Solving absolute value problems

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First of all, make sure you know how to solve basic inequalities. These types of problems are hard enough as is without being crippled by not knowing the basics. Knowing how to use the distributive law will keep your life easier too.

There are basically 2 types of absolute value problems:

| BLOB | < BLURB

and

| BLOB | > BLURB

We'll call the first one a "lesser-than" problem and the second one a "greater-than" problem. In both types of problems, there are 2 inequalities you have to solve, giving you 2 solution sets for x. The bottom line is the final solution set for x has to contain both solution sets for x. Wasn't that a mouthful to comprehend? Don't worry. Examples are on the way.

In the lesser-than problem, the 2 inequalities you have to solve are:

BLOB < BLURB (when BLOB > 0)
BLOB > -(BLURB) (when BLOB < 0)

In the greater-than problem, the 2 inequalities you have to solve are:

BLOB > BLURB (when BLOB > 0)
BLOB < -(BLURB) (when BLOB < 0)

Enough theory. Let's do some examples.

Example 1: Solve:

|x - 3| < 2x + 5

The 2 inequalities we have to solve are:

x - 3 < 2x + 5 (based on x - 3 > 0 or x > 3)
x - 3 > -(2x + 5) (based on x - 3 < 0 or x < 3)

When we solve the first inequality we get x > -8. However, since x > 3, the final solution for the first inequality is x > 3.
When we solve the second inequality we get x > -2/3. However, since x < 3, the final solution for the second inequality is -2/3 < x < 3. Because our final solution set has to contain both, our final answer is x > -2/3.

Example 2: Solve:

|x + 2| > -3x - 1

The 2 inequalities we have to solve are:

x + 2 > -3x - 1 (based on x + 2 > 0 or x > -2)
x + 2 < -(-3x - 1) (based on x + 2 < 0 or x < -2)

When we solve the first inequality we get x > -3/4. Since x > -2, the solution for the first inequality is x > -3/4.
When we solve the second inequality we get x > 1/2. However, this is based on the premise that x < -2. Since both conditions can't be true at the same time, there is no solution for the second inequality. As before, our final solution set has to contain both. So, our final answer is x > -3/4.

Sometimes, the solution set for x is sandwiched in between 2 numbers. For example, if one inequality solution is x > 1 and the other is x < 5, then the final solution set is 1 < x < 5.