**3.4 Graph Linear Inequalities in Two Variables**

Topics covered in this section are:

- Verify solutions to an inequality in two variables.
- Recognize the relation between the solutions of an inequality and its graph.
- Graph linear inequalities in two variables.
- Solve applications using linear inequalities in two variables.

**3.4.1 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.

**LINEAR INEQUALITY**

A **linear inequality** is an inequality that can be written in one of the following forms:

$\ \ \ \ \ \ \ Ax+By > C \ \ \ \ \ \ \ Ax+By≥C \ \ \ \ \ \ \ Ax+By<C \ \ \ \ \ \ \ Ax+By≤C \ \ \ \ \ \ \ \ $

Where $A$ and $B$ are not both zero.

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 $3$. See Figure 3.10.

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**.

**SOLUTION TO A LINEAR INEQUALITY**

An ordered pair $(x, y)$ is a **solution to a linear inequality** if the inequality is true when we substitute the values of $x$ and $y$.

**Example 1**

Determine whether each ordered pair is a solution to the inequality $y>x+4$:

- $(0, 0)$
- $(1, 6)$
- $(2, 6)$
- $(-5, -15)$
- $(-8, 12)$

**Solution**

**Part 1**

$(0, 0)$ | $y>x+4$ |

Subsitute $\textcolor{red}{0}$ for $x$ and $\textcolor{blue}{0}$ for $y$. | $\textcolor{blue}{0} > \textcolor{red}{0} +4$ |

Simplify. | $0 \cancel{>} 4$ |

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

**Part 2**

$(1, 6)$ | $y>x+4$ |

Subsitute $\textcolor{red}{1}$ for $x$ and $\textcolor{blue}{6}$ for $y$. | $\textcolor{blue}{6} > \textcolor{red}{1} +4$ |

Simplify. | $6 > 5$ |

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

**Part 3**

$(2, 6)$ | $y>x+4$ |

Subsitute $\textcolor{red}{2}$ for $x$ and $\textcolor{blue}{6}$ for $y$. | $\textcolor{blue}{6} > \textcolor{red}{2} +4$ |

Simplify. | $6 \cancel{>} 6$ |

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

**Part 4**

$(-5, -15)$ | $y>x+4$ |

Subsitute $\textcolor{red}{-5}$ for $x$ and $\textcolor{blue}{-15}$ for $y$. | $\textcolor{blue}{-15} > \textcolor{red}{-5} +4$ |

Simplify. | $-15 \cancel{>} -1$ |

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

**Part 5**

$(-8, 12)$ | $y>x+4$ |

Subsitute $\textcolor{red}{-8}$ for $x$ and $\textcolor{blue}{12}$ for $y$. | $\textcolor{blue}{12} > \textcolor{red}{-8} +4$ |

Simplify. | $12 > -4$ |

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

**3.4.2 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 in 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 $3$. See Figure 3.11.

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**.

**BOUNDARY LINE**

The line with equation $Ax+By=C$ is the **boundary line** that separates the region where $Ax+By>C$ from the region where $Ax+By<C$.

For an inequality in one variable, the endpoint is shown with a parenthesis or a bracket depending on whether or not $a$ 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 it the line is included in the solution.

$Ax+By<C$ | $Ax+By≤C$ |

$Ax+By>C$ | $Ax+By≥C$ |

Boundary line is $Ax+By=C$ | Boundary line is $Ax+By=C$ |

Boundary line is not included in solution. | Boundary line is included in solution. |

Boundary line is dashed. | Boundary line is solid. |

Now, let’s take a look at what we found in Example 1. We’ll start by graphing the line $y=x+4$, and then we’ll plot the five points we tested, as shown in the graph. See Figure 3.12.

In Example 1 we found that some of the points were solutions to the inequality $y>x+4$ and some were not.

Which of the points we plotted are solutions to the inequality $y>x+4$?

The points $(1, 6)$ and $(−8, 12)$ are solutions to the inequality $y>x+4$. 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 $y=x+4$, and they are not solutions to the inequality $y>x+4.$ 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 the inequality $y>x+4$. The point $(0, 10)$ clearly looks to above the boundary line, doesn’t it? Is it a solution to the inequality?

$\begin{align*} y&>x+4 \\ 10&>0+4 \\ 10&>4 \end{align*}$

So, $(0, 10)$ is a solution to $y>x+4$.

Any point you choose above the boundary line is a solution to the inequality $y>x+4$. All points above the boundary line are solutions.

Similarly, all points below the boundary line, the side with $(0, 0)$ and $(−5, −15)$, are not solutions to $y>x+4$, as shown in Figure 3.13.

The graph of the inequality $y>x+4$ is shown in below.

The line $y=x+4$ divides the plane into two regions. The shaded side shows the solutions to the inequality $y>x+4$.

