# How to solve absolute value equations and inequalities Canonical: https://duckyhelper.com/learn/algebra-1/absolute-value/ Updated: 2026-10-01 The absolute value \(|x|\) is a number's distance from zero, so it is never negative. To solve an absolute value equation, first get the absolute value alone, then split it into two equations: the inside equals the number, or the inside equals the opposite of the number. If an absolute value is set equal to a negative number, there is no solution. Check every answer in the original equation. ## The key idea Both 5 and \(-5\) are 5 steps from zero, so \(|5| = |-5| = 5\). That is why an absolute value equation usually has two answers: whatever is inside the bars can be the positive number or its opposite. $$ |u| = c \;\Longrightarrow\; u = c \;\text{ or }\; u = -c \qquad (c \ge 0) $$ **The four patterns (u is whatever is inside the bars, c is a positive number)** | Pattern | Rewrite as | What the answer looks like | | --- | --- | --- | | \(\|u\| = c\) | \(u = c\) or \(u = -c\) | two numbers | | \(\|u\| < c\) | \(-c < u < c\) | one segment (an "and" inequality) | | \(\|u\| > c\) | \(u < -c\) or \(u > c\) | two rays pointing apart (an "or" inequality) | | \(\|u\| = \) a negative number | nothing to rewrite | no solution, since distance is never negative | ## Worked examples **Example 1: the basic split** Problem: Solve \(|x - 4| = 9\). 1. The absolute value is already alone. Split into two cases: the inside is 9, or the inside is \(-9\). $$ x - 4 = 9 \quad\text{or}\quad x - 4 = -9 $$ 2. Add 4 in each case. $$ x = 13 \quad\text{or}\quad x = -5 $$ 3. Check: \(|13 - 4| = |9| = 9\) and \(|-5 - 4| = |-9| = 9\). Both work. On a number line, 13 and \(-5\) are both 9 steps from 4. Answer: \(x = 13\) or \(x = -5\) **Example 2: isolate first** Problem: Solve \(3|2x + 1| - 4 = 11\). 1. Do not split yet. Add 4 to both sides. $$ 3|2x + 1| = 15 $$ 2. Divide by 3. Now the absolute value is alone. $$ |2x + 1| = 5 $$ 3. Split into two cases. $$ 2x + 1 = 5 \quad\text{or}\quad 2x + 1 = -5 $$ 4. Solve each one. $$ x = 2 \quad\text{or}\quad x = -3 $$ Answer: \(x = 2\) or \(x = -3\) **Example 3: an absolute value inequality** Problem: Solve \(|2x - 3| \le 7\). 1. Less than means the inside is trapped between \(-7\) and 7. $$ -7 \le 2x - 3 \le 7 $$ 2. Add 3 to all three parts. $$ -4 \le 2x \le 10 $$ 3. Divide all three parts by 2. $$ -2 \le x \le 5 $$ 4. Graph: closed circles at \(-2\) and 5 with the segment between them shaded. Answer: \(-2 \le x \le 5\) **Example 4 (test-hard): a variable outside the bars** Problem: Solve \(|x - 3| = 2x\). 1. Split into two cases as usual. $$ x - 3 = 2x \quad\text{or}\quad x - 3 = -2x $$ 2. Solve each case. $$ x = -3 \quad\text{or}\quad x = 1 $$ 3. Check \(x = -3\): the left side is \(|-6| = 6\), but the right side is \(2(-3) = -6\). They do not match, so \(-3\) is extraneous. Throw it out. 4. Check \(x = 1\): the left side is \(|-2| = 2\) and the right side is \(2(1) = 2\). It works. Answer: \(x = 1\) only. ## Common mistakes - **Splitting before you isolate.** In \(3|2x + 1| - 4 = 11\), the 3 and the \(-4\) are outside the bars. Fix: undo them first, so the bars stand alone, then split. - **Solving \(|x + 2| = -4\) anyway.** No distance is negative, so there is no solution. Fix: after isolating, glance at the sign of the number on the other side. - **Using "and" for a greater-than inequality.** \(|u| > c\) means \(u\) is far from zero in either direction, so it is \(u < -c\) **or** \(u > c\). - **Skipping the check when x is outside the bars.** If the right side contains x, it can turn negative and create an extraneous answer, like \(-3\) in Example 4. - **Thinking \(|u|\) just means "drop the minus sign" of a variable.