Compare Your Answers
  1. Solve \(\left| {3x - 7} \right| = 8\) for the interval \(3x - 7 \ge 0\).

    Since \(3x - 7 \ge 0\), the contents of the absolute value are positive, so \(\left|3x - 7 \right| = 3x - 7\).

    \(\begin{align}
     \left| {3x - 7} \right| &= 8 \\
     3x - 7 &= 8 \\
     3x &= 15 \\
     x &= 5 \\
     \end{align}\)
    Verify for \(x = 5\).

    Left Side Right Side
    \[\begin{array}{r}
    \left| {3x - 7} \right| \\
    \left| {3(5) - 7} \right| \\
     \left| 8 \right| \\
      8 \end{array}\]
    		\(8\)
    LS = RS


  2. Solve \(\left| {x^2 - 3x} \right| = 2\) for the interval \(x^2 - 3x < 0\).

    Since \(x^2 - 3x \lt 0\), the contents of the absolute value are negative, so \(\left|x^2 - 3x \right| = -(x^2 - 3x)\).

    \(\begin{align}
     \left| {x^2 - 3x} \right| &= 2 \\
      - \left( {x^2 - 3x} \right) &= 2 \\
      - x^2 + 3x &= 2 \\
     0 &= x^2 - 3x + 2 \\
     0 &= \left( {x - 2} \right)\left( {x - 1} \right) \end{align}\)

    \(\begin{align}
     0 &= x - 2 \\
     2 &= x \\
     \end{align}\)
    \(\begin{align}
     0 &= x - 1 \\
     1 &= x \\
     \end{align}\)
    Verify for \(x = 2\).

    Left Side Right Side
    \[\begin{array}{r}
    \left| {x^2 - 3x} \right| \\
    \left| {2^2 - 3(2)} \right| \\
     \left| 4 - 6 \right| \\
     \left| -2 \right| \\
      2 \end{array}\]
    		\(2\)
    LS = RS

    Verify for \(x = 1\).

    Left Side Right Side
    \[\begin{array}{r}
    \left| {x^2 - 3x} \right| \\
    \left| {1^2 - 3(1)} \right| \\
     \left| 1 - 3 \right| \\
    \left| -2 \right| \\
      2 \end{array}\]
    		\(2\)
    LS = RS