In this section, you will....
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write and solve systems of linear equations that represent real world applications
Section 3.2 Applications of Systems of Linear Equations
If we translate an application to a mathematical setup using two variables, then we need to form a linear system with two equations. Setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear.
Supplemental Videos.
The main topics of this section are also presented in the following videos:
Example 3.18.
The sum of \(4\) times a larger integer and \(5\) times a smaller integer is \(7\text{.}\) When twice the smaller integer is subtracted from \(3\) times the larger, the result is \(11\text{.}\) Find the integers.
Solution.
Begin by assigning variables to the larger and smaller integer. Let \(x\) represent the larger integer, and let \(y\) represent the smaller integer.
When using two variables, we need to set up two equations. The first sentence describes a sum and the second sentence describes a difference. The phrase "\(4\) times a larger integer" translates to the expression \(4x\text{,}\) and "\(5\) times a smaller integer" translates to \(5y\text{.}\) Altogether, we get \(4x+5y=7\) for the first equation. Likewise, we get \(3x-2y=11\) for the second equation. This leads to the following system:
\begin{equation*}
\begin{cases} 4x+5y\amp =7\\ 3x-2y\amp =11 \end{cases}
\end{equation*}
We will solve this using the elimination method. To eliminate the variable \(y\) multiply the first equation by \(2\) and the second by \(5\text{.}\)
\begin{equation*}
\left\{\begin{array}{rrrl} (\alert{2})\amp (4x\amp +5y\amp =7)\\ (\alert{5})\amp (3 x\amp -2y\amp =11)\\ \end{array}\right.= \begin{cases} 8x+10y\amp =14\\ 15x-10y\amp =55 \end{cases}
\end{equation*}
Next we add the equations in the equivalent system and solve for \(x\text{.}\)
\begin{equation*}
\begin{aligned} 8x \amp\, \alert{+10y}\amp =14\\ 15x\amp\, \alert{-10y}\amp =55\\ \hline\\ 23x\amp\amp =69 \end{aligned}
\end{equation*}
This gives \(x=\frac{69}{23}=3\text{.}\) Now we use this for a substitution into an original equation to find \(y\text{.}\)
\begin{equation*}
\begin{aligned} 4x+5y\amp =7\\ 4(\alert{3})+5y\amp =7\\ 12+5y\amp =7\\ 5y\amp = -5\\ y\amp =-1 \end{aligned}
\end{equation*}
The larger integer is \(3\) and the smaller integer is \(-1\text{.}\)
Example 3.19.
An integer is \(1\) less than twice that of another. If their sum is \(20\text{,}\) find the integers.
Solution.
Let \(x\) be the first integer and let \(y\) be the second. We’re told that the first integer is \(1\) less than twice that of the other, so \(x=2y-1\text{.}\) Further, their sum is \(20\text{,}\) so \(x+y=20\text{.}\) Together, we have
\begin{equation*}
\begin{cases} x \amp= 2y-1 \\ x+y \amp= 20 \end{cases}
\end{equation*}
We can solve this system with substitution. Substituting the first equation into the second, we have
\begin{equation*}
\begin{aligned} (\alert{2y-1})+y \amp= 20 \\ 3y \amp= 21 \\ y \amp= 7 \end{aligned}
\end{equation*}
To find \(x\text{,}\) we simply plug the solution for \(y\) into the first equation.
\begin{equation*}
x=2(\alert{7})-1=13
\end{equation*}
Therefore the two integers are \(13\) and \(7\text{.}\)
Example 3.20.
A total of \(\$12,800\) was invested in two accounts. Part was invested in a CD at a \(3.125\)% annual interest rate and part was invested in a money market fund at a \(4.75\)% annual interest rate. If the total simple interest for one year was \(\$465\text{,}\) then how much was invested in each account?
Solution.
We begin by identifying two variables: let \(x\) represent the amount invested at \(3.125\)%=\(0.03125\text{,}\) and let \(y\) represent the amount invested at \(4.75\)%=\(0.0475\text{.}\)
The total amount in both accounts can be expressed as
\begin{equation*}
x+y=12,800
\end{equation*}
To set up a second equation, use the fact that the total interest was \(\$465\text{.}\) The interest for one year is the interest rate times the principal (\(I=prt=pr\cdot 1=pr\)). Use this to add the interest in both accounts. Be sure to use the decimal equivalents for the interest rates given as percentages.
\begin{equation*}
\begin{aligned} \text{interest from the CD }\, \amp + \, \text{interest from the fund } \amp =\amp \text{total interest}\\ 0.03125x \amp + 0.0475y \amp = \amp 465 \end{aligned}
\end{equation*}
These two equations together form the following linear system:
\begin{equation*}
\left\{\begin{array}{rrl} x\amp +y\amp =12,800\\ 0.03125x \amp + 0.0475y \amp = 465 \end{array}\right.
\end{equation*}
Eliminate \(x\) by multiplying the first equation by \(-0.03125\text{.}\)
\begin{align*}
\left\{\begin{array}{rrrl} (\alert{-0.03125})\amp ( x\amp +y\amp =12,800)\\ \amp 0.03125x \amp + 0.0475y \amp = 465\\ \end{array}\right.\\
= \left\{\begin{array}{rrl} -0.03125x\amp -0.03125y\amp = -400\\ 0.03125x \amp + 0.0475y \amp = 465 \end{array}\right.
