# 9.7: Systems of Linear Functions

**At Grade**Created by: CK-12

## Introduction

*A Second Trip to the Omni*

Kelly loved the Omni presentation on the rainforest so much that she decided to go and see it again. She asked Tyler if he wanted to go with her on Saturday afternoon.

“Do you want to go with me?” Kelly asked.

“Sure, but I have karate first, so I will have to meet you there. What time is the show?” Tyler asked.

“The show starts at 2:30 pm. I’m going to leave at one o’clock so I can look around,” Kelly said.

“Well, I don’t get done karate until then, so I probably won’t leave until 2:00 pm,” Tyler said.

On Saturday, the two went about their day and both left for the museum. Kelly’s Mom tends to drive slowly and cautiously, so she was traveling on city streets at 45 mph. Tyler’s karate class is downtown, so he could take the highway to get to the Omni theater and his Dad drove at an average of 55 mph. Will the two catch up to each other?

**This is a problem about equations and systems of equations. You will need to solve a system of equations to figure this one out. To figure this out, you would need to find a solution that will work for both equations. You will learn all about it in this lesson.**

*What You Will Learn*

In this lesson, you will learn how to do the following skills.

- Recognize a solution of a linear system in two variables as an ordered pair that is a solution of each equation in the system.
- Recognize linear systems with no solutions and linear systems with infinitely many solutions.
- Solve linear systems by graphing.
- Solve linear systems by substitution.
- Model and solve real - world problems using systems of linear equations.

*Teaching Time*

I. **Recognize a Solution of a Linear System in Two Variables as an Ordered Pair that is a Solution of Each Equation in the System**

As useful as you have seen linear functions to be, there are yet other applications of the idea. In a system of linear equations, you will see how linear equations can work together in a system to solve even more complex problems. Indeed, there are various ways that we can find solutions to these problems or find that there may be no solution at all.

If you add two numbers together, you get 13. Can you think of any ordered pairs that would work for this?

(1, 12), (3, 10), (-4, 17), (4.5, 8.5)

Hopefully you’ll agree that there are infinite pairs of numbers whose sum is 13. You might also agree that there are infinite pairs of numbers whose difference is 7.

(9, 2), (11, 4), (37, 30), (95.8, 88.8), (-3, -10)

However, which ordered pair is true for both conditions at once? Which pair has a sum of 13 and a difference of 7?

If you make a list of ordered pairs, you can check them to see which makes both equations true. This is a *system of equations***—two or more equations at the same time.** In the example above, the solution is (10, 3) because the sum of the two numbers is 13 and their difference is 7.

**The pair (10, 3) makes both equations true.**

Example

Which ordered pair makes both equations true?

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

a. (2, 6)

b. (3, 15)

c. (4, 4)

d. (1, 7)

Let’s test each pair and see which pair, if any, works:

a. \begin{align*}x+y &=8 \\ 2+6 &=8? \\ 8 &=8 \\ 4x-y &=-3 \\ 4 \cdot 2-6 &=-3? \\ 8-6 &=-3? \\ 2 &\ne -3 \end{align*}

The ordered pair (2, 6) makes the first equation true, but not the second. Because it is not true for both equations, it is not a solution to the system.

b. \begin{align*}x+y &=8 \\ 3+15 &=8? \\ 18 &\ne 8 \end{align*}

The ordered pair (3, 15) does not even make the first equation true. It cannot be a solution to the system.

c. \begin{align*}x+y &=8 \\ 4+4 &=8? \\ 8 &=8 \\ 4x-y &=-3 \\ 4 \cdot 4-4 &=-3? \\ 16-4 &=-3 \\ 12 &\ne -3\end{align*}

The ordered pair (4, 4) makes the first equation true, but not the second. Because it is not true for both equations, it is not a solution to the system.

d. \begin{align*}x+y &=8 \\ 1+7 &=8? \\ 8 &=8 \\ 4x-y &=-3 \\ 4 \cdot 1-7 &=-3? \\ 4-7 &=-3?\\ -3 & = -3\end{align*}

**The ordered pair (1, 7) makes both equations true. This is a solution to the system.**

Remember, the ordered pair must be a solution to both equations in the system.

II. **Recognize Linear Systems with No Solutions and Linear Systems with Infinitely Many Solutions**

A solution to a system of equations is an ordered pair that makes both equations true. Is there always a solution? Can there be more than one solution? Let’s investigate this with an example.

Example

Two numbers have a sum of 17. If you add two numbers together, their sum is 15. As you know, there are infinite ordered pairs whose sum is 17. There are also infinite ordered pairs whose sum is 15. But can a single ordered pair have a sum of both 17 and 15 at the same time?

