| Primary source | OpenStax College Algebra 2e, Section 3.6: "Absolute Value Functions" |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/3-6-absolute-value-functions |
| Supplementary | OpenStax College Algebra 2e, Section 2.7: "Linear Inequalities and Absolute Value Inequalities" |
| Supplementary link | https://openstax.org/books/college-algebra-2e/pages/2-7-linear-inequalities-and-absolute-value-inequalities |
Absolute value is distance from zero on the number line.
$|x - a|$ is the distance from $x$ to $a$.
This geometric interpretation makes every absolute value rule obvious rather than arbitrary.
If you learn only this -- that absolute value is distance -- you can derive every rule in this lesson from scratch.
Definition. \[ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \]
Key properties.
| Property | Statement |
|---|---|
| Non-negativity | $|x| \geq 0$; equals 0 only when $x = 0$ |
| Symmetry | $|-x| = |x|$ |
| Product | $|xy| = |x| \cdot |y|$ |
| Triangle inequality | $|x + y| \leq |x| + |y|$ |
Equations and inequalities.
| Form | Solution method | Solution set |
|---|---|---|
| $|x| = a$ ($a > 0$) | $x = a$ or $x = -a$ | Two values |
| $|x| = 0$ | $x = 0$ | One value |
| $|x| = a$ ($a < 0$) | No solution | $\emptyset$ |
| $|x| < a$ ($a > 0$) | $-a < x < a$ | Interval $(−a, a)$ |
| $|x| > a$ ($a \geq 0$) | $x < -a$ or $x > a$ | Union of two rays |
$|x|$ is the distance from $x$ to $0$ on the number line. Distance is always non-negative, so $|x| \geq 0$ for all real $x$.
The two-case definition: \[ |x| = \begin{cases} x & \text{if } x \geq 0 \text{ (already non-negative, no change)} \\ -x & \text{if } x < 0 \text{ (negating a negative gives a positive)} \end{cases} \]
Note: $-x$ does not mean "negative number." If $x = -5$, then $-x = -(-5) = 5$, which is positive.
Example 1. Evaluate: $|{-7}|$, $|3.2|$, $|0|$, $|{-\pi}|$.
$|-7| = 7$, $|3.2| = 3.2$, $|0| = 0$, $|-\pi| = \pi$.
$|x - a|$ is the distance between $x$ and $a$ on the number line.
Example 2. What is the distance between $x = 7$ and $a = 3$?
$|7 - 3| = |4| = 4$. Or: $|3 - 7| = |-4| = 4$. Distance is symmetric; order does not matter.
Example 3. For which values of $x$ is $|x - 3| = 5$?
The distance from $x$ to $3$ is 5. The two points at distance 5 from 3 are $3 + 5 = 8$ and $3 - 5 = -2$.
$x = 8$ or $x = -2$.
Case split method. If $|E| = a$ (with $a > 0$), split into: \[ E = a \qquad \text{or} \qquad E = -a \]
Solve each linear equation separately. Check all solutions.
Example 4. Solve $|2x - 5| = 9$.
Case 1: $2x - 5 = 9 \implies 2x = 14 \implies x = 7$.
Case 2: $2x - 5 = -9 \implies 2x = -4 \implies x = -2$.
Check: $|2(7) - 5| = |9| = 9$. $|2(-2) - 5| = |-9| = 9$. Both correct.
Example 5. Solve $|3x + 1| = -4$.
The absolute value is always $\geq 0$. It can never equal $-4$. No solution.
Example 6. Solve $|x - 4| = |x + 2|$.
This says: the distance from $x$ to $4$ equals the distance from $x$ to $-2$. The point equidistant from $4$ and $-2$ is the midpoint: $\dfrac{4 + (-2)}{2} = 1$. So $x = 1$.
Algebraically: square both sides (both sides are non-negative): $(x-4)^2 = (x+2)^2$
$x^2 - 8x + 16 = x^2 + 4x + 4$
$-12x = -12 \implies x = 1$. $\checkmark$
Always check solutions in absolute value equations. Squaring both sides or combining cases can introduce extraneous solutions.
