Compare Your Answers
Consider the inequality \(3x + 4y + 12 \gt 0\).

  1. Graph the inequality.

    First, use the corresponding equation to determine the boundary. The intercepts can be used to graph the boundary. The inequality is strict, so a dashed line is used.

    \(\begin{align}
     3x + 4y + 12 &= 0 \\
     3x + 4\left( 0 \right) + 12 &= 0 \\
     3x + 12 &= 0 \\
     3x &= - 12 \\
     x &= - 4 \end{align}\)
    \(\begin{align}
     3x + 4y + 12 &= 0 \\
     3\left( 0 \right) + 4y + 12 &= 0 \\
     4y + 12 &= 0 \\
     4y &= - 12 \\
     y &= - 3  \end{align}\)

    Next, interpret the inequality or use a test point to determine which side of the boundary represents the solution region. No variable is isolated, so a test point of \((0, 0)\) is used here.

    Left Side Right Side
    \(\begin{array}{r}
     3x + 4y + 12\\
     3(0) + 4(0) + 12 \\
     12  \end{array}\)
    		 \(0\)
    LS \(\gt\) RS\(\hspace{30pt}\)

    The point \((0, 0)\) satisfies the inequality, so it and all points on that side of the boundary are solutions.

    If you choose to interpret the inequality to decide which side of the boundary to shade, rearrange the inequality to isolate \(y\).

    \(\begin{align}
     3x + 4y + 12 &> 0 \\
     4y &> - 3x - 12 \\
     y &> - \frac{3}{4}x - 3  \end{align}\)

    The inequality is of the form \(y \gt\) ___, so values above the boundary should be shaded.


  2. Determine if the point \((5, -2)\) is part of the solution.

    The point \((5, -2)\) lies in the solution region, so it is part of the solution. This can also be checked algebraically.

    Left Side Right Side
    \(\begin{array}{r}
     3x + 4y + 12 \\
     3(5) + 4(-2) + 12 \\
     15 - 8 + 12 \\
     19  \end{array}\)
    		 \(0\)
    LS \(\gt\) RS\(\hspace{30pt}\)

    The point \((5, -2)\) satisfies the inequality, so it is part of the solution.