# Absolute Value Inequality

1. Jul 29, 2010

### Spectrum47

1. The problem statement, all variables and given/known data

Solve the inequality and sketch the graph of the solution on the real number line.

2. Relevant equations

|x - a|< or equal to b, b > 0

Let us imagine that the ">" and "<" signs also include "equal to" except for the condition, b > 0, in order to solve this question.

3. The attempt at a solution

My attempt in accordance with the solution that is within the textbook:

|x - a|< b
-b < x - a < b
a - b < x < a + b

Now this is confusing due to the fact that the condition, "b must be greater than 0", prohibits b from being negative in the second line of my attempt. Though this is the only way I know how to solve it. Can anyone explain as to why this is the case? Does the condition just deal with the real number line and placing these values in the positive direction and to the right of zero?

2. Jul 29, 2010

### QuarkCharmer

|x - a|< b
-b < x - a < b
a - b < x < a + b

b could still very well be positive could it not?

Last edited: Jul 29, 2010
3. Jul 29, 2010

### Spectrum47

Hello.

Do you mean to imply that:

|x - a| < b
(-)b < x - a < b
a - b < x < a + b

Therefore, there is a negative integer in front of b and it is actually positive? Is that so because we are looking at both cases (positive and negative) for an absolute value inequality but they must still be positive values?

4. Jul 29, 2010

### Staff: Mentor

No, there is NOT a negative integer in front of b. Let's look at a similar inequality with b replaced by a positive number, 2.

|x - a| <= 2
-2 <= x - a <= 2
a - 2 <= x <= a + 2

2 is positive, so it's opposite, -2, is negative.

In your problem b is positive, so -b is negative.

5. Jul 29, 2010

### HallsofIvy

Staff Emeritus
In fact, since any absolute value and, in particular, |x- a| can never be negative, the inequality |x- a|< b is only possible if b> 0- that is no restriction at all.
-b< x- a< b only makes sense if b> 0 since it implies -b< b which is not true if b< 0.

6. Jul 30, 2010

### Spectrum47

Alright, thank you all very much for your help. That clarified it for me.