Graphing Systems of Linear Inequalities

4.7 Graphing Systems of Linear Inequalities

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

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations.

SYSTEM OF LINEAR INEQUALITIES

Two or more linear inequalities grouped together form a system of linear inequalities.

A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here.

$\Bigg\{ \begin{align*} &x+4y≥10 \\ &3x-2y<12 \end{align*}$

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs $(x, y)$ that make both inequalities true.

SOLUTIONS OF A SYSTEM OF LINEAR INEQUALITIES

Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true.

The solution of a system of linear inequalities is shown as a shaded region in the $x$, $y$ coordinate system that includes all the points whose ordered pairs make the inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Example 1

Determine whether the ordered pair is a solution to the system $\Bigg\{ \begin{align*} &x+4y≥10 \\ &3x-2y<12 \end{align*}$

  • $(-2, 4)$
  • $(3, 1)$
Solution

Part 1– Is the ordered pair $(-2, 4)$ a solution?

We substitute $\textcolor{red}{x=-2}$ and $\textcolor{blue}{y=4}$ into both inequalities.

$\begin{align*} \textcolor{red}{x}+4\textcolor{blue}{y}&≥10 \\ \textcolor{red}{-2}+4(\textcolor{blue}{4})&≥10 \\ 14&≥10 \text{ true} \end{align*}$$\begin{align*} 3\textcolor{red}{x}-2\textcolor{blue}{y}&<12 \\ 3(\textcolor{red}{-2})-2(\textcolor{blue}{4})&<12 \\ -14&<12 \text{ true} \end{align*}$

The ordered pair $(-2, 4)$ made both inequalities true. Therefore $(-2, 4)$ is a solution to this system.

Part 2– Is the ordered pair $(3, 1)$ a solution?

We substitute $\textcolor{red}{x=3}$ and $\textcolor{blue}{y=1}$ into both inequalities.

$\begin{align*} \textcolor{red}{x}+4\textcolor{blue}{y}&≥10 \\ \textcolor{red}{3}+4(\textcolor{blue}{1})&≥10 \\ 7&≥10 \text{ false} \end{align*}$$\begin{align*} 3\textcolor{red}{x}-2\textcolor{blue}{y}&<12 \\ 3(\textcolor{red}{3})-2(\textcolor{blue}{1})&<12 \\ 7&<12 \text{ true} \end{align*}$

The ordered pair $(3, 1)$ made one inequality true, but the other one false. Therefore $(3, 1)$ is not a solution to this system.

4.7.2 Solve a System of Linear Inequalities by Graphing

The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

Example 2

Solve the system by graphing: $\Bigg\{ \begin{align*} y&≥2x-1 \\ y&<x+1 \end{align*}$

Solution
Step 1. Graph the first inequality.
•Graph the boundary line.
• Shade in the side of the boundary line where the inequality is true.
We will graph $\textcolor{red}{y≥2x-1}$.
We graph the line $y=2x-1$. It is a solid line because the inequality sign is $≥$.
We choose $(0, 0)$ as a test point. It is a solution to $y≥2x-1$, so we shade in above the boundary line.
$\Bigg\{\begin{align*} y&≥2x-1 \\ y&<x+1 \end{align*}$
Step 2. On the same grid, graph the second inequality.
•Graph the boundary line.
• Shade in the side of that boundary line where the inequality is true.
We will graph $\textcolor{blue}{y<x+1}$ on the same grid.
We graph the line $y=x+1$. It is a dashed line because the inequality sign is <.
Again, we use $(0, 0)$ as a test point. It is a solution, so we share in that side of the line $y=x+1$.
Step 3. The solution is the region where the shading overlaps.The point where the boundary lines intersect is not a solution because it is not a solution to $y<x+1$.The solution is all points in the area shaded twice—which appears as the darkest shaded region.
Step 4. Check by choosing a test point. We’ll use $(-1, -1)$ as a test point. Is $(\textcolor{red}{-1}, \textcolor{blue}{-1})$ a solution to $y≥2x-1$?
$\textcolor{blue}{-1}≥2(\textcolor{red}{-1})-1$
$-1≥-3$ true
Is $(\textcolor{red}{-1}, \textcolor{blue}{-1})$ a solution to $y<x+1$?
$\textcolor{blue}{-1}<(\textcolor{red}{-1})+1$
$-1<0$ true
The region containing $(-1, -1)$ is the solution to this system.

HOW TO: Solve a system of linear inequalities by graphing.

