Absolute Value Inequalities
Solve absolute value inequalities by converting to compound inequalities ("less than" becomes "and," "greater than" becomes "or"). Graph and interpret solution sets on a number line.
A lesson of Algebra 1, in Linear Inequalities & Absolute Value, about 35 minutes.
By the end of this lesson, you will be able to turn an absolute value inequality into a compound inequality, "and" for the less-than type and "or" for the greater-than type, solve it, and write the answer in interval notation.
Within, or Beyond
An absolute value equation asks which numbers are exactly so far from zero. An inequality asks a more useful question.
$$|x| < 5 \quad \text{which numbers are WITHIN 5 of zero?}$$
Everything between $-5$ and $5$. One unbroken stretch of the number line.
$$-5 < x < 5$$
$$|x| > 5 \quad \text{which numbers are FURTHER than 5 from zero?}$$
Everything out past $-5$ on the left, and everything out past $5$ on the right. Two separate pieces with a gap in the middle.
$$x < -5 \quad \text{or} \quad x > 5$$
That is the whole lesson, and it is worth committing to memory in this form:
Less than gives one piece, joined by "and". Greater than gives two pieces, joined by "or".
The Two Rules
For any expression $A$ and any positive $c$:
Less-than type
$$|A| < c \quad \text{becomes} \quad -c < A < c$$
Greater-than type
$$|A| > c \quad \text{becomes} \quad A < -c \quad \text{or} \quad A > c$$
The same holds with $\leq$ and $\geq$, keeping whichever sign you started with.
Notice that both rules produce compound inequalities, which you already know how to solve. Converting is the new skill. Everything after the conversion is last lesson's work.
Isolate the bars first. Neither rule can be applied while something is still attached to the outside of the absolute value.
Example 1: Less Than, So One Piece
Here is the whole method on one inequality, worked all the way through. Solve $|3x - 6| \leq 9$. Whatever $3x - 6$ is, it is within $9$ of zero. So it sits between $-9$ and $9$.
- Rewrite it as a three-part inequality -9 <= 3x - 6 <= 9
- Add 6 to all three parts -3 <= 3x <= 15
- Divide all three parts by 3 -1 <= x <= 5
- Write the solution in interval notation [-1, 5]
Worked Example 1: Less Than, Your Turn
Solve $|2x + 4| < 10$.