The "less than" case ($|E| < a$, distance constraint).
$|E| < a$ means $E$ is within $a$ units of $0$: $-a < E < a$.
Example 7. Solve $|x - 2| < 5$.
Distance from $x$ to $2$ is less than 5. Equivalently: $-5 < x - 2 < 5$.
Add 2 throughout: $-3 < x < 7$.
Solution: $(-3, 7)$.
The "greater than" case ($|E| > a$, distance constraint).
$|E| > a$ means $E$ is more than $a$ units from $0$: $E < -a$ or $E > a$.
Example 8. Solve $|2x + 3| \geq 7$.
Split: $2x + 3 \leq -7$ or $2x + 3 \geq 7$.
Left: $2x \leq -10 \implies x \leq -5$.
Right: $2x \geq 4 \implies x \geq 2$.
Solution: $(-\infty, -5] \cup [2, \infty)$.
Example 9. Solve $|x - 1| < -2$.
The absolute value is $\geq 0$, so it is never $< -2$. No solution.
Example 10. Solve $|4x - 3| + 2 < 9$.
Isolate the absolute value first: $|4x - 3| < 7$.
Split: $-7 < 4x - 3 < 7$.
Add 3: $-4 < 4x < 10$.
Divide by 4: $-1 < x < \dfrac{5}{2}$.
Solution: $\left(-1, \dfrac{5}{2}\right)$.
Non-negativity: $|x| \geq 0$ with equality only at $x = 0$.
Product rule: $|xy| = |x| \cdot |y|$.
Why: A positive times a positive is positive; a negative times a negative is positive. In both cases, the product's absolute value equals the product of the absolute values.
Triangle inequality: $|x + y| \leq |x| + |y|$.
Why (geometric): The distance from $0$ to $x + y$ is at most the distance from $0$ to $x$ plus the distance from $0$ to $y$. Think of $x$ and $y$ as steps: walking right $|x|$ then right $|y|$ gets you to $|x| + |y|$. But if you walk right and then left, you end up closer than $|x| + |y|$. The farthest you can get from $0$ in two steps is when both steps are in the same direction.
The triangle inequality appears directly in the formal definition of limit in calculus.
$f(x) = |x|$ is the parent function from the graphing lesson. Key features:
Transformations apply as in the transformations lesson:
| Error | Example | Correction |
|---|---|---|
| Treating $-x$ as negative | $|-3| = -(-3)= $ "negative" | $-(-3) = 3 > 0$; absolute value is always non-negative |
| Forgetting the second case in equations | $|2x-5| = 9 \to 2x-5=9$ only | Also solve $2x-5=-9$ |
| Wrong direction for "$>$" inequality | $|x| > 3 \to -3 < x < 3$ | Correct for "$<$"; for "$>$": $x < -3$ or $x > 3$ |
| Not isolating before splitting | $|x-1| + 2 < 9 \to (x-1+2<9)$ and ... | Isolate: $|x-1| < 7$ first |
| Missing no-solution check | Solving $|x| = -5$ and writing $x = 5$ | $|x| \geq 0$ always; $|x| = -5$ has no solution |
Problem 1. Solve: $|x + 7| = 12$.
$x + 7 = 12 \implies x = 5$.
$x + 7 = -12 \implies x = -19$.
Check: $|5+7| = 12$. $|-19+7| = |-12| = 12$. $\checkmark$
Solution: $x = 5$ or $x = -19$.
Problem 2. Solve: $|3x - 6| < 9$.
$-9 < 3x - 6 < 9$.
Add 6: $-3 < 3x < 15$.
Divide by 3: $-1 < x < 5$.
Solution: $(-1, 5)$.
Problem 3. Solve: $|2x + 1| \geq 5$.
$2x + 1 \leq -5$ or $2x + 1 \geq 5$.
$x \leq -3$ or $x \geq 2$.
Solution: $(-\infty, -3] \cup [2, \infty)$.
Problem 4. Solve: $|5 - 2x| = |x + 1|$.
Case 1: $5 - 2x = x + 1 \implies 4 = 3x \implies x = \frac{4}{3}$.
