Module 4: Systems of Linear Equations and Inequalities
4.4: Applications of 2×2 Systems of Equations
section 4.4 Learning Objectives
4.4: Applications of 2×2 Systems of Equations
Set up and solve a linear system of equations by direct translation
Use linear systems to solve value problems
Use linear systems to solve mixture problems
Use linear systems to solve motion problems
Write a system of linear equations from direct translations
Systems of equations are a very useful tool for modeling real-life situations and answering questions about them. If you can translate the application into two linear equations with two variables, then you have a system of equations that you can solve to find the solution. You can use any of the methods learned in this module to solve the system of equations.
To introduce the topic, let us begin with direct translations.
In Section 1.7, we learned how to translate many key phrases in English into mathematical equations. Similarly, we can translate from English to form a system of equations.
Example 1
One number is 3 less than 4 times another. If the sum of these numbers is 32, find the two numbers.
Show Solution
There are two numbers in the problem, which we can represent by the variables x and y.
Examining the first sentence, we recognize some familiar key phrases. Recall that “is” will correspond to equals while “less than” is subtraction, while being careful with the order as this implies 3 is being subtracted from 4 times the second number. So, the statement “one number is 3 less than 4 times another” results in the equation
[latex]x=4y-3[/latex]
The second sentence discusses the “sum,” which indicates addition. Thus, “the sum of these numbers is 32” translates to the equation
While we can use any of the techniques learned in this module to achieve this, substitution will work particularly nicely since the first equation already has one of the variables isolated. Substituting [latex]x=4y-3[/latex] into the second equation for [latex]x[/latex] yields
Write a system of linear equations representing a value problem
One application of systems of equations is known as value problems. Value problems are ones where each variable has a value attached to it. For example, the marketing team for an event venue wants to know how to focus their advertising based on who is attending specific events—children, or adults? They know the cost of a ticket to a basketball game is $25.00 for children and $50.00 for adults. Additionally, on a certain day, attendance at the game is 2,000 and the total gate revenue is $70,000. How can the marketing team use this information to find out whether to spend more money on advertising directed toward children or adults?
We will use a table to help us set up and solve this value problem. The basic structure of the table is shown below:
Number (usually what you are trying to find)
Value
Total
Item 1
Item 2
Total
The first column in the table is used for the number of things we have. Quite often, this will be our variables. The second column is used for the value each item has. The third column is used for the total value which we calculate by multiplying the number by the value.
Example 2
Find the total number of child and adult tickets sold given that the cost of a ticket to a basketball game is $25.00 for children and $50.00 for adults. Additionally, on a certain day, attendance at the game is 2,000 and the total gate revenue is $70,000.
Show Solution
Read and Understand: We want to find the number of child and adult tickets, we know the total number of tickets sold, the total revenue and the cost of a child and adult ticket.
Define and Translate: Let c = the number of children and a = the number of adults in attendance. Revenue comes from number of tickets sold multiplied by the price of the ticket. We will get revenue for adults by multiplying $50.00 times a. $25.00 times c will give the revenue from the number of child tickets sold.
Write and Solve: Although a table is not necessary, it can help you get started. For this problem, we labeled columns as number, value, and total revenue because that is the information we are given.
The total number of people is [latex]2,000[/latex].
Number
Value
Total Revenue
Child Tickets
c
$25.00
25c
Adult Tickets
a
$50.00
50a
Total Tickets
2000
$70,000
The total revenue is $70,000. We can use this and the revenue from child and adult tickets to write an equation for the revenue. [latex]25c+50a=70,000[/latex]
Number
Value
Total Revenue
Child Tickets
c
$25.00
25c
Adult Tickets
a
$50.00
50a
Total Tickets
2000
[latex]25c+50a=70,000[/latex]
The number of people at the game that day is the total number of child tickets sold plus the total number of adult tickets, [latex]c+a=2,000[/latex]
Number
Value
Total Revenue
Child Tickets
c
$25.00
25c
Adult Tickets
a
$50.00
50a
Total Tickets
[latex]c+a=2,000[/latex]
[latex]25c+50a=70,000[/latex]
We now have a system of linear equations in two variables.[latex]\begin{array}{r}c+a=2,000\,\,\,\\ 25c+50a=70,000\end{array}[/latex].
We can use any method of solving systems of equations to solve this system for a and c. Substitution looks easiest because we can solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].
We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the game that day. The marketing group may want to focus their advertising toward attracting young people.
This example showed you how to find two unknown values given information that connected the two unknowns. With two equations, you are able to find a solution for two unknowns. If you were to have three unknowns, you would need three equations to find them, and so on.
In the following video, you are given an example of how to use a system of equations to find the number of children and adults admitted to an amusement park based on entrance fees and total revenue. This example shows how to write equations and solve the system without a table.
In the next example, we will find the number of coins in a change jar given the total amount of money in the jar and the fact that the coins are either quarters or dimes.
Example 3
In a change jar there are 11 coins that have a value of $1.85. The coins are either quarters or dimes. How many of each kind of coin is in the jar?
