Absolute Value Equations and Inequalities

The absolute value of a number is its distance from zero on the number line — always positive or zero, never negative. The notation $|x|$ means “the absolute value of $x$.” For example, $|7| = 7$ and $|-7| = 7$ because both 7 and $-7$ are exactly 7 units from zero.

This concept of distance is the key to understanding every absolute value equation and inequality. When you see $|x| = 5$, you are asking: “What numbers are exactly 5 units from zero?” The answer is $x = 5$ or $x = -5$.

Solving Absolute Value Equations

The Core Principle

For any expression $E$ and positive number $c$:

$|E| = c \quad \implies \quad E = c \quad \text{or} \quad E = -c$

The expression inside the absolute value can equal the positive version or the negative version. This always produces two equations to solve.

Example 1: Solve $|2x - 3| = 7$

Set up two equations:

$2x - 3 = 7 \quad \text{or} \quad 2x - 3 = -7$

Solve the first: $2x = 10 \implies x = 5$

Solve the second: $2x = -4 \implies x = -2$

Answer: $x = 5$ or $x = -2$

Check both:

• $|2(5) - 3| = |10 - 3| = |7| = 7$- $|2(-2) - 3| = |-4 - 3| = |-7| = 7$

Example 2: Solve $|4x + 1| - 5 = 8$

First, isolate the absolute value:

$|4x + 1| = 13$

Set up two equations:

$4x + 1 = 13 \quad \text{or} \quad 4x + 1 = -13$

Solve the first: $4x = 12 \implies x = 3$

Solve the second: $4x = -14 \implies x = -3.5$

Answer: $x = 3$ or $x = -3.5$

Special Cases for Equations

No solution: If the absolute value equals a negative number, there is no solution. An absolute value can never be negative.

$|3x + 2| = -4 \quad \implies \quad \text{No solution}$

One solution: If the absolute value equals zero, there is exactly one solution.

$|3x + 2| = 0 \quad \implies \quad 3x + 2 = 0 \quad \implies \quad x = -\frac{2}{3}$

Absolute Value Inequalities: Less Than (Distance “Close To”)

When the absolute value is less than a number, you are asking: “What values are closer than $c$ units from zero?”

The Rule

$|E| < c \quad \implies \quad -c < E < c$

This produces an “and” compound inequality. The expression is trapped between $-c$ and $c$.

The same pattern works for $\leq$:

$|E| \leq c \quad \implies \quad -c \leq E \leq c$

Example 3: Solve $|x - 4| < 3$

Apply the rule:

$-3 < x - 4 < 3$

Add 4 to all three parts:

$1 < x < 7$

Answer: $x$ is between 1 and 7 (not including the endpoints).

Interval notation: $(1, 7)$

Interpretation: The values of $x$ that are less than 3 units away from 4 on the number line.

Example 4: Solve $|2x + 5| \leq 9$

Apply the rule:

$-9 \leq 2x + 5 \leq 9$

Subtract 5 from all parts:

$-14 \leq 2x \leq 4$

Divide all parts by 2:

$-7 \leq x \leq 2$

Answer: $[-7, 2]$

Absolute Value Inequalities: Greater Than (Distance “Far From”)

When the absolute value is greater than a number, you are asking: “What values are farther than $c$ units from zero?”

The Rule

$|E| > c \quad \implies \quad E < -c \quad \text{or} \quad E > c$

This produces an “or” compound inequality. The expression is either very negative or very positive — outside the interval $(-c, c)$.

The same pattern works for $\geq$:

$|E| \geq c \quad \implies \quad E \leq -c \quad \text{or} \quad E \geq c$

Example 5: Solve $|3x - 6| > 12$

Apply the rule:

$3x - 6 < -12 \quad \text{or} \quad 3x - 6 > 12$

Solve the first: $3x < -6 \implies x < -2$

Solve the second: $3x > 18 \implies x > 6$

Answer: $x < -2$ or $x > 6$

Interval notation: $(-\infty, -2) \cup (6, \infty)$

Example 6: Solve $|x + 1| \geq 5$

Apply the rule:

$x + 1 \leq -5 \quad \text{or} \quad x + 1 \geq 5$

Solve the first: $x \leq -6$

Solve the second: $x \geq 4$

Answer: $(-\infty, -6] \cup [4, \infty)$

Multi-Step Absolute Value Inequalities

Just like with equations, you must isolate the absolute value first before applying any rule.

