Absolute Value Equations and Inequalities

Absolute Value Means Distance

The absolute value of a number is its distance from zero on the number line, and distance is never negative. So $|7| = 7$ and $|-7| = 7$; both are 7 units from zero. This distance idea is the key to every absolute value problem on the SAT: whenever you see $|ax + b|$, think "how far is $ax + b$ from zero?" Because two different quantities can sit the same distance from zero — one positive, one negative — absolute value equations almost always produce two cases.

Solving $|ax + b| = c$ as Two Cases

If $|ax + b| = c$ and $c \ge 0$, then the expression inside can equal either $c$ or $-c$: $ax + b = c \quad \text{or} \quad ax + b = -c$

Solve each separately to get up to two solutions. One important guardrail: if $c$ is negative, there is no solution, because absolute value can never be negative.

Worked example: solve $|2x - 3| = 11$.

Step 1 — Split into two cases: $2x - 3 = 11 \quad \text{or} \quad 2x - 3 = -11$

Step 2 — Solve the first case: $2x - 3 = 11 \Rightarrow 2x = 14 \Rightarrow x = 7$

Step 3 — Solve the second case: $2x - 3 = -11 \Rightarrow 2x = -8 \Rightarrow x = -4$

Step 4 — Check both. For $x = 7$: $|2(7) - 3| = |11| = 11$. For $x = -4$: $|2(-4) - 3| = |-11| = 11$. Both work, so the solutions are $x = 7$ and $x = -4$.

Isolate the Absolute Value First

A common SAT twist buries the absolute value inside other operations, like $3|x + 1| - 4 = 11$. You cannot split into cases until the absolute value stands alone. Add 4 to both sides to get $3|x + 1| = 15$, then divide by 3 to get $|x + 1| = 5$. Now the two-case rule applies: $x + 1 = 5$ or $x + 1 = -5$, giving $x = 4$ or $x = -6$. Splitting too early is a frequent, avoidable error.

Absolute Value Inequalities: "Less Than" Traps Between, "Greater Than" Spreads Apart

The distance interpretation makes inequalities intuitive. There are two patterns, and knowing which is which is the whole game.

Form	Meaning (distance)	Rewrite as	Solution shape
$\lvert ax + b \rvert < c$	Within $c$ of zero	$-c < ax + b < c$	A single interval between two values
$\lvert ax + b \rvert > c$	Farther than $c$ from zero	$ax + b < -c$ or $ax + b > c$	Two intervals spreading outward

A handy memory hook: "less thAND, greatOR." A less-than absolute value inequality becomes an AND (a compound "between" statement), while a greater-than becomes an OR (two separate pieces).

Worked example: solve $|x + 2| < 5$.

Step 1 — Rewrite as a compound inequality because it is a "less than": $-5 < x + 2 < 5$

Step 2 — Subtract 2 from all three parts: $-7 < x < 3$

So the solution is every $x$ strictly between $-7$ and $3$. On a number line, that is an open interval with open circles at $-7$ and $3$.

Worked example: solve $|x - 1| \ge 4$.

Because it is a "greater than or equal to," split into two outward pieces: $x - 1 \le -4 \quad \text{or} \quad x - 1 \ge 4$ $x \le -3 \quad \text{or} \quad x \ge 5$

The solution is two rays pointing away from each other, with closed circles at $-3$ and $5$ because the inequality is inclusive.

When a Variable Sits on Both Sides

The hardest SAT absolute value problems put a variable on the other side too, like $|x - 2| = 2x - 7$. Split into cases as usual, but you must check every answer in the original equation, because a case can produce an extraneous solution — a value that satisfies the split equation but not the original (typically because the right side turned out negative). Never report a solution to this kind of problem without verifying it.

Common Mistakes to Avoid

• Dropping the negative case — an equation like $|2x - 3| = 11$ has two answers, not one.
• Splitting into cases before the absolute value is isolated.
• Declaring "no solution" too early: only $|ax+b| = c$ with a negative $c$ has no solution.
• Mixing up the inequality patterns — "less than" gives one interval between two numbers ("and"), while "greater than" gives two intervals ("or").
• Skipping the check when a variable appears on both sides, and reporting an extraneous solution.