1. ## Absolute Value Inequality

Solve |2x - 1| + 5 > 2

My Work:

This is an absolute value inequality. Thus, we are not solving for x.

|2x - 1| + 5 > 2

|2x - 1| > 2 - 5

|2x - 1| > -3

At this point, I created two cases.

2x - 1 < 3

2x < 3 + 1

2x < 4

x < 4/2

x < 2

or

2x - 1 > -3

2x > -3 + 1

2x > -2

x > -2/2

x > -1

The solution to |2x - 1| + 5 > 2 is found here -1 < x < 2.

Correct?

2. ## Re: Absolute Value Inequality

|2x-1|>-3

First Case:
(2x-1) is +ve
2x-1>=0
x>=1/2 -----------(i)

since supposed term to be +ve
just remove modulus.

(2x-1)>-3
x>-1 ---------------(ii)

On solving (i) and (ii):
x belongs to [1/2,infinity]

Second Case:

(2x-1) is -ve

Removing modulus:

(2x-1)<-3
x<-1

x belongs to (-infinity,-1)

so overall value (-infinity,-1)U[1/2,infinity)

3. ## Re: Absolute Value Inequality

There's no work to do: $|2x-1| > -3$ for all $x$.

4. ## Re: Absolute Value Inequality

Originally Posted by mak29
|2x-1|>-3

First Case:
(2x-1) is +ve
2x-1>=0
x>=1/2 -----------(i)

since supposed term to be +ve
just remove modulus.

(2x-1)>-3
x>-1 ---------------(ii)

On solving (i) and (ii):
x belongs to [1/2,infinity]

Second Case:

(2x-1) is -ve

Removing modulus:

(2x-1)<-3
x<-1

x belongs to (-infinity,-1)

so overall value (-infinity,-1)U[1/2,infinity)
Sorry for the solution.
As stated by Archie , no need to do this.

5. ## Re: Absolute Value Inequality

Thank you everyone.