Example 7: Solve $3|x - 2| + 4 \leq 19$

Step 1 — Subtract 4 from both sides:

$3|x - 2| \leq 15$

Step 2 — Divide both sides by 3:

$|x - 2| \leq 5$

Step 3 — Apply the less-than rule:

$-5 \leq x - 2 \leq 5$

Step 4 — Add 2 to all parts:

$-3 \leq x \leq 7$

Answer: $[-3, 7]$

Example 8: Solve $2|4x + 3| - 1 > 9$

Step 1 — Add 1: $2|4x + 3| > 10$

Step 2 — Divide by 2: $|4x + 3| > 5$

Step 3 — Apply the greater-than rule:

$4x + 3 < -5 \quad \text{or} \quad 4x + 3 > 5$

Solve the first: $4x < -8 \implies x < -2$

Solve the second: $4x > 2 \implies x > \frac{1}{2}$

Answer: $(-\infty, -2) \cup \left(\frac{1}{2}, \infty\right)$

Special Cases for Inequalities

Understanding special cases prevents errors on tests and in practice.

$|E| < 0$: No expression has an absolute value less than zero. No solution ($\emptyset$).

$|E| \leq 0$: The only way an absolute value is less than or equal to zero is if it equals zero. Solve $E = 0$ for a single solution.

$|E| > 0$: Every number except zero has an absolute value greater than zero. The solution is all real numbers except where $E = 0$.

$|E| \geq 0$: Always true. The solution is all real numbers.

$|E| > -3$ (or any negative number): Always true, since absolute values are always at least 0. The solution is all real numbers.

$|E| < -3$ (or any negative number): Never true. No solution.

Real-World Application: Electrician — Voltage Tolerance

An electrician is testing a residential circuit that should deliver 120 volts. The National Electrical Code allows a voltage tolerance of $\pm 5$%, meaning the actual voltage $V$ must be within 5% of the target.

Five percent of 120 is $0.05 \times 120 = 6$ volts. So the acceptable range is:

$|V - 120| \leq 6$

Apply the less-than-or-equal-to rule:

$-6 \leq V - 120 \leq 6$

Add 120 to all parts:

$114 \leq V \leq 126$

If the electrician measures a voltage of 112 volts, is it within tolerance?

$|112 - 120| = |-8| = 8$

Since $8 > 6$, the voltage is outside the acceptable range. The electrician needs to investigate — possible causes include a loose connection, undersized wiring, or excessive load on the circuit.

Nursing — Medication Concentration Tolerance

A pharmacist prepares an IV solution that should contain 250 mg of a drug per bag. Quality control requires that the actual amount $A$ (in mg) must be within 3% of the target:

$|A - 250| \leq 7.5$

(Since $0.03 \times 250 = 7.5$)

Apply the rule:

$-7.5 \leq A - 250 \leq 7.5$

$242.5 \leq A \leq 257.5$

A batch testing at 240 mg would be rejected: $|240 - 250| = 10 > 7.5$. The concentration is too low and could result in a subtherapeutic dose.

A Quick-Reference Decision Chart

When you see an absolute value problem, follow this decision process:

1. Isolate the absolute value expression on one side
2. Check the right side — if it is negative, apply the special case rules above
3. Identify the type:
   • Equals ($=$): Split into two equations
   • Less than ($<$ or $\leq$): Write as an “and” compound inequality
   • Greater than ($>$ or $\geq$): Write as an “or” compound inequality
4. Solve the resulting equation(s) or inequality
5. Check your answers in the original equation or inequality

Common Mistakes to Avoid

1. Forgetting to isolate the absolute value first. If the equation is $3|x + 2| - 5 = 10$, you must get $|x + 2| = 5$ before splitting. Do not split $3|x + 2| - 5$ into two parts.

2. Writing an “and” compound inequality for a greater-than problem. The inequality $|x| > 5$ does not mean $-5 > x > 5$ (which is impossible). It means $x < -5$ or $x > 5$.

3. Forgetting the negative case. The equation $|x - 3| = 7$ has two solutions: $x = 10$ and $x = -4$. Solving only $x - 3 = 7$ misses half the answer.

4. Claiming a solution exists when the absolute value equals a negative. The equation $|2x + 1| = -6$ has no solution. An absolute value is never negative — stop immediately and write “no solution.”

5. Not flipping the inequality sign when setting up the negative case for greater-than problems. For $|E| > c$, the two cases are $E > c$ or $E < -c$. The second case uses $<$, not $>$.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Absolute value represents distance from zero — it is always non-negative
• Equations ($|E| = c$): Split into $E = c$ or $E = -c$ to get two solutions
• Less-than inequalities ($|E| < c$): Convert to an “and” compound inequality $-c < E < c$ — the solution is an interval
• Greater-than inequalities ($|E| > c$): Convert to an “or” compound inequality $E < -c$ or $E > c$ — the solution is two rays
• Always isolate the absolute value before applying any rule
• Check for special cases: absolute value equal to a negative (no solution), absolute value greater than a negative (all real numbers)
• In trades, absolute value inequalities model tolerances — how far a measurement can deviate from a target and still be acceptable

Return to Algebra for more topics in this section.