** \(|x - 4|\) is not \(x + 4\). Fix: treat the whole inside as one quantity. ## Quick methods > **Tip: A memory aid for inequalities** > > "Less thAND, greatOR." Less than gives one "and" segment. Greater than gives two "or" rays. It is just a mnemonic, so if you forget it, test a number: for \(|x| < 3\), the number 0 works and 10 does not, so the answer is the middle segment. > **Note: Check with a graph** > > In a graphing calculator such as Desmos, graph \(y = |x - 3|\) and \(y = 2x\). They cross only once, at \(x = 1\), which confirms that \(-3\) in Example 4 is extraneous. ## Practice **5 practice questions** 1. Solve \(|x + 6| = 2\). A. \(x = -4\) or \(x = -8\) B. \(x = 4\) or \(x = -8\) C. \(x = -4\) only D. \(x = 4\) or \(x = 8\) Answer: \(x = -4\) or \(x = -8\). Split: \(x + 6 = 2\) gives \(x = -4\), and \(x + 6 = -2\) gives \(x = -8\). Both are 2 steps from \(-6\). Stopping at \(-4\) misses the second case. 2. Solve \(2|x - 1| + 3 = 1\). A. \(x = 0\) or \(x = 2\) B. \(x = 0\) C. \(x = 2\) D. No solution Answer: No solution. Subtract 3: \(2|x - 1| = -2\). Divide by 2: \(|x - 1| = -1\). An absolute value can never be negative, so no x works. The answer \(x = 0\) or \(x = 2\) comes from ignoring the negative sign. 3. Solve \(|x - 5| < 3\). A. \(2 < x < 8\) B. \(x < 2\) or \(x > 8\) C. \(-8 < x < -2\) D. \(x < 8\) Answer: \(2 < x < 8\). Less than gives an "and" inequality: \(-3 < x - 5 < 3\). Add 5 to all three parts: \(2 < x < 8\). These are the numbers less than 3 steps from 5. The "or" choice is the answer to \(|x - 5| > 3\). 4. Solve \(|3x + 3| \ge 12\). A. \(-5 \le x \le 3\) B. \(x \le -5\) or \(x \ge 3\) C. \(x \le -3\) or \(x \ge 3\) D. \(x \ge 3\) Answer: \(x \le -5\) or \(x \ge 3\). Greater than gives "or": \(3x + 3 \ge 12\) gives \(x \ge 3\), and \(3x + 3 \le -12\) gives \(3x \le -15\), so \(x \le -5\). The choice with \(-3\) just copies the 3 with a minus sign instead of solving the second case. 5. Which equation has no solution? A. \(|x| = 0\) B. \(|x - 2| = 3\) C. \(|x| + 4 = 1\) D. \(-|x| = -5\) Answer: \(|x| + 4 = 1\). Subtracting 4 gives \(|x| = -3\), which is impossible. \(|x| = 0\) has one solution (0), \(|x - 2| = 3\) has two (5 and \(-1\)), and \(-|x| = -5\) is the same as \(|x| = 5\), with two solutions. ## Frequently asked questions ### Why do absolute value equations have two answers? Absolute value measures distance from zero, and there are two numbers at any positive distance: one on each side. So \(|u| = 7\) means \(u\) is 7 or \(-7\). If the distance is 0, there is only one answer, and if it is negative, there is none. ### Can an absolute value equation have no solution? Yes. Once the absolute value is alone, if it equals a negative number, like \(|x - 1| = -1\), there is no solution, because a distance cannot be negative. Isolate first, though: \(|x| - 4 = -1\) looks negative but becomes \(|x| = 3\), which has two answers. ### What is an extraneous solution? It is an answer your algebra produces that does not actually work in the original equation. With absolute value, this happens when x also appears outside the bars, as in \(|x - 3| = 2x\). Plug each answer back in and drop any that fail. ### How do I know whether an absolute value inequality is "and" or "or"? Look at the direction once the absolute value is alone on the left. Less than (\(<\) or \(\le\)) means close to the center, so it becomes one "and" segment. Greater than (\(>\) or \(\ge\)) means far away, so it becomes two "or" pieces. ## Related - [How to solve and graph inequalities](https://duckyhelper.com/learn/algebra-1/inequalities/) - [How to solve multi-step equations](https://duckyhelper.com/learn/algebra-1/solving-equations/) - [Nonlinear equations and systems on the SAT](https://duckyhelper.com/learn/sat-math/nonlinear-equations-and-systems/) - [Algebra 1 study guides](https://duckyhelper.com/learn/algebra-1/) ## Try asking Ducky - "Why do I have to split this into two equations?" - "I got x = -3 and x = 1 but the book says only 1. What happened to -3?" - "Is |2x - 5| > 3 an and or an or problem? Walk me through it." ## Get DuckyHelper Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen. 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