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# 7.1: Linear Systems by Graphing

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Determine whether an ordered pair is a solution to a system of equations.
• Solve a system of equations graphically.
• Solve a system of equations graphically with a graphing calculator.
• Solve word problems using systems of equations.

## Introduction

In this lesson, we will discover methods to determine if an ordered pair is a solution to a system of two equations. We will then learn to solve the two equations graphically and confirm that the solution is the point where the two lines intersect. Finally, we will look at real-world problems that can be solved using the methods described in this chapter.

## Determine Whether an Ordered Pair is a Solution to a System of Equations

A linear system of equations consists of a set of equations that must be solved together.

Consider the following system of equations.

\begin{align*}y & =x+2 \\ y & =-2x+1\end{align*}

Since the two lines are in a system we deal with them together by graphing them on the same coordinate axes. The lines can be graphed using your favorite method. Let’s graph by making a table of values for each line.

Line 1 \begin{align*}y=x+2\end{align*}

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 2
1 3

Line 2 \begin{align*}y= -2x + 1\end{align*}

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 1
1 -1

A solution for a single equation is any point that lies on the line for that equation. A solution for a system of equations is any point that lies on both lines in the system.

For Example

• Point \begin{align*}A\end{align*} is not a solution to the system because it does not lie on either of the lines.
• Point \begin{align*}B\end{align*} is not a solution to the system because it lies only on the blue line but not on the red line.
• Point \begin{align*}C\end{align*} is a solution to the system because it lies on both lines at the same time.

In particular, this point marks the intersection of the two lines. It solves both equations, so it solves the system. For a system of equations, the geometrical solution is the intersection of the two lines in the system. The algebraic solution is the ordered paid that solves both equations.

You can confirm the solution by plugging it into the system of equations, and confirming that the solution works in each equation.

Example 1

Determine which of the points (1, 3), (0, 2) or (2, 7) is a solution to the following system of equations.

\begin{align*}y & =4x-1 \\ y & =2x+3\end{align*}

Solution

To check if a coordinate point is a solution to the system of equations, we plug each of the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values into the equations to see if they work.

Point (1, 3)

\begin{align*}y & = 4x-1 \\ 3^{?} & = \ ^{?}4(1)-1 \\ 3 & = 3\end{align*}

The solution checks.

\begin{align*}y & = 2x+3 \\ 3^{?} & = \ ^{?}2(1)+3 \\ 3 & \neq 5 \end{align*}

The solution does not check.

Point (1, 3) is on line \begin{align*}y=4x-1\end{align*} but it is not on line \begin{align*}y=2x+3\end{align*} so it is not a solution to the system.

Point (0, 2)

\begin{align*}y & = 4x-1 \\ 2^{?} & = \ ^{?}4(0)-1 \\ 2 & \neq -1\end{align*}

The solution does not check.

Point (0, 2) is not on line \begin{align*}y=4x-1\end{align*} so it is not a solution to the system. Note that it is not necessary to check the second equation because the point needs to be on both lines for it to be a solution to the system.

Point (2, 7)

\begin{align*}y & = 4x-1 \\ 7^{?} & = \ ^{?}4(2)-1 \\ 7 & = 7\end{align*}

The solution checks.

\begin{align*}y & =2x+3 \\ 7^{?} & = \ ^{?}2(2)+3 \\ 7 & = 7\end{align*}

The solution checks.

Point (2, 7) is a solution to the system since it lies on both lines.

Answer The solution to the system is point (2, 7).

## Determine the Solution to a Linear System by Graphing

The solution to a linear system of equations is the point which lies on both lines. In other words, the solution is the point where the two lines intersect.

We can solve a system of equations by graphing the lines on the same coordinate plane and reading the intersection point from the graph.

This method most often offers only approximate solutions. It is exact only when the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values of the solution are integers. However, this method it is a great at offering a visual representation of the system of equations and demonstrates that the solution to a system of equations is the intersection of the two lines in the system.

