Intermediate Algebra: Solving Absolute Value Equations
Practice with Solutions

Problem: Solve \( |x| = 5 \).

Solution:

The absolute value \( |x| \) represents the distance of \( x \) from zero on the number line, so \( x \) can be either 5 units to the right or 5 units to the left of zero.
Therefore, the solutions are: \[ x = 5 \quad \text{or} \quad x = -5 \]

Problem: Solve \( |x - 2| = 7 \).

Solution:

The expression \( |x - 2| \) measures the distance between \( x \) and 2 on the number line. This distance is 7, so: \[ x - 2 = 7 \quad \Rightarrow \quad x = 9 \] or \[ x - 2 = -7 \quad \Rightarrow \quad x = -5 \]

Problem: Solve \( |89x + 4| = -2 \).

Solution:

Absolute values are always non-negative, so they cannot equal a negative number.
No solution.

Problem: Solve \( |-11x - 9| = 0 \).

Solution:

The absolute value is zero only when the expression inside is zero: \[ -11x - 9 = 0 \quad \Rightarrow \quad x = -\dfrac{9}{11} \]

Problem: Solve \( 2|x + 5| = 10 \).

Solution:

First, divide both sides by 2: \[ |x + 5| = 5 \] So \( x + 5 \) is either 5 or -5: \[ x + 5 = 5 \quad \Rightarrow \quad x = 0 \] or \[ x + 5 = -5 \quad \Rightarrow \quad x = -10 \]

Problem: Solve \( 3|-9x - 7| - 2 = 13 \).

Solution:

Add 2 to both sides: \[ 3|-9x - 7| = 15 \] Divide both sides by 3: \[ |-9x - 7| = 5 \] So the inside equals 5 or -5: \[ -9x - 7 = 5 \quad \Rightarrow \quad -9x = 12 \quad \Rightarrow \quad x = -\dfrac{12}{9} = -\dfrac{4}{3} \] or \[ -9x - 7 = -5 \quad \Rightarrow \quad -9x = 2 \quad \Rightarrow \quad x = -\dfrac{2}{9} \]

Problem: Solve \( -5|-x + 2| + 8 = -12 \).

Solution:

Subtract 8 from both sides: \[ -5|-x + 2| = -20 \] Divide both sides by -5: \[ |-x + 2| = 4 \] So the inside is 4 or -4: \[ -x + 2 = 4 \quad \Rightarrow \quad -x = 2 \quad \Rightarrow \quad x = -2 \] or \[ -x + 2 = -4 \quad \Rightarrow \quad -x = -6 \quad \Rightarrow \quad x = 6 \]

Problem: Solve \( \dfrac{|-3x - 2|}{3} = 7 \).

Solution:

Multiply both sides by 3: \[ |-3x - 2| = 21 \] So the inside is 21 or -21: \[ -3x - 2 = 21 \quad \Rightarrow \quad -3x = 23 \quad \Rightarrow \quad x = -\dfrac{23}{3} \] or \[ -3x - 2 = -21 \quad \Rightarrow \quad -3x = -19 \quad \Rightarrow \quad x = \dfrac{19}{3} \]

Problem: Solve \( \left|\dfrac{x - 5}{4}\right| = 6 \).

Solution:

Simplify the fraction on the left side first using the quotient property of absolute values (\( \left|\dfrac{a}{b}\right| = \dfrac{|a|}{|b|} \)): \[ \dfrac{|x - 5|}{|4|} = \dfrac{|x - 5|}{4} = 6 \] Then multiply both sides by 4 and simplify: \[ |x - 5| = 24 \] Now, solve the resulting two cases: \[ x - 5 = 24 \quad \Rightarrow \quad x = 29 \] or \[ x - 5 = -24 \quad \Rightarrow \quad x = -19 \]

Problem: Solve \( |x + 1| = |-4x + 3| \).

Solution:

Since absolute values are equal, either the inner expressions are equal or one is the negative of the other: \[ x + 1 = -4x + 3 \quad \Rightarrow \quad 5x = 2 \quad \Rightarrow \quad x = \dfrac{2}{5} \] or \[ x + 1 = -(-4x + 3) \quad \Rightarrow \quad x + 1 = 4x - 3 \quad \Rightarrow \quad -3x = -4 \quad \Rightarrow \quad x = \dfrac{4}{3} \]

Problem: Solve \( |x + 3| = -|-x - 3| \).

Solution:

Simplify the right side using the property \( |-a| = |a| \): \[ -|-x - 3| = -|-(x + 3)| = -|x + 3| \] Substitute this back into the equation: \[ |x + 3| = -|x + 3| \] Add \( |x + 3| \) to both sides: \[ 2|x + 3| = 0 \quad \Rightarrow \quad |x + 3| = 0 \quad \Rightarrow \quad x + 3 = 0 \quad \Rightarrow \quad x = -3 \]