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6.2: Solving Absolute Value Equations

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    To solve absolute value equations, first consider the following two properties of absolute value:

    Definition: Properties of Absolute Value

    Property 1: For \(b > 0\), \(|a| = b\) if and only if \(a = b\) or \(a = −b\)

    Property 2: For any real numbers \(a\) and \(b\), \(|a| = |b|\) if and only if \(a = b\) or \(a = −b\)

    • Before Property 1 is applied, isolate the absolute value expression to either side of the equation.
    • Check the solutions by substituting them back into the original equation.
    • Solutions are presented as a solution set of the form \(\{p, q\}\), where \(p\) and \(q\) are any real numbers.
    • The solution set of an absolute value equation is graphed as points on a number line.
    Example 6.2.1

    Solve each equation and graph the solution set.

    1. \(|x| = 7\)
    2. \(|5x – 3| = 2\)
    3. \(|20 – x| = −80\)

    Solution

    1. To solve \(|x| = 7\), apply Property 1 with \(a = x\) and \(b = 7\).

    Therefore, the solutions are, \(x = −7\) and \(x = 7\), and the solution set is \(\{-7,7\}\). The graph of the solution set is as shown in the figure below.

    clipboard_e4e661622afb19f733118b1049d678a57.png

    1. The equation-solving method used in part a can be extended to the given equation in this part with \(a = 5x – 3\) and \(b = 2\).

    Thus, the absolute value equation \(|5x – 3| = 2\) is equivalent to:

    \(\begin{array} &&5x − 3 = 2 &\text{ or } &5x − 3 = −2 &\text{Property 1} \\ &5x = 5 &\text{ or } &5x = 1 &\text{Add \(3\) to both sides of the equations} \\ &x = 1 &\text{ or } &x = \dfrac{1}{5} &\text{Divide by \(5\) both sides of the equations} \end{array}\)

    Now, check if \(x = 1\) and \(x = \dfrac{1}{5}\) are solutions to the given absolute value equation.

    \(\begin{array} &&\text{For } x = 1 &\text{For } x = \dfrac{1}{5} &\\ &|5x − 3| = 2 &|5x − 3| = 2 &\text{Given} \\ &|5(1) − 3| \stackrel{?}{=} 2 &|5 \left( \dfrac{1}{5} \right) − 3| \stackrel{?}{=} 2 &\text{Substitute the \(x\)-values} \\ &|5 − 3| \stackrel{?}{=} 2 &|1 − 3| \stackrel{?}{=} 2 &\text{Simplify} \\ &|2| \stackrel{?}{=} 2 &|− 2| \stackrel{?}{=} 2 &\text{Apply the absolute value definition} \\ &2 = 2\; \checkmark &2 = 2\; \checkmark \end{array}\)

    Since the above equations are true, then, \(x = 1\) and \(x = \dfrac{1}{5}\) are solutions to the given absolute value equation. The solution set is \(\left\{\dfrac{1}{5} , 1\right\}\). The graph of the solution set is as shown in the figure below.

    clipboard_e78f4a9bbcf248134874230631b7dada3.png

    1. Since an absolute value can never be negative, there are no real numbers \(x\) that makes \(|20 – x| = −80\) true. The equation has no solution and the solution set is \(∅\).
    Example 6.2.2

    Solve and graph the solution set.

    1. \(\left| \dfrac{4}{3} x + 3 \right| + 8 = 18\)
    2. \(4 \left| \dfrac{1}{3}x − 6 \right| − 5 = −5\)
    3. \(|4x – 3| = |x + 6|\)

    Solution

    1. Notice that the absolute value expression is not isolated which means the properties cannot be applied. First, isolate \(\left| \dfrac{4}{3}x + 3 \right|\) on the left side of the equation, then, apply Property 1.

    \(\begin{array} &&\left| \dfrac{4}{3} x + 3 \right| + 8 = 18 &\text{Given equation} \\ & \left| \dfrac{4}{3} + 3 \right| = 10 &\text{Subtract \(8\) from both sides of the equation} \end{array}\)

    With the absolute value now isolated, solve\(\left| \dfrac{4}{3} + 3 \right| = 10\) using Property 1, with \(a = \dfrac{4}{3} x + 3\) and \(b = 10\) as follows,