Example 2

(The equations are in slope-intercept form)

Solve the following system of equations by graphing.

\begin{align*}y & =3x-5 \\ y & =-2x+5\end{align*}

Solution

Graph both lines on the same coordinate axis using any method you like.

In this case, let’s make a table of values for each line.

Line 1 \begin{align*} y=3x-5\end{align*}

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
1 -2
2 1

Line 2 \begin{align*}y=-2x+5\end{align*}

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
1 3
2 1

Answer The solution to the system is given by the intersection point of the two lines. The graph shows that the lines intersect at point (2, 1). So the solution is \begin{align*}x = 2, y = 1\end{align*} or (2, 1).

Example 3

(The equations are in standard form)

Solve the following system of equations by graphing

\begin{align*}2x+3y & = 6 \\ 4x-y & = -2\end{align*}

Solution

Graph both lines on the same coordinate axis using your method of choice.

Here we will graph the lines by finding the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts of each of the lines.

Line 1 \begin{align*}2x + 3y = 6 \end{align*}

\begin{align*}x-\end{align*}intercept set \begin{align*}y=0 \Rightarrow 2x=6 \Rightarrow x=3\end{align*} which results in point (3, 0).

\begin{align*}y-\end{align*}intercept set \begin{align*}x=0 \Rightarrow 3y=6 \Rightarrow y=2\end{align*} which results in point (0, 2).

Line 2 \begin{align*}-4x + y = 2\end{align*}

\begin{align*}x-\end{align*}intercept: set \begin{align*}y=0\Rightarrow -4x=2 \Rightarrow x=-\frac{1}{2}\end{align*} which results in point \begin{align*}\left(-\frac{1}{2},0\right)\end{align*}.

\begin{align*}y-\end{align*}intercept: set \begin{align*}x=0\Rightarrow y=2 \end{align*} which results in point (0, 2)

Answer The graph shows that the lines intersect at point (0, 2). Therefore, the solution to the system of equations is \begin{align*}x=0, \ y=2\end{align*}.

Example 4:

Solve the following system by graphing.

\begin{align*}y & = 3 \\ x+y & = 2\end{align*}

Line 1 \begin{align*}y=3\end{align*} is a horizontal line passing through point (0, 3).

Line 2 \begin{align*}x+y=2\end{align*}

\begin{align*}x-\end{align*}intercept: (2, 0)

\begin{align*}y-\end{align*}intercept: (0, 2)

Answer The graph shows that the solution to this system is (-1, 3) \begin{align*}x=-1, \ y=3\end{align*}.

These examples are great at demonstrating that the solution to a system of linear equations means the point at which the lines intersect. This is, in fact, the greatest strength of the graphing method because it offers a very visual representation of system of equations and its solution. You can see, however, that determining a solution from a graph would require very careful graphing of the lines, and is really only practical when you are certain that the solution gives integer values for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. In most cases, this method can only offer approximate solutions to systems of equations. For exact solutions other methods are necessary.

## Solving a System of Equations Using the Graphing Calculator

A graphing calculator can be used to find or check solutions to a system of equations. In this section, you learned that to solve a system graphically, you must graph the two lines on the same coordinate axes and find the point of intersection. You can use a graphing calculator to graph the lines as an alternative to graphing the equations by hand.

Example 6

Solve the following system of equations using a graphing calculator.

\begin{align*}x-3y & = 4 \\ 2x+5y & = 8\end{align*}

In order to input the equations into the calculator, they must be written in slope-intercept form (i.e., \begin{align*}y=mx+b\end{align*} form), or at least you must isolate \begin{align*}y\end{align*}.

\begin{align*} x - 3y & = 4 &&&& y = \frac{1}{3}x - \frac {4}{3} \\ && \Rightarrow && \\ 2x + 5y & = 8 &&&& y = \frac{-2}{5}x - \frac {8}{5} \end{align*}

Press the [y=] button on the graphing calculator and enter the two functions as:

\begin{align*}Y_1 & = \frac{x}{3}-\frac{4}{3} \\ T_2 & = -\frac{2x}{3}-\frac{8}{5}\end{align*}

Now press [GRAPH]. The window below is set to \begin{align*}-5 \leq x \leq 10\end{align*} and \begin{align*}-5 \leq x\leq 5\end{align*}.

