Absolute value questions look simple, but they hide a trap: one equation can secretly be two, and some "solutions" turn out to be fake. Whether you are in Matric, FSc, O/A Level or a university pre-calculus course, a clear method will help you get these right every time. Let's go through it step by step.
What Is Absolute Value?
The absolute value of a number is its distance from zero on the number line. Distance is never negative, so
$$ |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases} $$
For example, \(|7| = 7\), \(|-7| = 7\) and \(|0| = 0\). More generally, \(|a - b|\) is the distance between \(a\) and \(b\) on the number line.
Useful Properties
- \(|x| \geq 0\) for every real \(x\).
- \(|ab| = |a|\,|b|\) and \(\left|\frac{a}{b}\right| = \frac{|a|}{|b|}\) for \(b \neq 0\).
- \(|x|^2 = x^2\).
- \(|x| = |-x|\).
Solving Absolute Value Equations
The Basic Rule
If \(|A| = c\) where \(c > 0\), then
$$ A = c \quad \text{or} \quad A = -c $$
If \(c = 0\), then \(A = 0\) only. If \(c < 0\), there is no solution, because an absolute value can never be negative.
Worked Example 1: A Standard Equation
Solve \(|2x - 3| = 7\).
Step 1: Split into two cases.
- Case 1: \(2x - 3 = 7 \Rightarrow 2x = 10 \Rightarrow x = 5\)
- Case 2: \(2x - 3 = -7 \Rightarrow 2x = -4 \Rightarrow x = -2\)
Step 2: Check.
- \(x = 5\): \(|10 - 3| = |7| = 7\). Correct.
- \(x = -2\): \(|-4 - 3| = |-7| = 7\). Correct.
Answer: \(x = 5\) or \(x = -2\).
Worked Example 2: Variable on Both Sides
Solve \(|x + 1| = 2x - 4\).
This time the right side contains \(x\), so it might be negative for some values. Checking is essential.
Step 1: Case 1. \(x + 1 = 2x - 4\)
$$ 1 + 4 = 2x - x \quad \Rightarrow \quad x = 5 $$
Step 2: Case 2. \(x + 1 = -(2x - 4)\)
$$ x + 1 = -2x + 4 \quad \Rightarrow \quad 3x = 3 \quad \Rightarrow \quad x = 1 $$
Step 3: Check both.
- \(x = 5\): left side \(|6| = 6\); right side \(2(5) - 4 = 6\). Valid.
- \(x = 1\): left side \(|2| = 2\); right side \(2(1) - 4 = -2\). Since \(2 \neq -2\), this is extraneous (rejected).
Answer: \(x = 5\) only.
Worked Example 3: Two Absolute Values
Solve \(|x - 2| = |3x + 4|\).
When both sides are absolute values, the expressions inside are either equal or opposite.
Case 1: \(x - 2 = 3x + 4 \Rightarrow -6 = 2x \Rightarrow x = -3\)
Case 2: \(x - 2 = -(3x + 4) \Rightarrow x - 2 = -3x - 4 \Rightarrow 4x = -2 \Rightarrow x = -\frac{1}{2}\)
Check:
- \(x = -3\): \(|-5| = 5\) and \(|-9 + 4| = |-5| = 5\). Valid.
- \(x = -\frac{1}{2}\): \(\left|-\frac{5}{2}\right| = \frac{5}{2}\) and \(\left|-\frac{3}{2} + 4\right| = \frac{5}{2}\). Valid.
Answer: \(x = -3\) or \(x = -\frac{1}{2}\).
Solving Absolute Value Inequalities
Think about distance from zero:
- \(|A| < c\) means \(A\) is within \(c\) of zero:
$$ -c < A < c $$
- \(|A| > c\) means \(A\) is more than \(c\) away from zero:
$$ A < -c \quad \text{or} \quad A > c $$
A handy memory aid: "less than" gives an "and" (one interval); "greater than" gives an "or" (two pieces). The same rules apply with \(\leq\) and \(\geq\), including endpoints.
Worked Example 4: A "Less Than" Inequality
Solve \(|3x - 2| < 7\).
Step 1: Write as a compound inequality:
$$ -7 < 3x - 2 < 7 $$
Step 2: Add 2 to all three parts:
$$ -5 < 3x < 9 $$
Step 3: Divide all parts by 3:
$$ -\frac{5}{3} < x < 3 $$
Answer: \(x \in \left(-\frac{5}{3}, 3\right)\).
Quick check: \(x = 0\) gives \(|-2| = 2 < 7\), true. \(x = 4\) gives \(|10| = 10\), not less than 7, so 4 is correctly excluded.
Worked Example 5: A "Greater Than or Equal" Inequality
Solve \(|2x + 1| \geq 5\).
Step 1: Split into two pieces:
$$ 2x + 1 \leq -5 \quad \text{or} \quad 2x + 1 \geq 5 $$
Step 2: Solve each:
- \(2x \leq -6 \Rightarrow x \leq -3\)
- \(2x \geq 4 \Rightarrow x \geq 2\)
Answer: \(x \in (-\infty, -3] \cup [2, \infty)\).
Check: \(x = 0\) gives \(|1| = 1\), which is not \(\geq 5\), so 0 is correctly excluded.
Common Mistakes
- Forgetting the second case. \(|A| = c\) almost always gives two equations.
- Not checking answers when \(x\) appears outside the absolute value.
- Writing \(|A| > c\) as \(-c > A > c\). That is impossible; it must be two separate pieces joined by "or".
- Forgetting to flip the sign when multiplying or dividing an inequality by a negative number.
- Accepting solutions to \(|A| = -3\). An absolute value is never negative, so there is no solution.
Key Takeaways
- Absolute value is distance from zero, so it is never negative.
- \(|A| = c\) (with \(c > 0\)) splits into \(A = c\) or \(A = -c\); check for extraneous roots.
- \(|A| < c\) becomes \(-c < A < c\), a single interval.
- \(|A| > c\) becomes \(A < -c\) or \(A > c\), two separate intervals.
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