Lesson Objectives
• Demonstrate an understanding of how to solve a linear system in two variables
• Demonstrate an understanding of how to solve a linear system in three variables
• Learn the six-step method used to solve a word problem with linear systems

## How to Solve Word Problems with Linear Systems

Most word problems involve more than one unknown quantity. In some situations, we can solve a problem with more than one unknown with the use of a single variable. We saw problems of this type when we were working with applications of linear equations. When we solve a problem with two or more unknowns, it is usually easier to represent each unknown with its own variable. In order to get a solution, we must have the same number of equations as we have variables. For example, if we represent four unknowns with four variables, we need to create four equations that relate the unknown quantities in order to obtain a solution. We previously learned a six-step method for solving word problems. We will modify this procedure slightly for use with applications of linear systems.

### Applications of Linear Systems

1. Read the problem and determine the main objective
2. Assign a variable to represent each unknown
• For each variable, we need an equation
3. Write a system of linear equations
4. Solve the system
6. Check the result
• Read back through the problem and determine if the answer is reasonable
Let's look at a few examples.
Example 1: Solve each word problem
Jen and Mya are selling snacks for a school fundraiser. Customers can buy packages of figs and packages of cookies. Jen sold 12 packages of figs and 5 packages of cookies for a total of $287. Mya sold 1 package of figs and 10 packages of cookies for a total of$206. Find the sale price for one package of figs and one package of cookies.
Step 1) Read the problem and determine the main objective.
Our goal is to determine the sale price for each package of figs and cookies.
Step 2) Assign a variable to represent each unknown.
Let x = the price in dollars for one package of figs
Then y = the price in dollars for one package of cookies
Step 3) Write a system of linear equations.
While it isn't necessary, it is often useful to organize our information using a table.
Jen 125287
Mya 110206
We can set up the following system:
12x + 5y = 287
1x + 10y = 206
The first equation comes from Jen's sales activity. She sells 12 packages of figs, which are x dollars each. She also sells 5 packages of cookies which are y dollars each. If we sum the sales from figs (12x) and the sales from cookies (5y), we get her total sales of 287.
The second equation is set up in the exact same way. We look at Mya's sales activity and find that she sells 1 package of figs for x dollars each. She also sells 10 packages of cookies which are y dollars each. If we sum the sales from figs (1x) and the sales form cookies (10y), we get her total sales of 206.
Step 4) Solve the system.
Let's begin by labeling our equations as 1 and 2.
1) 12x + 5y = 287
2) 1x + 10y = 206
Since equation 2 has a variable with a coefficient of 1, the substitution method would likely be easiest. Let's solve equation 2 for x:
2) x = -10y + 206
Now we can plug in for x in equation 1:
12(-10y + 206) + 5y = 287
-120y + 2472 + 5y = 287
-115y = -2185
y = 19
Let's plug in a 19 for y in equation 2:
2) x + 10y = 206
x + 10(19) = 206
x + 190 = 206
x = 16
Since x represents the price of each package of figs and y represents the price of each package of cookies, we know that a package of figs sells for $16 and a package of cookies sells for$19.
Let's write this as a nice clean sentence.
The price of a package of figs is $16 and the price of a package of cookies is$19.
Step 6) Check the result.
To check our result, we read back through our problem. We see that Jen sold 12 packages of figs and 5 packages of cookies for $287. 12(16) + 5(19) = 287 192 + 95 = 287 287 = 287 Additionally, we see that Mya sold 1 package of figs and 10 packages of cookies for$206.
1(16) + 10(19) = 206
16 + 190 = 206
206 = 206
Example 2: Solve each word problem
A passenger jet traveled 1100 miles each way to Canada and back. The trip there was with the wind and took 10 hours. The trip back was into the wind and took 22 hours. What is the speed of the plane in still air? What is the speed of the wind?
Step 1) Read the problem and determine the main objective.
Our goal is to determine the speed of the plane in still air and the speed of the wind.
Step 2) Assign a variable to represent each unknown.
Let x = speed of the plane in still air
Then y = speed of the wind
Step 3) Write a system of linear equations.
Again, let's use a table to organize our information. Remember with a motion word problem, we use the distance formula:
d = r • t
Distance Rate Time
There 1100x + y10
Back 1100x - y22
We can set up the following system:
10(x + y) = 1100
22(x - y) = 1100
The first equation represents the trip there. We are following the formula: d = r • t. For the distance, we are given a value of 1100 miles for each scenario. Since the trip there involves traveling "with" the wind, our rate is the speed of the wind (y) plus the speed of the plane in still air (x). The time traveled is given to us as 10 hours. We put our equation together as the rate: (x + y), multiplied by the time 10, which is set equal to the distance of 1100.
The second equation represents the trip back. We are following the formula: d = r • t. For the distance, we are given a value of 1100 miles for each scenario. Since the trip back involves traveling "against" the wind, our rate is the speed of the plane in still air (x) minus the speed of the wind (y). The time traveled is given to us as 22 hours. We put our equation together as the rate: (x - y), multiplied by the time 22, which is set equal to the distance of 1100.
