Graph Linear Inequalities in Two Variables

Verify solutions to an inequality in two variables

Previously we learned to solve inequalities with only one variable. We will now learn about inequalities containing two variables. In particular we will look at linear inequalities in two variables which are very similar to linear equations in two variables.

Linear inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs—so that your business made a profit.

Recall that an inequality with one variable had many solutions. For example, the solution to the inequality $x > 3$ is any number greater than $3$. We showed this on the number line by shading in the number line to the right of $3$, and putting an open parenthesis at $3$.

Similarly, linear inequalities in two variables have many solutions. Any ordered pair $(x,y)$ that makes an inequality true when we substitute in the values is a solution to a linear inequality.

Example. Determine whether each ordered pair is a solution to the inequality $y > x + 4$: (a) $(0,0)$ (b) $(1,6)$ (c) $(2,6)$ (d) $(-5,-15)$ (e) $(-8,12)$.

(a) Substitute $0$ for $x$ and $0$ for $y$.

So, $(0,0)$ is not a solution to $y > x+4$.

(b) Substitute $1$ for $x$ and $6$ for $y$.

So, $(1,6)$ is a solution to $y > x+4$.

(c) Substitute $2$ for $x$ and $6$ for $y$.

So, $(2,6)$ is not a solution to $y > x+4$.

(d) Substitute $-5$ for $x$ and $-15$ for $y$.

So, $(-5,-15)$ is not a solution to $y > x+4$.

(e) Substitute $-8$ for $x$ and $12$ for $y$.

So, $(-8,12)$ is a solution to $y > x+4$.

For the inequality $y > x-3$, determine whether each ordered pair is a solution.

For the inequality $y < x+1$, determine whether each ordered pair is a solution.

Recognize the relation between the solutions of an inequality and its graph

Now, we will look at how the solutions of an inequality relate to its graph.

Let’s think about the number line shown previously again. The point $x=3$ separated that number line into two parts. On one side of $3$ are all the numbers less than $3$. On the other side of $3$ all the numbers are greater than $3$.

Similarly, the line $y=x+4$ separates the plane into two regions. On one side of the line are points with $y<x+4$. On the other side of the line are the points with $y>x+4$. We call the line $y=x+4$ a boundary line.

For an inequality in one variable, the endpoint is shown with a parenthesis or a bracket depending on whether or not it is included in the solution. Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to show whether or not the line is included in the solution.

Now, let’s take a look at what we found in the preceding example. We’ll start by graphing the line $y=x+4$, and then we’ll plot the five points we tested.

Some of the points were solutions to $y>x+4$ and some were not. The points $(1,6)$ and $(-8,12)$ are solutions. Notice that they are both on the same side of the boundary line $y=x+4$.

The two points $(0,0)$ and $(-5,-15)$ are on the other side of the boundary line, and they are not solutions to $y>x+4$. For those two points, $y<x+4$.

What about the point $(2,6)$? Because $6=2+4$, the point is a solution to the equation $y=x+4$, but not a solution to the inequality $y>x+4$. So the point $(2,6)$ is on the boundary line.

Let’s take another point above the boundary line and test whether or not it is a solution to $y>x+4$. The point $(0,10)$ clearly looks to be above the boundary line. Is it a solution to the inequality?

So, $(0,10)$ is a solution to $y>x+4$. Any point you choose above the boundary line is a solution to the inequality. All points above the boundary line are solutions. Similarly, all points below the boundary line are not solutions to $y>x+4$.

The line $y=x+4$ divides the plane into two regions. The shaded side shows the solutions to $y>x+4$. The points on the boundary line, those where $y=x+4$, are not solutions, so the line itself is not part of the solution. We show that by making the line dashed, not solid.

Example. The boundary line shown in this graph is $y=2x-1$. Write the inequality shown by the graph. The boundary line is solid, and the side containing $(0,0)$ is shaded.

The line $y=2x-1$ is the boundary line. On one side of the line are the points with $y>2x-1$ and on the other side are the points with $y<2x-1$. Let’s test the point $(0,0)$ and see which inequality describes its position relative to the boundary line.

At $(0,0)$, which inequality is true: $y>2x-1$ or $y<2x-1$?

Since $y>2x-1$ is true, the side of the line with $(0,0)$ is the solution. The shaded region shows the solution of $y>2x-1$. Since the boundary line is graphed with a solid line, the inequality includes the equal sign. The graph shows the inequality $y\geq2x-1$.

We could use any point as a test point, provided it is not on the line. We chose $(0,0)$ because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that $y<2x-1$.

Example. The boundary line shown in this graph is $2x+3y=6$. Write the inequality shown by the graph. The boundary line is dashed, and the side containing $(0,0)$ is shaded.

The line $2x+3y=6$ is the boundary line. On one side are the points with $2x+3y>6$ and on the other side are the points with $2x+3y<6$. Let’s test the point $(0,0)$ and see which inequality describes its side.

So the side with $(0,0)$ is the side where $2x+3y<6$. You may want to pick a point on the other side and check that $2x+3y>6$. Since the boundary line is dashed, the inequality does not include an equal sign. The shaded region shows the solution to $2x+3y<6$.

Graph linear inequalities in two variables

Now that we know what the graph of a linear inequality looks like and how it relates to a boundary equation we can use this knowledge to graph a given linear inequality.

