Systems of Linear Inequalities

Learning Objectives

  • Define solutions to systems of linear inequalities
    • 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
  • Solutions from graphs of linear inequalities
    • Solve systems of linear inequalities by graphing the solution region
    • Graph solutions to a system that contains a compound inequality
  • Applications of systems of linear inequalities
    • Write and graph a system that models the quantity that must be sold to achieve a given amount of sales
    • Write a system of inequalities that represents the profit region for a business
    • Interpret the solutions to a system of cost/ revenue inequalities

 

Graph a system of two inequalities

Remember from the module on graphing 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]. An upward-sloping dotted line with the region below it shaded. The shaded region is labeled y is less than 2x+5. A is equal to (-1,1). B is equal to (3,1).

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’ll 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 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 are not true.

Downward-sloping dotted line with the region above it shaded. The shaded region is y is greater than negative x. Point M=(-2,3). Point N=(4,-1).

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 two previous graphs combined. A blue dotted line with the region above shaded and labeled y is greater than negative x. A red dotted line with the region below it shaded and labeled y is less than 2x+5. The region where the shaded areas overlap is labeled y is greater than negative x and y is less than 2x+5. The point M equals (-2,3) and is in the blue shaded region. The point A equals (-1,-1) and is in the red shaded region. The point B equals (3,1) and is in the purple overlapping region. The point N equals (4,-1) and is also in the purple overlapping region.

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 following video examples, we show how to graph a system of linear inequalities, and define the solution region.

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

image014

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

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 dotted lines, one red and one blue. The region below the blue dotted line is shaded and labeled 2x+y is less than 8. The region above the dotted red line is shaded and labeled x+y is greater than 1. The overlapping shaded region is purple and is labeled x+y is greater than 1 and 2x+y is less than 8. The point (2,1) is in the overlapping purple region.

Example

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

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).

A downward-sloping bue dotted line with the region below shaded and labeled 3x+y is less than 4. A downward-sloping red dotted line with the region above it shaded and labeled x+y is greater than 1. An overlapping purple shaded region is labeled x+y is greater than 1 and 3x+y is less than 4. A point (2,1) is in the red shaded region, but not the blue or overlapping purple shaded region.

In the following video we show another example of determining whether a point is in the solution of a system of linear inequalities.

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

Systems with no solutions

In the next example, we will show 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, we found that there was no solution to the system.  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]

In the following examples, we will continue to practice graphing the solution region for systems of linear inequalities.  We will also graph the solutions to a system that includes a compound inequality.

Example

Shade the region of the graph that represents solutions for both inequalities.  [latex]x+y\geq1[/latex] and [latex]y–x\geq5[/latex].

The videos that follow show more examples of graphing the solution set of a system of linear inequalities.

The system in our last example includes a compound inequality.  We will see that you can treat a compound inequality like two lines when you are graphing them.

Example

Find the solution to the system 3x + 2y < 12 and −1 ≤ y ≤ 5.

In the video that follows, we show how to solve another system of inequalities that contains a compound inequality.

Applications

In our first example we will show how to write and graph a system of linear inequalities that models the amount of sales needed to obtain a specific amount of money.

Example

Cathy is selling ice cream cones at a school fundraiser. She is selling two sizes: small (which has 1 scoop) and large (which has 2 scoops). She knows that she can get a maximum of 70 scoops of ice cream out of her supply. She charges $3 for a small cone and $5 for a large cone.

Cathy wants to earn at least $120 to give back to the school. Write and graph a system of inequalities that models this situation.

In a previous example for finding a solution to a system of linear equations, we introduced a manufacturer’s cost and revenue equations:

Cost: [latex]y=0.85x+35,000[/latex]

Revenue: [latex]y=1.55x[/latex]

The cost equation is shown in blue in the graph below, and the revenue equation is graphed in orange.The point at which the two lines intersect is called the break-even point, we learned that this is the solution to the system of linear equations that in this case comprise the cost and revenue equations.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The region to the left represents quantities for which the company suffers a loss.

In the next example, you will see how the information you learned about systems of linear inequalities can be applied to answering questions about cost and revenue.

A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.

Note how the blue shaded region between the Cost and Revenue equations is labeled Profit. This is the “sweet spot” that the company wants to achieve where they produce enough bike frames at a minimal enough cost to make money. They don’t want more money going out than coming in!

Example

Define the profit region for the skateboard manufacturing business using inequalities, given the system of linear equations:

Cost: [latex]y=0.85x+35,000[/latex]

Revenue: [latex]y=1.55x[/latex]

In the following video you will see an example of how to find the break even point for a small sno-cone business.

And here is one more video example of solving an application using a sustem of linear inequalities.

We have seen that systems of linear equations and inequalities can help to define market behaviors that are very helpful to businesses.  The intersection of cost and revenue equations gives the break even point, and also helps define the region for which a company will make a profit.

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