Step 4) Solve the system.
Let's begin by labeling our equations as 1 and 2.
1) 10(x + y) = 1100
2) 22(x - y) = 1100
For equation 1, let's divide each side by 10:
1) x + y = 110
For equation 2, let's divide each side by 22:
2) x - y = 50
Since we have +1y and -1y, it is easier to use elimination. $$\require{cancel}x + \cancel{y}=110$$ $$\underline{x + \cancel{-y}=50}$$ $$2x=160$$ $$x=80$$ We can plug an 80 in for x in either equation 1 or 2. Let's use equation 1:
10(x + y) = 1100
10(80 + y) = 1100
80 + y = 110
y = 30
Since x represents the speed of the plane in still air, and y represents the wind speed, we know the plane traveled at an average speed of 80 miles per hour in still air and the average wind speed was 30 miles per hour.
Let's write this as a nice clean sentence.
The speed of the plane was 80 miles per hour, while the wind speed was 30 miles per hour.
Step 6) Check the result.
To check our result, we read back through our problem. We see that on the trip there, the plane traveled at an average speed of 110 miles per hour. This accounts for the wind speed plus the plane speed in still air. If we multiply this by the time traveled of 10 hours, we do get a distance of 1100 miles.
10(80 + 30) = 1100
10(110) = 1100
1100 = 1100
Additionally, we see that on the trip back, the plane traveled at an average speed of 50 miles per hour. This accounts for the speed of the plane in still air minus the wind speed. If we multiply this by the time traveled of 22 hours, we do get a distance of 1100 miles.
22(80 - 30) = 1100
22(50) = 1100
1100 = 1100
Example 3: Solve each word problem
Nobot Robotics Company produces three toy robots, models Dot, Lea, and Liz. To manufacture each toy robot, the company must use labor hours for coding, assembly, and painting. The amounts for each robot are shown in the table below:
Dot Lea Liz
Coding 451
Assembly 792
Painting 421
There are 165 labor hours available for coding, 295 labor hours available for assembly, and 150 labor hours available for painting each day. How many of each model should be produced in a day, if all labor hours are used?
Step 1) Read the problem and determine the main objective.
Our goal is to determine how many of each type of toy robot is being produced on a daily basis.
Step 2) Assign a variable to represent each unknown.
Let x = number of Dot models produced in a day
Then y = number of Lea models produced in a day
Then z = number of Liz models produced in a day
Step 3) Write a system of linear equations.
Again, let's use a table to organize our information. We will take our above table and add in the total daily hours given for coding, assembly, and painting.
Dot Lea Liz Totals
Coding 451165
Assembly 792295
Painting 421150
We can set up the following system:
4x + 5y + z = 165
7x + 9y + 2z = 295
4x + 2y + z = 150
For the first equation, we think about the hours allotted for coding. Each model Dot requires 4 hours for coding, where Lea requires 5, and Liz requires 1. Since we have 165 hours available for coding, we can multiply the number of hours it takes to code each model by the number of models produced. If we sum these amounts for the three models, it will be equal to the total amount of hours spent on coding, which is 165.
Similarly, for the second equation, we think about the hours allotted for assembly. Each Dot requires 7 hours for assembly, where Lea requires 9, and Liz requires 2. Since we have 295 hours available for assembly, we can multiply the number of hours it takes to assemble each model by the number of models produced. If we sum these amounts for the three models, it will be equal to the total amount of hours spent on code, which is 295.
Lastly, for the third equation, we think about the hours allotted for painting. Each Dot requires 4 hours for painting, where Lea requires 2, and Liz requires 1. Since we have 150 hours available for painting, we can multiply the number of hours it takes to paint each model by the number of models produced. If we sum these amounts for the three models, it will be equal to the total amount of hours spent on painting, which is 150.
Step 4) Solve the system.
Let's begin by labeling our equations as 1, 2, and 3.
1) 4x + 5y + z = 165
2) 7x + 9y + 2z = 295
3) 4x + 2y + z = 150
It seems easiest to eliminate the variable z. Let's multiply equation 1 by (-1) and add this result to equation 3:
1) -4x - 5y - z = -165
3) 4x + 2y + z = 150 $$\cancel{-4x}- 5y + \cancel{-z}-165$$ $$\underline{\cancel{4x}+ 2y + \cancel{z}=150}$$ $$-3y=-15$$ $$y=5$$ Let's eliminate z from equations 2 and 3. We can multiply equation 3 by (-2) and add the result to equation 2:
3) -8x - 4y - 2z = -300
2) 7x + 9y + 2z = 295
$$-8x - 4y + \cancel{-2z}=-300$$ $$\underline{7x + 9y + \cancel{2z}=295}$$ $$-x + 5y=-5$$ Since we already know that y is 5, let's plug this in and find the value for x:
-x + 5y = -5
-x + 5(5) = -5
-x + 25 = -5
-x = -30
x = 30
At this point, we know that x is 30 and y is 5. Let's plug into equation 1 and find the value for z:
1) 4x + 5y + z = 165
4(30) + 5(5) + z = 165
120 + 25 + z = 165
145 + z = 165
z = 20
Since x represents the number of Dot models produced in a day, y represents the number of Lea models produced in a day, and z represents the number of Liz models produced in a day, we can state our answer as:
On a daily basis, Nobot produces 30 Dot model robots, along with 5 Lea Robots, and 20 Liz Robots.
Step 6) Check the result.
Coding:
4(30) + 5(5) + 1(20) = 165
120 + 25 + 20 = 165
165 = 165
Assembly:
7(30) + 9(5) + 2(20) = 295
210 + 45 + 40 = 295
295 = 295
Painting:
4(30) + 2(5) + 1(20) = 150
120 + 10 + 20 = 150
150 = 150