The first screen below shows the screen of a TI-83 family graphing calculator with these lines graphed.

There are a few different ways to find the intersection point.

Option 1 Use [TRACE] and move the cursor with the arrows until it is on top of the intersection point. The values of the coordinate point will be on the bottom of the screen. The second screen above shows the values to be \begin{align*}X = 4.0957447\end{align*} and \begin{align*}Y =.03191489\end{align*}.

Use the [ZOOM] function to zoom into the intersection point and find a more accurate result. The third screen above shows the system of equations after zooming in several times. A more accurate solution appears to be \begin{align*}X = 4\end{align*} and \begin{align*}Y = 0\end{align*}.

Option 2 Look at the table of values by pressing [2nd] [GRAPH]. The first screen below shows a table of values for this system of equations. Scroll down until the values for and are the same. In this case this occurs at \begin{align*}X = 4\end{align*} and \begin{align*}Y = 0\end{align*}.

Use the [TBLSET] function to change the starting value for your table of values so that it is close to the intersection point and you don’t have to scroll too long. You can also improve the accuracy of the solution by taking smaller values of Table 1.

Option 3 Using the [2nd] [TRACE] function gives the screen in the second screen above.

Scroll down and select intersect.

The calculator will display the graph with the question [FIRSTCURVE]? Move the cursor along the first curve until it is close to the intersection and press [ENTER].

The calculator now shows [SECONDCURVE]?

Move the cursor to the second line (if necessary) and press [ENTER].

The calculator displays [GUESS]?

Press [ENTER] and the calculator displays the solution at the bottom of the screen (see the third screen above).

The point of intersection is \begin{align*}X = 4\end{align*} and \begin{align*}Y = 0\end{align*}.

Notes:

• When you use the “intersect” function, the calculator asks you to select [FIRSTCURVE]? and [SECONDCURVE]? in case you have more than two graphs on the screen. Likewise, the [GUESS]? is requested in case the curves have more than one intersection. With lines you only get one point of intersection, but later in your mathematics studies you will work with curves that have multiple points of intersection.
• Option 3 is the only option on the graphing calculator that gives an exact solution. Using trace and table give you approximate solutions.

## Solve Real-World Problems Using Graphs of Linear Systems

Consider the following problem

Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?

Draw a sketch

At time, \begin{align*}t = 0\end{align*}:

Formulas

Let’s define two variables in this problem.

\begin{align*}t =\end{align*} the time from when Nadia starts running

\begin{align*}d=\end{align*} the distance of the runners from the starting point.

Since we have two runners we need to write equations for each of them. This will be the system of equations for this problem.

Here we use the formula distance = speed \begin{align*}\times\end{align*} time

Nadia’s equation \begin{align*}d = 6t\end{align*}

Peter’s equation \begin{align*}d = 5t + 20 \end{align*}

(Remember that Peter was already 20 feet from the starting point when Nadia started running.)

Let’s graph these two equations on the same coordinate graph.

Time should be on the horizontal axis since it is the independent variable.

Distance should be on the vertical axis since it is the dependent variable.

We can use any method for graphing the lines. In this case, we will use the slope-intercept method since it makes more sense physically.

To graph the line that describes Nadia’s run, start by graphing the \begin{align*}y-\end{align*}intercept (0, 0). If you do not see that this is the \begin{align*}y-\end{align*}intercept, try plugging in the test-value of \begin{align*}x = 0\end{align*}.

