Solve Equations with Absolute Value

Learn how to solve equations involving absolute value with step-by-step examples, detailed solutions, and explanations.

Review of Absolute Value

Here are the key rules to solve absolute value equations:

1. \(|x| = 0\) if \(x = 0\)
2. \(|x| = x\) if \(x > 0\)
3. \(|x| = -x\) if \(x < 0\)
4. The equation \(|x| = k\) with \(k < 0\) has no real solutions.
5. The equation \(|x| = k\) with \(k \ge 0\) is equivalent to \(x = k\) or \(x = -k\).

Examples with Solutions

Example 1

Solve the equation and check the answers:

\(|x + 6| = 7\)

• By rule 5: \[ x + 6 = 7 \quad \text{or} \quad x + 6 = -7 \]
• Solve the first equation: \[ x + 6 = 7 \implies x = 1 \]
• Solve the second equation: \[ x + 6 = -7 \implies x = -13 \]

• \(x = 1\): \(|1 + 6| = |7| = 7\), matches right-hand side.
• \(x = -13\): \(|-13 + 6| = |-7| = 7\), matches right-hand side.

The solutions are \(x = 1\) and \(x = -13\).

Matched Exercise 1

\(|-x - 8| = 10 \)

Solution to Matched Exercise

Example 2

Solve the equation:

\(-2 |x/2 + 3| - 4 = -10\)

• Rewrite in the form \(|A| = B\): \[ -2 |x/2 + 3| = -6 \implies |x/2 + 3| = 3 \]
• Split into two equations: \[ x/2 + 3 = 3 \quad \text{or} \quad x/2 + 3 = -3 \]
• Solve first: \(x/2 + 3 = 3 \implies x = 0\)
• Solve second: \(x/2 + 3 = -3 \implies x = -12\)

• \(x = 0\): LHS = \(-2|0/2 + 3| - 4 = -10\), matches RHS
• \(x = -12\): LHS = \(-2|-6 + 3| - 4 = -10\), matches RHS

The solutions are \(x = 0\) and \(x = -12\).

Example 3

Solve the equation:

\(|2x - 2| = x + 1\)

• Case 1: \(2x - 2 \ge 0 \implies x \ge 1\), then \(|2x-2| = 2x-2\): \[ 2x - 2 = x + 1 \implies x = 3 \]
• Case 2: \(2x - 2 < 0 \implies x < 1\), then \(|2x-2| = -(2x-2)\): \[ -(2x-2) = x + 1 \implies x = \frac{1}{3} \]

• \(x = 3\): \(|2*3-2| = 4 = 3+1\)
• \(x = 1/3\): \(|2*(1/3)-2| = 4/3 = 1/3 + 1\)

The solutions are \(x = 3\) and \(x = 1/3\).

Example 4

Solve the equation:

\(|x^2 - 4| = x + 2\)

• Case 1: \(x^2 - 4 \ge 0 \implies x^2 \ge 4\), then \(|x^2-4| = x^2-4\): \[ x^2 - 4 = x + 2 \implies (x+2)(x-3) = 0 \implies x = -2, 3 \]
• Case 2: \(x^2 - 4 < 0 \implies x^2 < 4\), then \(|x^2-4| = -(x^2-4)\): \[ -(x^2-4) = x + 2 \implies (x+2)(x-1) = 0 \implies x = 1 \]

• \(x = -2\): \(|(-2)^2 - 4| = 0 = -2 + 2\)
• \(x = 3\): \(|3^2 - 4| = 5 = 3 + 2\)
• \(x = 1\): \(|1^2 - 4| = 3 = 1 + 2\)

The solutions are \(x = -2, 1, 3\).

Solutions to Matched Exercises

Matched Exercise 1

\(|-x - 8| = 10\)

Solutions: \(x = 2, -18\)

Matched Exercise 2

\(4 |x + 2| - 30 = -10\)

Solutions: \(x = 3, -7\)

Matched Exercise 3

\(-4 |x + 2| = x - 8\)

Solutions: \(x = 0, -16/3\)

Matched Exercise 4

\(|x^2 - 16| = x - 4\)

Solution: \(x = 4\)

More Exercises with Answers

• \(|x-4|=9 \implies x = -5, 13\)
• \(|x^2 + 4| = 5 \implies x = -1, 1\)
• \(|x^2 - 9| = x + 3 \implies x = -3, 2, 4\)
• \(|x + 1| = x - 3 \implies \text{no real solutions} \)
• \(|-x| = 2 \implies x = -2, 2\)

More References and Links