

- #5

- 2

- 0

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?

[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?

Last edited:

- #6

symbolipoint

Homework Helper

Education Advisor

Gold Member

- 6,054

- 1,128

- Last Post

- Replies
- 8

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 12

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 5

- Views
- 2K

- Replies
- 8

- Views
- 8K

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 2K

- Replies
- 10

- Views
- 2K

- Last Post

- Replies
- 11

- Views
- 2K