  1. Graph the first inequality.
    • Graph the boundary line.
    • Shade in the side of the boundary line where the inequality is true.
  2. On the same grid, graph the second inequality.
    • Graph the boundary line.
    • Shade in the side of that boundary line where the inequality is true.
  3. The solution is the region where the shading overlaps.
  4. Check by choosing a test point.

Example 3

Solve the system by graphing: $\Bigg\{ \begin{align*} &x-y>3 \\ &y<-\frac{1}{5}x+4 \end{align*}$

Solution
$\Bigg\{ \begin{align*} &x-y>3 \\ &y<-\frac{1}{5}x+4 \end{align*}$
Graph $x-y>3$, by graphing $x-y=3$ and testing a point.

The intercepts are $x=3$ and $y=-3$ and the boundary line will be dashed.

Test $(0, 0)$ which makes the inequality false so shade (red) the side that does not contain $(0, 0)$.
Graph $y<-\frac{1}{5}x+4$ by graphing $y=-\frac{1}{5}x+4$ using the slope $m=-\frac{1}{5}$ and $y$-intercept $b=4$.
The boundary line will be dashed.
Test $(0, 0)$ which makes the inequality true, so shade (blue) the side that contains $(0, 0)$.
Choose a test point in the solution and verify that it is a solution to both inequalities.

The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region.

Example 4

Solve the system by graphing: $\Bigg\{ \begin{align*} &x-2y<5 \\ &y>-4 \end{align*}$

Solution
$\Bigg\{ \begin{align*} &x-2y<5 \\ &y>-4 \end{align*}$
Graph $x-2y<5$, by graphing $x-2y=5$ and testing a point. The intercepts are $x=5$ and $y=-2.5$ and the boundary line will be dashed.

Test $(0, 0)$ which makes the inequality true, so shade (red) the side that contains $(0, 0)$.
Graph $y>-4$ by graphing $y=-4$ and recognizing that it is a horizontal line through $y=-4$. The boundary line will be dashed.
Test $(0, 0)$ which makes the inequality true, so shade (blue) the side that contains $(0, 0)$.

The point $(0, 0)$ is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed.

The solution is the area shaded twice—which appears as the darkest shaded region.

Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in the next example.

Example 5

Solve the system by graphing: $\Bigg\{ \begin{align*} &4x+3y≥12 \\ &y<-\frac{4}{3}x+1 \end{align*}$

Solution
$\Bigg\{ \begin{align*} &4x+3y≥12 \\ &y<-\frac{4}{3}x+1 \end{align*}$
Graph $4x+3y≥12$, by graphing $4x+3y=12$ and testing a point. The intercepts are $x=3$ and $y=4$ and the boundary line will be solid.

Test $(0, 0)$ which makes the inequality false, so shade (red) the side that does not contain $(0, 0)$.
Graph $y<-\frac{4}{3}x+1$ by graphing $y=-\frac{4}{3}x+1$ using the slope $m=-\frac{4}{3}$ and $y$-intercept $b=1$. The boundary line will be dashed.
Test $(0, 0)$ which makes the inequality true, so shade (blue) the side that contains $(0, 0)$.

There is no point in both shaded regions, so the system has no solution.

Some systems of linear inequalities where the boundary lines are parallel will have a solution. We’ll see this in the next example.

Example 6

Solve the system by graphing: $\Bigg\{ \begin{align*} &y>\frac{1}{2}x-4 \\ &x-2y<-4 \end{align*}$

Solution
$\Bigg\{ \begin{align*} &y>\frac{1}{2}x-4 \\ &x-2y<-4 \end{align*}$
Graph $y>\frac{1}{2}x-4$, by graphing $y=\frac{1}{2}x-4$ using the slope $m=\frac{1}{2}$ and the $y$-intercept $b=-4$. The boundary line will be dashed.

Test $(0, 0)$ which makes the inequality true, so shade (red) the side that contains $(0, 0)$.
Graph $x-2y<-4$ by graphing $x-2y=-4$ using and testing a point. The intercepts are $x=-4$ and $y=2$ and the boundary line will be dashed.
Test $(0, 0)$ which makes the inequality false, so shade (blue) the side that does not contain $(0, 0)$.

No point on the boundary lines is included in the solution as both lines are dashed.

The solution is the region that is shaded twice which is also the solution to $x-2y<-4$.

4.7.3 Solve Applications of Systems of Inequalities

The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so we add inequalities to the system as additional requirements.