The slope tells us that Nadia runs 6 feet every one second so another point on the line is (1, 6). Connecting these points gives us Nadia’s line.

To graph the line that describes Peter’s run, again start with the \begin{align*}y-\end{align*}intercept. In this case, this is the point (0, 20).

The slope tells us that Peter runs 5 feet every one second so another point on the line is (1, 25). Connecting these points gives us Peter’s line.

In order to find when and where Nadia and Peter meet, we will graph both lines on the same graph and extend the lines until they cross. The crossing point is the solution to this problem.

The graph shows that Nadia and Peter meet 20 seconds after Nadia starts running and 120 feet from the starting point.

## Review Questions

Determine which ordered pair satisfies the system of linear equations.

1. \begin{align*}y = 3x -2\!\\ y = -x\end{align*}
1. (1, 4)
2. (2, 9)
3. \begin{align*}\left(\frac{1}{2}, -\frac{1}{2}\right)\end{align*}
2. \begin{align*}y = 2x -3\!\\ y = x + 5\end{align*}
1. (8, 13)
2. (-7, 6)
3. (0, 4)
3. \begin{align*}2x + y = 8\!\\ 5x + 2y = 10\end{align*}
1. (-9, 1)
2. (-6, 20)
3. (14, 2)
4. \begin{align*}3x + 2y = 6\!\\ y = \frac{x}{2} - 3\end{align*}
1. \begin{align*}\left(3, -\frac{3}{2}\right)\end{align*}
2. \begin{align*}(-4, 3) \end{align*}
3. \begin{align*}\left(\frac{1}{2}, 4\right)\end{align*}

Solve the following systems using the graphing method.

1. \begin{align*}y = x+ 3\!\\ y = -x + 3\end{align*}
2. \begin{align*}y = 3x- 6\!\\ y = -x + 6\end{align*}
3. \begin{align*}2x = 4\!\\ y=-3\end{align*}
4. \begin{align*}y = -x + 5\!\\ -x + y = 1\end{align*}
5. \begin{align*}x + 2y = 8\!\\ 5x + 2y = 0\end{align*}
6. \begin{align*}3x +2y = 12\!\\ 4x-y = 5\end{align*}
7. \begin{align*}5x + 2y = -4\!\\ x- y = 2\end{align*}
8. \begin{align*}2x + 4 = 3y\!\\ x- 2y + 4 = 0\end{align*}
9. \begin{align*}y = \frac{x}{2}- 3\!\\ 2x-5y = 5\end{align*}
10. \begin{align*}y = 4\!\\ x = 8 -3y\end{align*}

Solve the following problems by using the graphing method.

1. Mary’s car is 10 years old and has a problem. The repair man indicates that it will cost her $1200 to repair her car. She can purchase a different, more efficient car for$4500. Her present car averages about $2000 per year for gas while the new car would average about$1500 per year. Find the number of years for when the total cost of repair would equal the total cost of replacement.
2. Juan is considering two cell phone plans. The first company charges $120 for the phone and$30 per month for the calling plan that Juan wants. The second company charges $40 for the same phone, but charges$45 per month for the calling plan that Juan wants. After how many months would the total cost of the two plans be the same?
3. A tortoise and hare decide to race 30 feet. The hare, being much faster, decided to give the tortoise a head start of 20 feet. The tortoise runs at 0.5 feet/sec and the hare runs at 5.5 feet per second. How long will it be until the hare catches the tortoise?

## Review Answers

1. (c)
2. (a)
3. (b)
4. (a)
5. (0, 3)
6. (3, 3)
7. (2, -3)
8. (2, 3)
9. (-2, 5)
10. (2, 3)
11. (0, -2)
12. (4, 4)
13. (20, 7)
14. (-4, 4)
15. 6.6 years
16. 5.33 months
17. 4.0 seconds

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CK.MAT.ENG.SE.1.Algebra-I.7.1