Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
4. Solving Absolute Value Inequalities
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Exercise 2 Page 485

The solutions for |x| ≤ a are the numbers whose distance from 0 is less than or equal to a. The solutions for |x| ≥ a are the numbers whose distance from 0 is greater than or equal to a.

Solutions for |x|+2≤5: x ≥ - 3 and x ≤ 3
Solutions for |x|+2≥ 5: x ≥ 3 or x ≤ - 3

Practice makes perfect

We want to describe the solutions to the inequalities |x|+2 ≤ 5 and |x|+2 ≥ 5 without using absolute value. Let's do it one at a time.

Solutions of |x|+2 ≤ 5

We will start by isolating the absolute value expression.
|x|+2 ≤ 5
|x| ≤ 3
Recall now that the absolute value of a number x is its distance from 0. Therefore, the above means that the distance from 0 to x is less than or equal to 3. The numbers whose distance from 0 is less than or equal to 3 are those numbers that are greater than or equal to - 3, and less than or equal to 3.

Using the above graph, we can write two inequalities to represent the situation. x ≥ - 3 AND x ≤ 3 Note that the word and is used because x must be both greater than or equal to - 3 and less than or equal to 3.

Solutions of |x|+2 ≥ 5

Let's now solve the second inequality. Again, we will start by isolating the absolute value expression.
|x|+2 ≥ 5
|x| ≥ 3
One more time, recall that the absolute value of a number x is its distance from 0. Therefore, the above means that the distance from 0 to x is greater than or equal to 3. The numbers whose distance from 0 is greater than or equal to 3 are those numbers that are greater than or equal to 3, or less than or equal to - 3.

Using the above graph, we can write two inequalities to represent the situation. x ≥ 3 OR x ≤ - 3 Note that the word or is used because x must be either greater than or equal to 3 or less than or equal to - 3.