Example 7

Christy sells her photographs at a booth at a street fair. At the start of the day, she wants to have at least $25$ photos to display at her booth. Each small photo she displays costs her $\$4$ and each large photo costs her $\$10$. She doesn’t want to spend more than $\$200$ on photos to display.

  • Write a system of inequalities to model this situation.
  • Graph the system.
  • Could she display 10 small and 20 large photos?
  • Could she display 20 large and 10 small photos?
Solution

Part 1

To find the system of equations, translate the information.Let $x=$ the number of small photos and
$y=$ the number of large photos.
She wants to have at least $25$ photos. So the number of small photos plus the number of large should be at least $25$$x+y≥25$
$\$4$ for each small and $\$10$ for each large must be no more than $\$200$.$4x+10y≤200$
The number of small photos must be greater than or equal to $0$.$x≥0$
The number of large photos must be greater than or equal to $0$.$y≥0$
We have our system of equations.$\left\{ \begin{align*} &x+y≥25 \\ &4x+10y≤200 \\ &x≥0 \\
&y≥0 \end{align*}\right. $

Part 2

Since $x≥0$ and $y≥0$ (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.

To graph $x+y≥25$, graph $x+y=25$ as a solid line.

Choose $(0, 0)$ as a test point. Since it does not make the inequality true, shade (red) the side that does not include the point $(0, 0)$.

To graph $4x+10y≤200$, graph $4x+10y=200$ as a solid line.
Choose (0, 0) as a test point. Since it does make the inequality true, shade (blue) the side that includes the point (0, 0).

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly-shaded section are included in the solution as are the points on the $x$-axis from $(25, 0)$ to $(55, 0)$.

Part 3

To determine if $10$ small and $20$ large photos would work, we look at the graph to see if the point $(10, 20)$ is in the solution region. We could also test the point to see if it is a solution of both equations.It is not, so Christy would not display $10$ small and $20$ large photos.

Part 4

To determine if $20$ small and $10$ large photos would work, we look at the graph to see if the point $(20, 10)$ is in the solution region. We could also test the point to see if it is a solution of both equations. It is, so Christy could choose to display $20$ small and $10$ large photos.

Notice that we could also test the possible solutions by substituting the values into each inequality.

When we use variables other than $x$ and $y$ to define an unknown quantity, we must change the names of the axes of the graph as well.

Example 8

Omar needs to eat at least $800$ calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than $\$5$. At the hamburger restaurant near his college, each hamburger has $240$ calories and costs $\$1.40$. Each cookie has $160$ calories and costs $\$0.50$.

  • Write a system of inequalities to model this situation.
  • Graph the system.
  • Could he eat $3$ hamburgers and $1$ cookie?
  • Could he eat $2$ hamburgers and $4$ cookies?
Solution

Part 1

To find the system of equations translate the information.Let $h=$ the number of hamburgers and
$c=$ the number of cookies
The calories from hamburgers at $240$ calories each, plus the calories from cookies at $160$ calories each must be more than $800$.$240h+160c≥800$
The amount spent on hamburgers at $\$1.40$ each, plus the amount spent on cookies at $\$0.50$ each must be no more than $\$5.00$.$1.40h+0.50c≤5$
The number of hamburgers must be greater than or equal to $0$.$h≥0$
The number of cookies must be greater than or equal to $0$.$c≥0$
We have our system of equations.$\left\{ \begin{align*} &240h+160h≥800\\ &1.40h+0.50c≤5 \\ &h≥0 \\ &c≥0 \end{align*} \right.$

Part 2

Since $h>=0$ and $c>=0$ (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.

To graph $240h+160c≥800$, graph $240h+160c=800$ as a solid line.
Choose $(0, 0)$ as a test point.
Since it does not make the
inequality true, shade (red) the side
that does not include the point (0, 0).
To graph $1.40h+0.50c≤5$, graph
the boundary line $1.40h+0.50c=5$.
We test $(0, 0)$ and it makes the
inequality true. We shade the side
of the line that includes $(0, 0)$.

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the $x$-axis from $(5, 0)$ to $(10, 0)$.

Part 3

To determine if $3$ hamburgers and $1$ cookie would meet Omar’s criteria, we see if the point $(3, 1)$ is in the solution region. It is, so Omar might choose to eat $3$ hamburgers and $1$ cookie.

Part 4

To determine if $2$ hamburgers and $4$ cookies would meet Omar’s criteria, we see if the point $(2, 4)$ is in the solution region. It is, Omar might choose to eat $2$ hamburgers and $4$ cookies.

We could also test the possible solutions by substituting the values into each inequality.

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