Intermediate Algebra: Solving Absolute Value Equations - Practice with Solutions

This set of intermediate algebra problems focuses on solving equations that include absolute value expressions. Each problem is followed by its complete solution, making this page an excellent resource for self-study, review, or teaching. See the solutions section at the bottom of the page. You may also find it helpful to review the step-by-step tutorial on How to Solve Equations with Absolute Value.

Solve the following absolute value equations.

1. \(|x| = 5\)
2. \(|x - 2| = 7\)
3. \(|89x + 4| = -2\)
4. \(|-11x - 9| = 0\)
5. \(2|x + 5| = 10\)
6. \(3|-9x - 7| - 2 = 13\)
7. \(-5|-x + 2| + 8 = -12\)
8. \(\dfrac{|-3x - 2|}{3} = 7\)
9. \(\left|\dfrac{x - 5}{4}\right| = 6\)
10. \(|x + 1| = |-4x + 3|\)
11. \(|x + 3| = -|-x - 3|\)

Answers to the Above Questions

1. \( |x| = 5 \)
   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:
   \[ x = 5 \quad \text{or} \quad x = -5 \]

2. \( |x - 2| = 7 \)
   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 \[ x - 2 = -7 \quad \Rightarrow \quad x = -5 \]



3. \( |89x + 4| = -2 \)
   Absolute values are always non-negative, so they cannot equal a negative number.
   No solution.


4. \( |-11x - 9| = 0 \)
   The absolute value is zero only when the expression inside is zero:
   \[ -11x - 9 = 0 \quad \Rightarrow \quad x = -\dfrac{9}{11} \]


5. \( 2|x + 5| = 10 \)
   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 \[ x + 5 = -5 \quad \Rightarrow \quad x = -10 \]


6. \( 3|-9x - 7| - 2 = 13 \)
   Add 2 to both sides:
   \[ 3|-9x -7| = 15 \] Divide both sides by 3:
   \[ |-9x - 7| = 5 \] So the inside equals 5 or -5:
   \[ -9x - 7 = 5 \quad \Rightarrow \quad -9x = 12 \quad \Rightarrow \quad x = -\dfrac{4}{3} \] or \[ -9x - 7 = -5 \quad \Rightarrow \quad -9x = 2 \quad \Rightarrow \quad x = -\dfrac{2}{9} \]


7. \( -5|-x + 2| + 8 = -12 \)
   Subtract 8:
   \[ -5|-x + 2| = -20 \] 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 \[ -x + 2 = -4 \quad \Rightarrow \quad -x = -6 \quad \Rightarrow \quad x = 6 \]


8. \( \dfrac{|-3x - 2|}{3} = 7 \)
   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 \[ -3x - 2 = -21 \quad \Rightarrow \quad -3x = -19 \quad \Rightarrow \quad x = \dfrac{19}{3} \]


9. \( \left| \dfrac{x - 5}{4} \right| = 6 \)
   Multiply both sides by 4:
   \[ |x - 5| = 24 \] So: \[ x - 5 = 24 \quad \Rightarrow \quad x = 29 \] or \[ x - 5 = -24 \quad \Rightarrow \quad x = -19 \]


10. \( |x + 1| = |-4x + 3| \)
    The absolute values are equal, so either:
    \[ x + 1 = -4x + 3 \quad \Rightarrow \quad 5x = 2 \quad \Rightarrow \quad x = \dfrac{2}{5} \] or \[ x + 1 = -(-4x + 3) \quad \Rightarrow \quad -3x = -4 \quad \Rightarrow \quad x = \dfrac{4}{3} \]


11. \( |x + 3| = -|-x - 3| \)
    The right side is negative absolute value, but absolute values are always non-negative.
    Thus:
    \[ |x + 3| = -| -x - 3| \leq 0 \] But absolute values are \( \geq 0 \), so the only way this is true is if both sides are zero:
    \[ |x + 3| = 0 \quad \Rightarrow \quad x = -3 \]

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