Case 2: $5 - 2x = -(x + 1) \implies 5 - 2x = -x - 1 \implies 6 = x \implies x = 6$.
Check $x = \frac{4}{3}$: $|5 - \frac{8}{3}| = |\frac{7}{3}|$ and $|\frac{4}{3} + 1| = |\frac{7}{3}|$. $\checkmark$
Check $x = 6$: $|5 - 12| = 7$ and $|6 + 1| = 7$. $\checkmark$
Problem 5. Solve: $2|x - 3| + 5 = 11$.
Isolate: $2|x-3| = 6 \implies |x-3| = 3$.
$x - 3 = 3 \implies x = 6$, or $x - 3 = -3 \implies x = 0$.
Problem 6. Find all $x$ satisfying $|x^2 - 4| = 5$.
Case 1: $x^2 - 4 = 5 \implies x^2 = 9 \implies x = \pm 3$.
Case 2: $x^2 - 4 = -5 \implies x^2 = -1$. No real solution.
Check $x = 3$: $|9-4| = 5$. $\checkmark$ Check $x = -3$: $|9-4| = 5$. $\checkmark$
Solution: $x = 3$ or $x = -3$.
Problem 7. Prove the triangle inequality for two terms: $|x + y| \leq |x| + |y|$.
We use the fact that $-|a| \leq a \leq |a|$ for any real $a$ (since $|a| = a$ or $|a| = -a$).
So: $-|x| \leq x \leq |x|$ and $-|y| \leq y \leq |y|$.
Add: $-(|x| + |y|) \leq x + y \leq |x| + |y|$.
By definition of absolute value, $|x + y| \leq |x| + |y|$. $\square$
Problem 8. The condition $|x - 3| < 0.01$ is an example of the form used in epsilon-delta limits. Describe in plain words what values of $x$ satisfy this, and write the equivalent interval.
"The distance from $x$ to $3$ is less than $0.01$." In other words, $x$ is within $0.01$ of $3$.
Expanding: $-0.01 < x - 3 < 0.01$, so $2.99 < x < 3.01$.
Interval: $(2.99, 3.01)$.
In epsilon-delta notation, $0.01$ plays the role of $\delta$, and the condition $|x - 3| < \delta$ describes inputs $x$ in an interval of radius $\delta$ around the point $a = 3$.
Problem 9. Show that $|x - y| \geq |x| - |y|$ for all real $x, y$.
Apply the triangle inequality to $|x| = |(x - y) + y|$:
$|x| = |(x-y) + y| \leq |x - y| + |y|$.
Rearrange: $|x - y| \geq |x| - |y|$. $\square$
This is the reverse triangle inequality. It says the length of one side of a triangle is at least the difference of the other two sides.
Absolute value is a ruler that always shows non-negative distance.
On the number line, every point $x$ is some distance from $0$. That distance is $|x|$. The ruler cannot show negative distance; it always reads $\geq 0$.
$|x - a|$ is the same ruler, but now measuring distance from $a$ instead of from $0$. Think of $a$ as the origin of a shifted ruler.
Inequalities become radius questions:
This radius picture is exactly what appears in the epsilon-delta definition of a limit: $|x - a| < \delta$ is a circle of radius $\delta$ around $a$ on the number line. Once you see absolute value as distance, limits feel natural instead of foreign.
For students who find math intimidating: One image to hold: a ruler that always shows non-negative distance. $|x - a|$ is the reading on the ruler when one end is at $a$ and the other is at $x$. Every rule in this lesson follows from that image.
For students interested in proof: The triangle inequality is the algebraic version of the geometric fact that the shortest path between two points is a straight line. It holds not just for real numbers but for any metric space -- a set with a distance function satisfying non-negativity, symmetry, and the triangle inequality. Understanding this lesson means you are ready to meet metric spaces.
For students interested in careers: Absolute value inequalities describe tolerance bands in engineering (a component is acceptable if its diameter is within $0.01$ mm of the target). Signal processing uses absolute values to measure amplitude and error. Statistical loss functions often use $|y - \hat{y}|$ (the $L^1$ loss) as an alternative to squared error that is less sensitive to outliers.