**First, let’s write two equations to help us to sort out the information in this system of equations. There are two equations and both have a different sum.**

\begin{align*} x+y &=17 \\ x+y &=15\end{align*}

**If we think about these two equations, you will see that there aren’t any values that will work for both of these equations.**

**Therefore, this system has no solutions.**

Example

Two numbers have a sum of -8. Twice the first number plus twice the second number is -16.

**First, let’s write the two equations described above. Then we can investigate possible solutions.**

\begin{align*} x+y &= -8 \\ 2x+2y &=-16 \end{align*}

**Does this system have a solution? Think of a solution for the first equation. How about (-3, -5)? Does it work for the second? Yes. Think of another solution like (9, -1). This one is also true in both equations.**

**This equation has an infinite number of solutions.**

**Some systems of equations have infinite solutions because all ordered pairs that make one equation true also make the other true.**

III. **Solve Linear Systems by Graphing**

When we have a system of equations, we can graph the equations and solve the system by graphing. When you graph two linear equations, two lines are formed. These two lines will intersect. The point where the lines intersect is the solution for the system. Let’s look at an example.

Example

Solve the following system by graphing.

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

**To work on this system, we will graph both of these lines and look for the point of intersection. That point of intersection will be the solution to the system.**

**The point of intersection is (3, 2).**

**This is the solution to the system of equations.**

**Sometimes you might have two equations that do not intersect. If this happens, then you know that this is a system with no solutions. Parallel lines are one possible system that would not have a solution. Parallel lines have the same slope, so you can recognize a system with no solutions.**

IV. **Solve Linear Systems by Substitution**

Graphing systems of equations is an excellent way of finding their solutions. It is also an easy way to see if a system has no solutions or has infinitely many. However, as you saw in the last example, graphing is not always practical, specifically when the solution is not an integer because fractional numbers can be difficult to read on a graph. Or imagine if your \begin{align*}y\end{align*}-intercept is 5250 and your slope is \begin{align*}\frac{85}{3}\end{align*}. These types of numbers would make graphing a particular challenge. Solving the system algebraically—using properties of equations—is our next best bet.

Example

Let’s recall that in a system of equations, we are looking for the same \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values that make all equations true. So, the solution to the system

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

is the ordered pair \begin{align*}(x, y)\end{align*} that makes the first and the second equation true. In other words, \begin{align*}x\end{align*} in the first equation equals \begin{align*}x\end{align*} in the second equation and \begin{align*}y\end{align*} in the first equation equals \begin{align*}y\end{align*} in the second equation. Now, look at the second equation, \begin{align*}y-x=4\end{align*}. It is simple to solve for the \begin{align*}y\end{align*} value—add \begin{align*}x\end{align*} to both sides. So \begin{align*}y=x+4\end{align*}. Well, if \begin{align*}y\end{align*} is the same in both equations and \begin{align*}y=x+4\end{align*}, then we can substitute \begin{align*}x + 4\end{align*} in the place of \begin{align*}y\end{align*} in the first equation. That is why this is called the *substitution method*.

Now that we have substituted, we can solve this equation because it has a single variable.

\begin{align*}3(x+4) &= x-2 \\ 3x+12 &=x-2 \\ 2x+12 &=-2 \\ 2x &=-14 \\ x &=-7 \end{align*}

If \begin{align*}x = -7\end{align*}, substitute again to find \begin{align*}y\end{align*}:

\begin{align*} y &=x+4 \\ y &=-7+4 \\ y &=-3 \end{align*}

**Our solution, then, is (-7, -3).**

V. **Model and Solve Real – World Problems Using Systems of Linear Equations**

Two methods for solving systems of equations are graphing and substitution. There are many situations in the real world where these strategies can be useful.

Example

Two trains leave the station going in the same direction. One train leaves two hour before the other. The first train’s average speed is 65mph while the second train’s average speed is 90mph. How long will it take for the second train to catch up to the first?

**First, let’s think about functions with regard to this problem. The second train’s distance \begin{align*}d\end{align*} is a function of time \begin{align*}t\end{align*}. It can be found with the equation \begin{align*} d=90t \end{align*}.**

**Since the first train left two hours before the first train, it had two hours more than the second train to travel. Its speed was 65mph, though. Its distance can be found with the equation \begin{align*}d=65(t+2)\end{align*}.**

Solve the system of equations using a graph.

**From the graph you can see that the trains will meet shortly after 5 hours.**

**Now let’s go back and look at the problem from the introduction.**

## Real-Life Example Completed

*A Second Trip to the Omni*

**Here is the problem from the introduction. Reread it and then solve it for the solution.**

Kelly loved the Omni presentation on the rainforest so much that she decided to go and see it again. She asked Tyler if he wanted to go with her on Saturday afternoon.