Show Solution
Read and Understand: We want to find the number of quarters and the number of dimes in the jar. We know that dimes are $0.10 and quarters are $0.25, and the total number of coins is 11.
Define and Translate: We will call the number of quarters q and the number of dimes d. The part of the total $1.85 that comes from quarters will be determined by how many quarters and the fact that each one is worth $0.25, so $0.25q represents the amount of $1.85 that is quarters. The same idea can be used for dimes, so $0.10d represents the amount of $1.85 that is dimes.
Write and Solve: We can label a new table with the information we are given.
Number
Value
Total
Quarters
q
$0.25
$0.25q
Dimes
d
$0.10
$0.10d
Total number of coins
q+d=11
$0.25q+$0.10d=$1.85
We can write our two equations, remember that we need two to solve for two unknowns.
In the following video, you will see an example similar to the previous one, except that the equations are written and solved without the use of a table.
In the next example we see a variation on the value problem, where it will make more sense to label the first column “Amount” rather than “Number.”
Example 4
In a candy shop, chocolate which sells for $4.00 per pound is mixed with nuts which are sold for $2.50 per pound to produce a chocolate-nut candy which sells for $3.50 per pound. How much of each is used to make 30 pounds of the mixture?
Show Solution
Let c represent the pounds of chocolate and n the pounds of nuts.
We begin by making a table similar to those in the previous example. One important difference, however, is that this time, we have a “value” for the bottom row. For each row, multiply the amount by the value to obtain the total (in this case, signifying the total cost of the chocolate, nuts, and the mixture).
Amount
Value
Total
Chocolate
[latex]c[/latex]
[latex]4.00[/latex]
[latex]4c[/latex]
Nuts
[latex]n[/latex]
[latex]2.50[/latex]
[latex]2.5n[/latex]
Mixture
[latex]30[/latex]
[latex]3.50[/latex]
[latex]105[/latex]
Adding the amounts and total costs of the chocolate and nuts to those for the mixture produces the following system of equations:
[latex]c+n=30[/latex]
[latex]4c+2.5n=105[/latex]
As usual, we can use any method we have learned to solve this. Let us multiply the top equation by [latex]-2.5[/latex] and add the equations to eliminate the n terms.
We now plug this into either equation to solve for n.
[latex]20+n=30[/latex]
[latex]\hspace{.39in}n=10[/latex]
Answer
The candy shop should use 20 pounds of chocolate and 10 pounds of nuts.
Write a system of linear equations representing a mixture problem, solve the system and interpret the results
One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. There are many other disciplines that use solutions as well.
The concentration or strength of a liquid solution describes the amount of a substance that is in a specific volume of liquid. For example, if you have 30 grams of salt dissolved in water to create a 100mL saline solution, you have a concentration given by the following ratio:
[latex]\frac{30\text{ grams }}{100\text{ mL }}=0.30\frac{\text{ grams }}{\text{ mL }}=30\text{%}[/latex]
Concentration is often given by a percentage. In the above example, this can be achieved since 1 mL of water has a mass of 1 gram. Often, we will mix two liquids directly. For example, if 20mL of ethanol is in a 50mL solution, the concentration of ethanol is given by
[latex]\frac{20\text{ mL }}{50\text{ mL }}=0.4=40\text{ % }[/latex]
Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.
We will use the following table to help us solve mixture problems:
Amount of Solution
Concentration
Amount of Solute
Solution 1
Solution 2
Final Solution
To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex] solution, and the other is 75mL of a 0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex] solution. If we mix both of these solutions together we will have a new volume and a new mass of solute, and with those we can find a new concentration.
First, find the total mass of solids for each solution by multiplying the volume by the concentration.
Amount of Solution
Concentration
Amount of Solute
Solution 1
120 mL
0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex]
Solution 2
75 mL
0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex]
Final Solution
Next we add the new volumes and new masses.
Amount of Solution
Concentration (%)
Amount of Solute
Solution 1
120 mL
0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex]
Solution 2
75 mL
0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex]
Final Solution
195 mL
[latex]\frac{28.05\text{ grams }}{ 195 \text{ mL }}=0.14=14\text{ % }[/latex]
We have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.
Example 5
A chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?
Show Solution
Let’s use the problem solving process outlined in Module 1 to help us work through a solution to the problem.
Read and Understand: We are looking for a new amount – in this case a volume – based on the words “how much”. We know two starting concentrations and the final concentration, as well as one volume.
Define and Translate: Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane. We can call our unknown amount x.
Write and Solve: Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.
Amount of Solution
Concentration
Amount of Methane
Solution 1
70
0.5
Solution 2
x
0.8
Final Solution
70+x
0.6
Multiply the amount of solution by the concentration to get the amount of methane. Be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]. Add the entries in the solution amount column to get the final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.