#### Skills Check:

Example #1

Solve each word problem.

Olivia’s school is selling tickets to the annual dance competition. On the first day of ticket sales the school sold 2 adult tickets and 11 child tickets for a total of $155. The school took in$62 on the second day by selling 6 adult tickets and 2 child tickets. Find the price of an adult ticket and the price of a child ticket.

A
Adult Ticket: $3, Child Ticket:$20
B
Adult Ticket: $6, Child Ticket:$13
C
Adult Ticket: $8, Child Ticket:$11
D
Adult Ticket: $13, Child Ticket:$6
E
Adult Ticket: $4, Child Ticket:$2

Example #2

Solve each word problem.

Julio and Jenna each improved their yards by planting rose bushes and ornamental grass. They bought their supplies from the same store. Julio spent $90 on 5 rose bushes and 5 bunches of ornamental grass. Jenna spent$187 on 11 rose bushes and 10 bunches of ornamental grass. What is the cost of one rose bush and the cost of one bunch of ornamental grass?

A
rose bush: $4, ornamental grass:$17
B
rose bush: $5, ornamental grass:$7
C
rose bush: $7, ornamental grass:$11
D
rose bush: $11, ornamental grass:$7
E
rose bush: $20, ornamental grass:$15

Example #3

Solve each word problem.

The indoor climbing gym is a popular field trip destination. This year the senior class at Madison High School and the senior class at North Park High School both planned trips there. The senior class at Madison High School rented and filled 9 vans and 7 buses with 454 students. North Park High School rented and filled 11 vans and 14 buses with 789 students. Each van and each bus carried the same number of students. How many students can a van carry? How many students can a bus carry?

A
Van : 14, Bus : 55
B
Van : 12, Bus : 50
C
Van : 17, Bus : 43
D
Van : 13, Bus : 60
E
Van : 19, Bus : 57