# Define Solutions to Systems of Linear Inequalities

### Learning Outcomes

- Graph a system of linear inequalities and define the solutions region
- Verify whether a point is a solution to a system of inequalities
- Identify when a system of inequalities has no solution

## Graph a System of Two Inequalities

Recall 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]. 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. If you doubt that, try substituting the

*x*and

*y*coordinates of Points A and B into the inequality; you will see that they work. 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 do not satisfy the inequality. If the inequality had been [latex]y\leq2x+5[/latex], then the boundary line would have been solid. Now 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 are not true. To create a system of inequalities, you need to graph two or more inequalities together. Let us 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]. Watch the following video examples to see how to graph a system of linear inequalities and define the solution region. https://youtu.be/ACTxJv1h2_c https://youtu.be/cclH2h1NurM In the next section, we will see that points can be solutions to systems of equations and inequalities. We will verify algebraically whether a point is a solution to a linear equation or 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

Is the point [latex](2, 1)[/latex] a solution of the system [latex]x+y>1[/latex] and [latex]2x+y<8[/latex]?Answer:
Check the point with each of the inequalities. Substitute [latex]2[/latex] for *x* and [latex]1[/latex] for *y*. Is the point a solution of both inequalities?
[latex-display]\begin{array}{r}x+y>1\\2+1>1\\3>1\\\text{TRUE}\end{array}[/latex-display]
[latex](2, 1)[/latex] is a solution for [latex]x+y>1[/latex].
[latex-display]\begin{array}{r}2x+y<8\\2\left(2\right)+1<8\\4+1<8\\5<8\\\text{TRUE}\end{array}[/latex-display]
[latex-display](2, 1)[/latex] is a solution for [latex]2x+y<8.[/latex-display]
Since [latex](2, 1)[/latex] is a solution of each inequality, it is also a solution of the system.

### Example

Is the point [latex](2, 1)[/latex] a solution of the system [latex]x+y>1[/latex] and [latex]3x+y<4[/latex]?Answer:
Check the point with each of the inequalities. Substitute [latex]2[/latex] for *x* and [latex]1[/latex] for *y*. Is the point a solution of both inequalities?
[latex-display]\begin{array}{r}x+y>1\\2+1>1\\3>1\\\text{TRUE}\end{array}[/latex-display]
[latex](2, 1)[/latex] is a solution for [latex]x+y>1[/latex].
[latex-display]\begin{array}{r}3x+y<4\\3\left(2\right)+1<4\\6+1<4\\7<4\\\text{FALSE}\end{array}[/latex-display]
[latex](2, 1)[/latex] is *not *a solution for [latex]3x+y<4[/latex].
Since [latex](2, 1)[/latex] is *not *a solution of one of the inequalities, it is not a solution of the system.

- 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.

## Systems with No Solutions

In the next example, you'll see the solution to a system of two inequalities whose boundary lines are parallel to each other. When the graphs of a system of two linear equations are parallel to each other there is no solution to the system. That is, there is no point where the lines intersect. We will get a similar result for the following system of linear inequalities.### Examples

Graph the system [latex]\begin{array}{c}y\ge2x+1\\y\lt2x-3\end{array}[/latex]Answer: The boundary lines for this system are parallel to each other. Note how they have the same slopes.

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

Plotting the boundary lines will give the graph below. Note that the inequality [latex]y\lt2x-3[/latex] requires that we draw a dashed line, while the inequality [latex]y\ge2x+1[/latex] requires a solid line. Now we need to shade 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. Test [latex]\left(0,0\right)[/latex] to make the task less challenging. 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: Now 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 test to make our work less challenging. 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: This system of inequalities has no points in common so has no solution.## Summary

- Solutions to systems of linear inequalities are entire regions of points.
- You can verify whether a point is a solution to a system of linear inequalities in the same way you verify whether a point is a solution to a system of equations.
- Systems of inequalities can have no solutions when boundary lines are parallel.

## Licenses & Attributions

### CC licensed content, Shared previously

- Ex 1: Graph a System of Linear Inequalities.
**Authored by:**James Sousa (Mathispower4u.com) .**License:**CC BY: Attribution. - Ex 2: Graph a System of Linear Inequalities.
**Authored by:**James Sousa (Mathispower4u.com).**License:**CC BY: Attribution. - Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program.
**Provided by:**Monterey Institute of Technology**Located at:**https://www.nroc.org/.**License:**CC BY: Attribution.