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.

 

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.

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?

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?

 

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] [latex]10.8\text{ grams }+17.25\text{ grams }=28.05\text{ grams }[/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?

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?

 

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:

[latex]\mbox{distance}=\mbox{rate}\times\mbox{time}[/latex]

[latex]d=r\cdot t[/latex]

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.

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.

 

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.