Example. Graph the linear inequality $y\geq\tfrac{3}{4}x-2$.

Step 1. Identify and graph the boundary line. Replace the inequality sign with an equal sign to find the boundary line. Graph the boundary line $y=\tfrac{3}{4}x-2$. The inequality sign is $\geq$, so we draw a solid line.

Step 2. Test a point that is not on the boundary line. Is it a solution of the inequality? We’ll test $(0,0)$.

Since $0\geq-2$, $(0,0)$ is a solution.

Step 3. Shade in one side of the boundary line. The test point $(0,0)$ is a solution to $y\geq\tfrac{3}{4}x-2$, so we shade in that side. All points in the shaded region and on the boundary line represent the solutions.

Example. Graph the linear inequality $x-2y<5$.

First, we graph the boundary line $x-2y=5$. The inequality is $<$ so we draw a dashed line.

Then, we test a point. We’ll use $(0,0)$ again because it is easy to evaluate and it is not on the boundary line.

The point $(0,0)$ is a solution of $x-2y<5$, so we shade in that side of the boundary line. All points in the shaded region, but not those on the boundary line, represent the solutions.

What if the boundary line goes through the origin? Then, we won’t be able to use $(0,0)$ as a test point. No problem—we’ll just choose some other point that is not on the boundary line.

Example. Graph the linear inequality $y\leq-4x$.

First, we graph the boundary line $y=-4x$. It is in slope-intercept form, with $m=-4$ and $b=0$. The inequality is $\leq$ so we draw a solid line.

Now we need a test point. We can see that the point $(1,0)$ is not on the boundary line. Is $(1,0)$ a solution of $y\leq-4x$?

The point $(1,0)$ is not a solution, so we shade in the opposite side of the boundary line. All points in the shaded region and on the boundary line represent the solutions.

Some linear inequalities have only one variable. They may have an $x$ but no $y$, or a $y$ but no $x$. In these cases, the boundary line will be either a vertical or a horizontal line. Recall that $x=a$ is a vertical line and $y=b$ is a horizontal line.

Example. Graph the linear inequality $y>3$.

First, we graph the boundary line $y=3$. It is a horizontal line. The inequality is $>$ so we draw a dashed line. We test the point $(0,0)$. Since $0\not>3$, $(0,0)$ is not a solution. So we shade the side that does not include $(0,0)$.

All points in the shaded region, but not those on the boundary line, represent the solutions to $y>3$.

Solve applications using linear inequalities in two variables

Many fields use linear inequalities to model a problem. While our examples may be about simple situations, they give us an opportunity to build our skills and to get a feel for how they might be used.

Example. Hilaria works two part time jobs in order to earn enough money to meet her obligations of at least $240 a week. Her job in food service pays $10 an hour and her tutoring job on campus pays $15 an hour. How many hours does Hilaria need to work at each job to earn at least $240?

(a) Let $x$ be the number of hours she works at the job in food service and let $y$ be the number of hours she works tutoring. Write an inequality that would model this situation.

We let $x$ be the number of hours she works at the job in food service and let $y$ be the number of hours she works tutoring. She earns $10 per hour at the job in food service and $15 an hour tutoring. At each job, the number of hours multiplied by the hourly wage will give the amount earned at that job. The amount earned at the food service job plus the amount earned tutoring is at least $240:

(b) Graph the inequality. To graph it, we put it in slope-intercept form.

(c) From the graph, we see that the ordered pairs $(15,10)$, $(0,16)$, $(24,0)$ represent three of infinitely many solutions. Check the values in the inequality.

For Hilaria, it means that to earn at least $240, she can work 15 hours tutoring and 10 hours at her fast-food job, earn all her money tutoring for 16 hours, or earn all her money while working 24 hours at the job in food service.

Hugh works two part time jobs. One at a grocery store that pays $10 an hour and the other is babysitting for $13 hour. Between the two jobs, Hugh wants to earn at least $260 a week. How many hours does Hugh need to work at each job to earn at least $260?

Three ordered pairs that are solutions are $(0,20)$, $(13,10)$, and $(26,0)$. They mean Hugh can earn at least $260 by babysitting 20 hours, working 13 hours at each job, or working 26 hours at the grocery store.

Veronica works two part time jobs in order to earn enough money to meet her obligations of at least $280 a week. Her job at the day spa pays $10 an hour and her administrative assistant job on campus pays $17.50 an hour. How many hours does Veronica need to work at each job to earn at least $280?

Three ordered pairs that are solutions are $(0,16)$, $(14,8)$, and $(28,0)$. They mean Veronica can earn at least $280 by working 16 hours as an administrative assistant, working 14 hours at the day spa and 8 hours as an administrative assistant, or working 28 hours at the day spa.

Key terms

linear inequality — an inequality that can be written as $Ax+By>C$, $Ax+By\geq C$, $Ax+By<C$, or $Ax+By\leq C$, where $A$ and $B$ are not both zero. solution to a linear inequality — an ordered pair $(x,y)$ that makes the inequality true when the values are substituted. boundary line — the line $Ax+By=C$ that separates the region where $Ax+By>C$ from the region where $Ax+By<C$.

This section is adapted from Intermediate Algebra 2e, Section 3.4: Graph Linear Inequalities in Two Variables by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated number-line and coordinate-plane figures as accessible interactive graphics; omitted the Be Prepared quiz, Media link, Self Check, and Section Exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.