Amount of Solution
Concentration
Amount of Methane
Solution 1
70
0.5
35
Solution 2
x
0.8
0.8x
Final Solution
70+x
0.6
[latex]42+0.6x[/latex]
Add the amounts of methane for solution 1 and solution 2 to get the total amount of methane for the 60% solution. This is our equation for finding the unknown volume.
35mL must be added to the original 70 mL to gain a solution with a concentration of 60%
The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.
Example 6
A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons that is 20% butterfat?
Show Solution
Read and Understand: We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.
Define and Translate: We will call the unknown volume of the 24% solution x, and the unknown volume of the 18% solution y.
Write and Solve: Fill in the table with the information we know.
Amount of Solution (Milk)
Concentration (%)
Amount of Butterfat
Solution 1
x
0.24
Solution 2
y
0.18
Final Solution
42
0.2
Find the amount of butterfat by multiplying the amount of each solution by the concentration.
Amount
Concentration (%)
Total Mass
Solution 1
x
0.24
0.24x
Solution 2
y
0.18
0.18y
Final Solution
x+y=42
0.2
0.24x+0.18y=8.4
When you sum the amount column you get one equation: x+ y = 42
When you sum the total column you get a second equation: 0.24x + 0.18y = 8.4
Use elimination to find a value for x, and y.
Multiply the first equation by -0.18
-0.18(x+y) = (42)(-0.18)
-0.18x – 0.18y = -7.56
Now our system of equations looks like this:
-0.18x – 0.18y = -7.56
0.24x + 0.18y = 8.4
Adding the two equations together to eliminate the y terms gives this equation:
0.06x = 0.84
Divide by 0.06 on each side:
x = 14
Now substitute the value for x into one of the equations in order to solve for y.
(14) + y = 42
y = 28
Answer
This can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.
In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.
Write a system of linear equations representing a distance, rate, and time problem
Distance, rate, and time applications are common in algebra. To solve this type of problem, we will use the following formula:
A problem that will typically result in a system of linear equations arises when there are multiple rates that contribute to the overall rate of an object. Before we jump into a full [latex]d=r\cdot t[/latex] problem, let us first explore an example that focuses solely on the rates.
Example 7
A boat traveling upstream on the Big River averages 5 mph, while traveling downstream with the same current, it averages 7 mph. Find the speed of the current and the speed of the boat in still water.
Show Solution
There are two factors contributing to the overall rate of the boat. The boat is trying to travel at a certain speed, the “speed of the boat in still water,” but the current will either add to or subtract from this speed.
Let b represent the speed of the boat in still water and c the speed of the current.
When the boat travels upstream, it is fighting against the current, which slows the boat down. Given that the overall rate is 5 mph, this is given by the equation
[latex]b-c=5[/latex]
When the boat travels downstream, the current instead speeds the boat up. With he overall rate of 5 mph, this is given by
[latex]b+c=7[/latex]
Thus, we now have a system of equations. This particular system can be quickly solved with the elimination method since adding the original equations together will immediately lead to the c terms canceling each other out.
[latex]b-c=5[/latex]
[latex]\underline{b+c=7}[/latex]
[latex]\hspace{.22in}2b=12[/latex]
[latex]\hspace{.23in}b=6[/latex]
We can substitute this result into either equation to solve for c.
[latex]6+c=7[/latex]
[latex]\hspace{.28in}c=1[/latex]
.
Answer
The speed of the current is 1 mph and the speed of the boat in still water is 6 mph.
We can now utilize this idea to deal with a distance, rate, and time problem. As with the earlier types of problems, you may find it helpful to organize the information in a table.
Example 8
A boat can travel 20 miles downstream in 2 hours, while it takes 5 hours to make the same trip upstream. Find the speed of the boat in still water and the speed of the current.
Show Answer
First, note that since the boat makes the “same trip” upstream, this implies that it travels the same distance of 20 miles.
Let b represent the speed of the boat in still water and c the speed of the current. Using what we learned about rate from the previous example helps us to fill in the following table:
Rate
Time
Distance
Downstream
[latex]b+c[/latex]
[latex]2[/latex]
[latex]20[/latex]
Upstream
[latex]b-c[/latex]
[latex]5[/latex]
[latex]20[/latex]
Now we can form our system of equations by applying the formula [latex]r\cdot t=d[/latex] to each row of the table. To indicate that we are multiplying the entire rate times the time, we use parentheses.
While your first instinct might be to distribute (and this would work), we can form a much simpler system if we divide both sides by the time in each equation.
This results in the following system, which can be quickly solved using the elimination method.
[latex]b+c=10[/latex]
[latex]\underline{b-c=4}[/latex]
[latex]\hspace{.23in} 2b=14[/latex]
[latex]\hspace{.25in}b=7[/latex]
For simplicity, we can substitue this result into either equation in our modified system to solve for c.
[latex]7+c=10[/latex]
[latex]\hspace{.32in}c=3[/latex]
Answer
The speed of the current is 3 mph and the speed of the boat in still water is 7 mph.
To summarize the section, the following video provides several examples of applications of linear systems, including a ticket price problem, a mixture problem, and a motion problem.