Overview
Absolute value represents distance from zero on the number line and is always non-negative. On the SAT, absolute value problems appear 1–2 times per test and require recognizing which of three scenarios applies: an equation (=), a less-than inequality (<), or a greater-than inequality (>). Each scenario produces a different algebraic structure — two separate equations, an AND compound inequality, or an OR compound inequality. Always isolate the absolute value expression before splitting into cases.
Key Points
1. The Fundamental Definition
Absolute value equals the distance a number is from zero, so it is never negative.
|a| = a if a ≥ 0
|a| = -a if a < 0
|a| ≥ 0 always
|a| = 0only whena = 0|a| = negative number→ No solution (impossible)
2. Solving Absolute Value Equations: |ax + b| = c
Step 1. Confirm c ≥ 0. If c < 0, stop — there is no solution. Step 2. Split into two cases:
Case 1: ax + b = c → solve for x₁
Case 2: ax + b = -c → solve for x₂
Step 3. Check both answers by substituting back into the original equation.
Example: |2x − 1| = 5
- Case 1: 2x − 1 = 5 → x = 3
- Case 2: 2x − 1 = −5 → x = −2
- Both check out:
|2(3)−1| = 5✓ and|2(−2)−1| = 5✓
3. Absolute Value Inequalities: Less Than
|ax + b| < c (where c > 0) → AND compound inequality (single connected interval)
-c < ax + b < c
Solve the compound inequality in one step by isolating the middle expression.
Example: |x − 3| < 4 → −4 < x − 3 < 4 → −1 < x < 7
Number line interpretation: all points between two values (within a distance of c from center).
4. Absolute Value Inequalities: Greater Than
|ax + b| > c (where c > 0) → OR compound inequality (two separate rays)
ax + b > c OR ax + b < -c
Solve each branch independently and write the solution as a union.
Example: |x − 3| > 4 → x − 3 > 4 OR x − 3 < −4 → x > 7 OR x < −1
Number line interpretation: all points outside a distance of c from center.
5. Summary Table
| Form | Solution Type | Algebraic Rule |
|---|---|---|
|ax+b| = c (c > 0) | Two solutions | ax+b = c and ax+b = −c |
|ax+b| = 0 | One solution | ax+b = 0 |
|ax+b| = c (c < 0) | No solution | Impossible |
|ax+b| < c (c > 0) | AND interval | −c < ax+b < c |
|ax+b| > c (c > 0) | OR two rays | ax+b > c OR ax+b < −c |
6. Graphical / Number Line Interpretation
|x − a| = b means “the distance from x to a on the number line equals b.”
This gives two points: x = a + b and x = a − b.
On the SAT, Desmos can confirm: graph y = |ax + b| and y = c and read the x-coordinates of the intersection points.
Pitfalls and Common Mistakes
Pitfall 1: Trying to solve when no solution exists
Seeing |2x + 3| = −5 and proceeding to split into cases produces garbage answers. Since absolute value is always ≥ 0, it can never equal a negative number.
Fix: Before splitting, check whether the right side is negative. If so, write “no solution” immediately.
Pitfall 2: Solving only one case Finding x = 3 from Case 1 and forgetting to solve Case 2 (ax + b = −c). This costs a point on multiple-choice and a full credit on grid-in. Fix: Always write out both Case 1 and Case 2 explicitly before solving either one.
Pitfall 3: Not isolating the absolute value first
Attempting to split 2|x + 1| − 4 = 6 without first simplifying produces wrong cases. The absolute value must stand alone on one side before applying the two-case rule.
Fix: Isolate to get |x + 1| = 5, then split.
Pitfall 4: Mixing up the AND / OR directions
Using an OR structure for |x| < c or an AND structure for |x| > c produces opposite solution sets. The mnemonic: “Less than = within a distance” (AND / between); “Greater than = outside a distance” (OR / two pieces).
Fix: Sketch the number line first — shade “between” for < and “outside” for >.
Pitfall 5: Not checking for extraneous solutions Splitting into two cases can occasionally produce a value that does not satisfy the original equation. This is especially likely when the absolute value expression involves a variable in a denominator or after squaring. Fix: Always substitute each solution back into the original absolute value equation before marking it as final.
Related Entries
- Linear_Equations
- Linear_Inequalities
- Linear_Inequalities
- Quadratic_Equations
- Systems_Linear_Equations
Quick Reference Card
| Rule | Formula / Key Fact |
|---|---|
| Definition | |a| = distance from 0; always ≥ 0 |
| Equation (c > 0) | |expr| = c → expr = c or expr = −c |
| Equation (c < 0) | |expr| = c → No solution |
| Less-than | |expr| < c → −c < expr < c (AND) |
| Greater-than | |expr| > c → expr > c or expr < −c (OR) |
| Isolate first | Must get |...| alone before splitting |
| Check solutions | Sub back into original to catch extraneous answers |
| SAT frequency | 1–2 questions per test |
| Desmos shortcut | Graph y = |ax+b| and y = c, read intersections |