“Do you want to go with me?” Kelly asked.

“Sure, but I have karate first, so I will have to meet you there. What time is the show?” Tyler asked.

“The show starts at 2:30 pm. I’m going to leave at one o’clock so I can look around,” Kelly said.

“Well, I don’t get done karate until then, so I probably won’t leave until 2:00 pm,” Tyler said.

On Saturday, the two went about their day and both left for the museum. Kelly’s Mom tends to drive slowly and cautiously, so she was traveling on city streets at 45 mph. Tyler’s karate class is downtown, so he could take the highway to get to the Omni theater and his Dad drove at an average of 55 mph. Will the two catch up to each other?

**This is a problem about equations and systems of equations. You will need to solve a system of equations to figure this one out.**

*Now write two equations to represent the system of equations for this problem. Then see if there is a solution for both of them.*

*Solution to Real – Life Example*

**The first equation we can write is to represent Tyler’s time. His Dad is traveling 55 mph. Therefore his distance is a function of speed and time.**

\begin{align*}d = 55t\end{align*}

**Kelly left one hour before Tyler did. She is traveling 45 miles per hour. Therefore, the speed times Tyler’s time plus one hour equals Kelly’s time.**

\begin{align*}d = 45(t + 1)\end{align*}

**Now see if there is a solution that will work for both equations. We can try to solve this by using substitution.**

\begin{align*}55t &= 45(t + 1) \\ 55t &= 45t + 1 \\ 55t - 45t &= 1 \\ 10t &= 1 \\ t &= \frac{1}{10} \end{align*}

**Now we go to Tyler.**

\begin{align*}d &= 55\left ( \frac{1}{10} \right ) \\ d &= 5.5 \end{align*}

**The solution could be the following values for \begin{align*}d\end{align*}** *and***\begin{align*}t\end{align*}. However, when you substitute those values into both equations, the solution does not work. Therefore there isn’t a solution for this system, and the Kelly and Tyler won’t meet up while driving.**

## Vocabulary

Here are the vocabulary words that are found in this lesson.

- System of Equations
- two or more equations at the same time. The solution will be the ordered pair that works for both equations.

## Time to Practice

Directions: Figure out which pair is a solution for each given system.

- Which ordered pair is a solution of the following system?

\begin{align*} x-3y &=9 \\ 3x+y &=7\end{align*}

(a) \begin{align*}(6, -1)\end{align*}

(b) \begin{align*}(-1, -4)\end{align*}

(c) \begin{align*}(0, 7)\end{align*}

(d) \begin{align*}(3, -2)\end{align*}

- Which ordered pair is a solution of the following system?

\begin{align*} y &=3x-7 \\ 5x-3y &=13\end{align*}

(a) \begin{align*} \left (3, \frac{2}{3} \right) \end{align*}

(b) \begin{align*}(2, -1)\end{align*}

(c) \begin{align*}(4, 7)\end{align*}

(d) \begin{align*}(5, 8)\end{align*}

- Figure out which ordered pair is a solution for the following system.

\begin{align*} \frac{1}{2}x +3y &= -6 \\ 2x + 4y &= 0 \end{align*}

Directions: Determine whether each system has infinite solutions or no solutions.

\begin{align*} x +y &= 10 \\ y &= -x +10 \end{align*}

\begin{align*} 3x -6y &= -24 \\ x -2y &= -8 \end{align*}

\begin{align*} \frac{3}{4}x &= \frac{2}{3}y-1 \\ 9x &= 8y-12 \end{align*}

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

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

Directions: Answer each question true or false.

- Parallel lines have the same slope.
- Parallel lines have infinite solutions.
- Perpendicular lines have one solution.
- Lines with an infinite number of solutions are not parallel.

Directions: Graph the following systems of equations. Identify the solution or write no solution or infinitely many solutions, if appropriate.

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

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

\begin{align*} y &= -3x + 1 \\ 3x - y &= -7 \end{align*}

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

Directions: Solve the following systems using the substitution method.

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

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

\begin{align*} y &= 2x \\ 6x - y &= 8 \end{align*}

\begin{align*} x + 4y &= -3 \\ 2x + 8y &= -6 \end{align*}

- Anglica’s mother leaves to visit her grandmother for her birthday. Her grandmother lives 450 miles away and her mother drives at an average of 50mph. Three hours later, Angelica’s step-father notices that her mother forgot the gift for her grandmother so he decides to try to catch up to her. If he drives at an average of 80mph, will he catch up to her before she gets to her grandmother’s house? Write a system of equations to model the situation and solve using the substitution method.

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