    \(\begin{array} && &\left| \dfrac{4}{3} + 3 \right| = 10 & & \\ &\dfrac{4}{3} + 3 = 10 &\text{ or } & \dfrac{4}{3} + 3 = -10 &\text{Property 1} \\ &\dfrac{4}{3} x = 7 &\text{ or } &\dfrac{4}{3}x = −13 &\text{Subtract \(3\) from both sides} \\ &x = \dfrac{21}{4} &\text{ or } &x = −\dfrac{39}{4} &\text{Multiply both sides by \(\dfrac{3}{4}\)} \end{array}\)

    Check the solutions \(x = −\dfrac{39}{4}\) and \(x = \dfrac{21}{4}\) by substituting them into the original absolute value equation. The solution set is \(\left\{ −\dfrac{39}{4}, \dfrac{21}{4} \right\}\) and the graph of the solution set is as shown in the figure below.

    clipboard_e42b1bbc90f71c52e8a95664a185e2c67.png

    1. Similar to part a, isolate the absolute value expression. So, first isolate \(\left| \dfrac{1}{3} x − 6 \right|\) on the left side of the equation and apply Property 1.

    \(\begin{array} &&4 \left| \dfrac{1}{3}x − 6 \right| − 5 = −5 &\text{Given equation} \\ &4 \left| \dfrac{1}{3}x − 6 \right| = 0 &\text{Add \(5\) to both sides of the equation} \\ &\left| \dfrac{1}{3}x − 6 \right| = 0 &\text{Divide by \(4\) both sides of the equation} \end{array}\)

    The absolute value is isolated. Since \(0\) is the only number whose absolute value is \(0\), the expression \(\dfrac{1}{3}x − 6\) must be equal to \(0\). So,

    \(\begin{array} &&\dfrac{1}{3}x − 6 = 0 & \\ &\dfrac{1}{3}x − 6 &\text{Add \(6\) to both sides of the equation} \\ &x = 18 &\text{Multiply both sides by \(3\)}\end{array}\)

    The solution is \(18\) and the solution set is \(\{18\}\). Verify that it satisfies the original equation. The graph of the solution set is as shown in the figure below.

    clipboard_e2acd5153df84bfef5569962926db627b.png

    1. \(|4x − 7| = |x + 14|\) Notice that to solve \(|4x − 7| = |x + 14|\), use Property 2 with \(a = 4x − 7\) and \(b = x + 14\).

    \(\begin{array} && &|4x − 7| = |x + 14| & &\text{Given} \\ &4x−7 = x+14 &\text{ or } &4x − 7 = −(x + 14) &\text{Property 2} \\ &4x−7 = x+14 &\text{ or } &4x − 7 = −x − 14 &\text{Distribute \(−1\) to simplify the right equation} \\ &4x = x + 21 &\text{ or } &4x = −x − 7 &\text{Add \(7\) to both sides of each equality} \\ &3x = 21 &\text{ or } &5x = −7 &\text{Simplify} \\ &x = 7 &\text{ or } &x = −\dfrac{7}{5} &\text{Divide each equation by the \(x\)-coefficient} \end{array}\)

    Check the solutions \(x = −\dfrac{7}{5}\) and \(x = 7\) by substituting them into the original absolute value equation. The solution set is \(\left\{ −\dfrac{7}{5}, 7\right\}\). The graph of the solution is as shown in the figure below.

    clipboard_eb0f0ab26e578678046e463bf1d8ac854.png

    Exercise 6.2.1

    Solve each equation, check the solution and graph the solution set.

    1. \(|x| = 19\)
    2. \(|x − 4| = 10\)
    3. \(|2x − 5| = 12\)
    4. \(\left|\dfrac{x}{11} \right| = 2.5\)
    5. \(|x − 3.8| = −2.7\)
    6. \(|3x − 4.5| = 9.3\)
    7. \(\dfrac{8}{3} |x − 6| = 14\)
    8. \(|x + 15| − 19 = 7\)
    9. \(|11x + 3| + 28 = 16\)
    10. \( \left| \dfrac{8}{7} x + 9 \right| − 2 = 8\)
    11. \( −3|2x − 7| + 13 = 13\)
    12. \( 8 − 5|10x + 6| = 5\)
    13. \( |5x − 14| = |3x − 9|\)
    14. \( |15x| = |x − 21|\)
    15. \( |4x − 7| = |5(2x + 3)|\)
    16. \( \dfrac{7}{8} = \dfrac{3x}{2} + \dfrac{2x}{5}\)

    This page titled 6.2: Solving Absolute Value Equations is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Victoria Dominguez, Cristian Martinez, & Sanaa Saykali (ASCCC Open Educational Resources Initiative) .