The points on the boundary line, those where $y=x+4$, are not solutions to the inequality $y>x+4$, so the line itself is not part of the solution. We show that by making the line dashed, not solid.

**Example 2**

The boundary line shown in this graph is $y=2x−1$. Write the inequality shown by the graph.

**Solution**

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 of the line 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 $y<2x-1$?

$y>2x-1$ | $y<2x-1$ |

$0>2 \cdot 0-1$ | $0<2 \cdot 0 – 1$ |

$0 > -1$ True | $0 < -1$ False |

Since, $y>2x−1$ is true, the side of the line with $(0, 0)$, is the solution. The shaded region shows the solution of the inequality $y>2x−1$.

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality $y≥2x−1$.

We could use any point as a test point, provided it is not on the line. Why did we choose $(0, 0)$? 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 3**

The boundary line shown in this graph is $2x+3y=6$. Write the inequality shown by the graph.

**Solution**

The line $2x+3y=6$ is the boundary line. On one side of the line are the points with $2x+3y>6$ and on the other side of the line are the points with $2x+3y<6$.

Let’s test the point $(0, 0)$ and see which inequality describes its side of the boundary line.

At $(0, 0)$, which inequality is true: $2x+3y>6$ or $2x+3y<6$?

$2x+3y>6$ | $2x+3y<6$ |

$2(0)+3(0)>6$ | $2(0)+3(0)<6$ |

$0>6$ False | $0 < 6$ True |

So the side with $(0, 0)$ is the side where $2x+3y<6$.

(You may want to pick a point on the other side of the boundary line and check that $2x+3y>6$.)

Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.

The shaded region shows the solution to the inequality $2x+3y<6$.

**3.4.3 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 4**

Graph the linear inequality $y≥\frac{3}{4}x-2$.

**Solution**

The steps we take to graph a linear inequality are summarized here.

**HOW TO: Graph a linear inequality in two variables.**

- Identify and graph the boundary line.
- If the inequality is $≤$ or $≥$, the boundary line is solid.
- If the inequality is $<$ or $>$, the boundary line is dashed.

- Test a point that is not on the boundary line. Is it a solution of the inequality?
- Shade in one side of the boundary line.
- If the test point is a solution, shade in the side that includes the point.
- If the test point is not a solution, shade in the opposite side.

**Example 5**

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

**Solution**

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

Is $(0, 0)$ a solution of $x-2y<5$?

$\begin{align*} \textcolor{blue}{0} -2(\textcolor{blue}{0}) &< 5 \\ 0 – 0 &<5 \\ 0&<5 \end{align*}$

The point $(0, 0)$ is a solution of $x-2y<5$, so we shade in that side of the boundary line.

All the points in the shaded region, but not those on the boundary line, represent the solution to $x-2y<5$.

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 6**

Graph the linear inequality: $y≤-4x$.

**Solution**

First, we graph the boundary line $y=-4x$. It is in slope-intercept form, with $m=-4$ and $b=0$. The inequality is $≤$ 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 $y≤-4x$?

$\textcolor{blue}{0} ≤-4(1)$

$0 \cancel{≤} 4$

The point $(1, 0)$ is not a solution to $y≤-4x$, so we shade the opposite side of the boundary line.

All points in the shaded region and on the boundary line represent the solutions to $y≤-4x$.

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:

$x=a$ | vertical line |

$y=b$ | horizontal line |

**Example 7**

Graph the linear inequality: $y>3$.

**Solution**

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 $(0, 0)$.

$y>3$

$0\cancel{>}3$

So, $(0, 0)$ is not a solution to $y>3$.

We shade the side the does not include $(0, 0)$ as shown in this graph.

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

**3.4.4 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 8**

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 $\$240$?

- 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.
- Graph the inequality.
- Find three ordered pairs $(x, y)$ that would be solutions to the inequality. Then, explain what that means for Hilaria.

**Solution**

**Part 1 **

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 gives the amount earned at that job.

**Part 2**

To graph the inequality, we put it in slope-intercept form.

$\begin{align*} 10x+15y&≥240 \\ 15y&≥-10x+240 \\ y&≥-\frac{2}{3}x+16 \end{align*}$

**Part 3**

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

$(15, 10)$ | $(0, 16)$ | $(24, 0)$ |

$10x+15y≥240$ | $10x+15y≥240$ | $10x+15y≥240$ |

$10(\textcolor{red}{15})+ 15(\textcolor{red}{10})≥240$ | $10(\textcolor{red}{0}) + 15(\textcolor{red}{16})≥240$ | $10(\textcolor{red}{24}) + 15(\textcolor{red}{0})≥240$ |

$300≥240$ True | $240≥240$ True | $240≥240$ True |

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.

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*Marecek, L., & Mathis, A. H. (2020). Graph Linear Inequalities in Two Variables. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/3-4-graph-linear-inequalities-in-two-variables*.*License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction*