# Solving an inequality

1. Jan 8, 2008

### 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)

2. Jan 8, 2008

### symbolipoint

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. Jan 8, 2008

### arildno

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. Jan 8, 2008

### epenguin

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. Jan 8, 2008

### SengNee

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

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

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

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

Can I do like that?

Last edited: Jan 9, 2008
6. Jan 8, 2008

### symbolipoint

You could always take the three different regions around the two critical points in case you feel confused.