Example  1
  1. Solve \(\left|2x + 1\right| = 7\) for the interval \(2x + 1 \ge 0\).

    Since \(2x + 1 \ge 0\), the contents of the absolute value are positive, so \(\left|2x + 1\right| = 2x + 1\).

    \(\begin{align}
     \left| {2x + 1} \right| &= 7 \\
     2x + 1 &= 7 \\
     2x &= 6 \\
     x &= 3 \\
     \end{align}\)

    Verify for \(x = 3\).

    Left Side Right Side
    \[\begin{array}{r}
    \left| {2x + 1} \right| \\
    \left| {2(3) + 1} \right| \\
     \left| 7 \right| \\
     7 \end{array}\]
    		\(7\)
    LS = RS


  2. Solve \(\left|2x + 1\right| = 7\) for the interval \(2x + 1 \lt 0\).

    Since \(2x + 1 \lt 0\), the contents of the absolute value are negative, so \(\left|2x + 1\right| = -(2x + 1)\).

    \(\begin{align}
     \left| {2x + 1} \right| &= 7 \\
      -(2x + 1) &= 7 \\
      -2x - 1 &= 7 \\
      -2x &= 8 \\
      x &= -4 \\
     \end{align}\)

    Verify for \(x = -4\).

    Left Side Right Side
    \[\begin{array}{r}
    \left| {2x + 1} \right| \\
    \left| {2(-4) + 1} \right| \\
     \left| -7 \right| \\
     7 \end{array}\]
    		\(7\)
    LS = RS