4.5: Systems of Linear Inequalities

section 4.5 Learning Objective

4.5: Systems of Linear Inequalities

Graph systems of two-variable linear inequalities

Graph a system of two inequalities

Remember from Module 3 on graphing linear inequalities that the graph of a single linear inequality splits the coordinate plane into two regions. On one side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality [latex]y<2x+5[/latex]. Coordinate plane with dotted line y is less than 2 x + 5, shaded below. Points A(negative 1, negative 1) and B(3,1) marked within shading. x and y axes from negative 4 to 4.

The dashed line is [latex]y=2x+5[/latex]. Every ordered pair in the shaded area below the line is a solution to [latex]y<2x+5[/latex], as all of the points below the line will make the inequality true. Try substituting the x and y coordinates of Points A and B into the inequality—you’ll see that they work. On the other hand, if you plug in the coordinates of any point on the opposite side of the line into the inequality, they will result in a false statement. So, the shaded area shows all of the solutions for this inequality.

The boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line don’t satisfy the inequality. If the inequality had been [latex]y\leq2x+5[/latex], then the boundary line would have been solid.

Let’s graph another inequality: [latex]y>−x[/latex]. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line do not satisfy the inequality.

Downward-sloping dotted line on coordinate system with region above shaded, y is greater than negative x. Points M = (negative 2, 3) and N = (4, negative 1) marked within shading.

To create a system of inequalities, you need to graph two or more inequalities together. Let’s use [latex]y<2x+5[/latex] and [latex]y>−x[/latex] since we have already graphed each of them.

The purple area shows where the solutions of the two inequalities overlap. This area is the solution to the system of inequalities. Any point within this purple region will be true for both [latex]y>−x[/latex] and [latex]y<2x+5[/latex]. In the next example, you are given a system of two inequalities whose boundary lines are parallel to each other.

Example 1

Graph the system [latex]\begin{array}{c}y\ge2x+1\\y\lt2x-3\end{array}[/latex]

Show Solution

The boundary lines for this system are:

[latex]\begin{array}{c}y=2x+1\\y=2x-3\end{array}[/latex]

The inequality [latex]y\lt2x-3[/latex] requires that we draw a dashed line (shown in blue), while the inequality [latex]y\ge2x+1[/latex] will require a solid line (in red). The graphs will look like this:

Two lines on coordinate plane grid. Solid red line with slope of 2 and y intercept of 1. Second parallel dashed blue line with y-intercept of negative 4.

Now we need to add the regions that represent the inequalities. For the inequality [latex]y\ge2x+1[/latex] we can test a point on either side of the line to see which region to shade. Let’s test [latex]\left(0,0\right)[/latex] to make it easy.

Substitute [latex]\left(0,0\right)[/latex] into [latex]y\ge2x+1[/latex]

[latex]\begin{array}{c}y\ge2x+1\\0\ge2\left(0\right)+1\\0\ge{1}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality [latex]y\ge2x+1[/latex]. The graph will now look like this:

Two lines on coordinate plane grid. Solid red line with slope of 2 and y intercept of 1. Second parallel solid blue line with y-intercept of negative 3. Shading above red line.

Now let’s shade the region that shows the solutions to the inequality [latex]y\lt2x-3[/latex]. Again, we can pick [latex]\left(0,0\right)[/latex] to make it easy.

Substitute [latex]\left(0,0\right)[/latex] into [latex]y\lt2x-3[/latex]

[latex]\begin{array}{c}y\lt2x-3\\0\lt2\left(0,\right)x-3\\0\lt{-3}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\lt2x-3[/latex]. The graph will now look like this:

Two lines on coordinate plane grid. Solid red line with slope of 2 and y intercept of 1. Second parallel dashed blue line with y-intercept of negative 3. Shading above red line and below blue dashed line.

The graph above has been used to help us identify that this system of inequalities shares no points in common. So, the system of inequalities has no solutions. If we were to graph the solution set, it would be an empty graph.

Consider This

What would the graph look like if the system had looked like this? (Notice the direction of one of the inequalities has changed).

[latex]\begin{array}{c}y\ge2x+1\\y\gt2x-3\end{array}[/latex].

Testing the point [latex]\left(0,0\right)[/latex] would result in a true statement for the inequality [latex]y\gt2x-3[/latex], and the graph would then look like this:

Two lines on coordinate plane grid. Solid red line with slope of 2 and y-intercept of 1. Second parallel dashed blue line with y-intercept of negative 3. Shading is above red line and above blue dashed line, creating an overlapping purple region above red line.

The purple region is the region of overlap for both inequalities.

In other words, the solution set to this system of inequalities is only the region shown below.

Purple line with slope of 2 and y intercept of 1. Purple shading above line.

In a previous section, we determined whether or not a point was a solution to a system of equations. Next, we will verify algebraically whether a point is a solution to a linear inequality.

Determine whether an ordered pair is a solution to a system of linear inequalities

On the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements.

In contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality [latex]y>−x[/latex] and point A is a solution for the inequality [latex]y<2x+5[/latex], neither point is a solution for the system. The following example shows how to test a point to see whether it is a solution to a system of inequalities.

Example 2

Is the point (2, 1) a solution of the system [latex]x+y>1[/latex] and [latex]2x+y<8[/latex]?

Show Solution

Here is a graph of the system in the example above. Notice that (2, 1) lies in the purple area, which is the overlapping area for the two inequalities.

Two dashed lines on coordinate system: one with region above shaded and labeled x+y is greater than 1, one with region below shaded and labeled 2 x plus y is less than 8. Overlapping region is shaded purple and point (2,1) is marked within overlap.

Example 3

Is the point (2, 1) a solution of the system [latex]x+y>1[/latex] and [latex]3x+y<4[/latex]?

Show Solution

Here is a graph of this system. Notice that (2, 1) is not in the purple area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).

Coordinate plane grid with x and y axes from negative 8 to 8. Two dotted lines, one labeled '3x + y is less than 4' with region below shaded, and another labeled 'x + y is greater than 1' with region above shaded. Overlapping region is purple. Point (2,1) is marked outside the overlap.

As shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities. The general steps are outlined below:

Graph each inequality as a line and determine whether it will be solid or dashed
Determine which side of each boundary line represents solutions to the inequality by testing a point on each side
Shade the region that represents solutions for both inequalities