Example 1
Completion requirements
| Example 1 |
Multimedia |
Graph the inequality \(3x + y \lt 4\). Give an example of an ordered pair that is part of the solution set and an example of an ordered pair that is not part of the solution set.
Start by graphing the corresponding equality \(3x + y = 4\). Rearrange the equality to slope-intercept form by isolating \(y\).
\(\begin{align}
3x + y &= 4 \\
y &= - 3x + 4
\end{align}\)
The inequality is strict, so graph the boundary with a dashed line.
A test point can be used to determine which region to shade. Select a point that is not on the boundary and substitute it into the original inequality. If the test point satisfies the original inequality, the point is part of the solution region. If the test point does not satisfy the original inequality, the point is not part of the solution region. The test point \((0, 0)\) is selected here because it includes values that are simple to work with and it is not on the boundary.
The point \((0, 0)\) satisfies the inequality, so it and all other points on that side of the boundary are solutions. Shade the region that includes the test point.
The ordered pair \((–3, 3)\) is in the solution region, so it is part of the solution. The ordered pair \((3, 3)\) is not in the solution region, so it is not part of the solution.
\(\begin{align}
3x + y &= 4 \\
y &= - 3x + 4
\end{align}\)
The inequality is strict, so graph the boundary with a dashed line.
A test point can be used to determine which region to shade. Select a point that is not on the boundary and substitute it into the original inequality. If the test point satisfies the original inequality, the point is part of the solution region. If the test point does not satisfy the original inequality, the point is not part of the solution region. The test point \((0, 0)\) is selected here because it includes values that are simple to work with and it is not on the boundary.
| Left Side | Right Side |
|---|---|
|
\(\begin{array}{r} 3x + y \\ 3\left( 0 \right) + 0 \\ 0 \end{array}\) |
\(4\) |
| LS \(\lt\) RS | |
The point \((0, 0)\) satisfies the inequality, so it and all other points on that side of the boundary are solutions. Shade the region that includes the test point.
The ordered pair \((–3, 3)\) is in the solution region, so it is part of the solution. The ordered pair \((3, 3)\) is not in the solution region, so it is not part of the solution.