What if the absolute value is less than a negative number?

Absolute value represents distance, which can never be negative. If you end up with something like $|x| < -3$, there is no solution. You cannot have a distance less than zero.

What if the absolute value is greater than a negative number?

Since absolute value is always zero or positive, it is always greater than any negative number. An inequality like $|x| > -5$ is true for all real numbers.

Do I flip the inequality sign when removing the absolute value bars?

You don't flip the sign on the original positive side, but you do reverse the symbol for the negative case in an 'OR' inequality. For example, $|x| > 4$ becomes $x > 4$ or $x < -4$.

How do you solve an absolute value inequality?

Q: What if the absolute value is greater than a negative number?

Since absolute value is always zero or positive, it is always greater than any negative number. An inequality like $|x| > -5$ is true for all real numbers.

To solve an absolute value inequality, you first isolate the absolute value expression, then rewrite it as a compound inequality without the bars. The way you rewrite it depends on whether the inequality is a 'less than' or a 'greater than' symbol. <br><br> Think of absolute value as distance from zero. If someone says 'stay within 5 miles of home,' your distance is less than 5, meaning you are trapped anywhere between 5 miles west and 5 miles east. If they say 'go more than 5 miles away,' your distance is greater than 5, meaning you are flying outward in opposite directions.

Solving 'Less Than' Inequalities (AND)

When you have an inequality like $|x| < a$ , it means the distance from zero is less than $a$ . To solve this, you trap the expression between the negative and positive values of $a$ . You write this as a compound 'AND' inequality: $-a < x < a$ . This tells you that $x$ must be greater than $-a$ AND less than $a$ at the same time. The same rule applies for $\leq$ .

Solving 'Greater Than' Inequalities (OR)

If you have $|x| > a$ , the distance from zero is strictly greater than $a$ . The expression is pushed outward, away from zero. You rewrite this as two separate inequalities joined by 'OR': $x < -a$ or $x > a$ . This means the solution is either far to the left or far to the right. The same rule applies for $\geq$ . Remember, you cannot write an 'OR' inequality as a single continuous statement like $-a > x > a$ .

Where Students Slip Up

A common mistake is forgetting to isolate the absolute value before splitting the inequality. If you have $2|x| + 3 < 9$ , you must subtract 3 and divide by 2 first to get $|x| < 3$ . Another slip is forgetting to flip the inequality symbol when multiplying or dividing by a negative number inside the algebra steps.

Worked through

Solve the inequality $|3x - 2| \leq 7$ .

First, recognize that the absolute value is already isolated and it is a 'less than or equal to' inequality. This means it is an 'AND' compound inequality. We trap the expression inside between -7 and 7: $-7 \leq 3x - 2 \leq 7$ . Next, we solve for $x$ in the middle. Add 2 to all three parts: $-5 \leq 3x \leq 9$ . Finally, divide all three parts by 3: $-5/3 \leq x \leq 3$ . The solution is all numbers between $-5/3$ and 3, inclusive.

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Where this comes from: OpenStax College Algebra, Chapter 2: Linear Inequalities and Absolute Value Inequalities · Khan Academy, Algebra 1: Absolute Value Equations and Inequalities