\end{align*}
Next, we add the equations together.
\begin{equation*}
\begin{aligned} \alert{-0.03125x}\amp -0.03125y\amp = -400\\ \alert{0.03125x} \amp + 0.0475y \amp = 465\\ \hline\\ \amp \, 0.01625y\amp=65 \end{aligned}
\end{equation*}
This gives \(y=\frac{65}{0.01625}=4,000\text{.}\) We use this for a substitution to find \(x\text{.}\)
\begin{equation*}
\begin{aligned} x+y\amp = 12,800\\ x+4,000\amp = 12,800\\ x\amp =8,800 \end{aligned}
\end{equation*}
Of the money invested, \(\$4,000\) was invested at \(4.75\)% and \(\$8,800\) was invested at \(3.125\)%.
Example 3.21.
A jar consisting of only nickels and dimes contains \(58\) coins. If the total value is \(\$4.20\text{,}\) how many of each coin is in the jar?
Solution.
We begin by identifying two variables: let \(n\) represent the number of nickels in the jar and let \(d\) represent the number of dimes in the jar.
The total number of coins in the jar can be expressed as
\begin{equation*}
n+d=58
\end{equation*}
Next, we use the value of each coin to determine the total value \(\$4.20\text{.}\) A nickel is worth \(\$0.05\) and a dime is worth \(\$0.10\text{.}\) So, when we multiply \(0.05\) by the number of nickels \(n\text{,}\) we will get the value of all our nickels combined, and likewise with the dimes. Together, the value of the nickels added together with the value of the dimes will give us the total value:
\begin{equation*}
0.05n+0.10d=4.20
\end{equation*}
This leads us to following linear system:
\begin{equation*}
\left\{\begin{array}{rrl} n\amp +d\amp =58\\ 0.05n \amp + 0.10d\amp = 4.20 \end{array}\right.
\end{equation*}
Here we will solve using the substitution method. In the first equation, we can solve for \(n\text{.}\)
\begin{equation*}
\begin{aligned} n+d\amp =58\\ n\amp = 58-d \end{aligned}
\end{equation*}
Now we will substitute \(n=58-d\) into the second equation and solve for \(d\text{.}\)
\begin{equation*}
\begin{aligned} 0.05(\alert{58-d})+0.10d\amp =4.20\\ 2.9-0.05d+0.10d\amp = 4.20\\ 2.9+0.05d\amp =4.20\\ 0.05d\amp = 1.3\\ d\amp =26 \end{aligned}
\end{equation*}
Now we use this for a substitution to find the number of nickels.
\begin{equation*}
\begin{aligned} n\amp = 58-d\\ \amp= 58-26\\ \amp = 32 \end{aligned}
\end{equation*}
The jar contains \(32\) nickels and \(26\) dimes.
Example 3.22.
Chen has a jar full of \(40\) coins consisting of only quarters and nickels. If the total value is \(\$5.00\text{,}\) how many of each coin does Chen have?
Solution.
Let \(q\) be the number of quarters that Chen has and let \(n\) be the number of nickels. Since he has \(40\) coins,
\begin{equation*}
q+n=40
\end{equation*}
Since the total value is \(\$5.00\text{,}\)
\begin{equation*}
0.25q+0.05n=5
\end{equation*}
Together we have the following system:
\begin{equation*}
\begin{cases} q+n \amp= 40 \\ 0.25q+0.05n \amp= 5 \end{cases}
\end{equation*}
We will solve by elimination. Multiplying the second equation by \(-20\text{,}\) we have the equivalent system
\begin{equation*}
\begin{cases} q+n \amp= 40 \\ -5q-n \amp= -100 \end{cases}
\end{equation*}
Adding the equations yields
\begin{equation*}
\begin{aligned} -4q \amp= -60 \\ q \amp= 15 \end{aligned}
\end{equation*}
To find \(n\text{,}\) we substitute \(q\) into the first equation.
\begin{equation*}
\begin{aligned} \alert{15}+n \amp= 40 \\ n \amp= 25 \end{aligned}
\end{equation*}
Chen has \(25\) nickels and \(15\) quarters.
Subsection 3.2.1 Mixture Applications
Mixture problems often include a percentage and some total amount. It is important to make a distinction between these two types of quantities. For example, if a problem states that a 20-ounce container is filled with a \(2\)% saline (salt) solution, then this means that the container is filled with a mixture of salt and water as follows:
| \(\) | \(Percentage\) | \(Amount\) |
| \(Salt\) | \(2\)%\(=0.02\) | \(0.02\) (\(20\) ounces)\(=0.4\) ounces |
| \(Water\) | \(98\)%\(=0.98\) | \(0.98\) (\(20\) ounces)\(=19.6\) ounces |
In other words, we multiply the percentage times the total to get the amount of each part of the mixture.
Example 3.23.
A \(1.8\)% saline solution is to be combined and mixed with a \(3.2\)% saline solution to produce \(35\) ounces of a \(2.2\)% saline solution. How much of each is needed?
Solution.
We begin by identifying two variables: let \(x\) represent the volume in ounces of \(1.8\)% saline solution needed and let \(y\) represent the volume in ounces of \(3.2\)% saline solution needed.
The total amount of saline solution needed is \(35\) ounces. This leads to one equation:
\begin{equation*}
x+y=35
\end{equation*}
The second equation adds up the amount of salt in the correct percentages. The amount of salt is obtained by multiplying the percentage times the amount, where the variables \(x\) and \(y\) represent the amounts of the solutions. The amount of salt in the end solution is \(2.2\)% of the \(35\) ounces, or \(0.022(35)\text{.}\)
\begin{equation*}
0.018x+0.032y=0.022(35)
\end{equation*}
This leads us to the following linear system:
\begin{equation*}
\left\{\begin{array}{rrl} x\amp +y\amp =35\\ 0.018x \amp + 0.032y\amp = 0.022(35) \end{array}\right.
\end{equation*}
Here we will solve using the elimination method.
\begin{align*}
\left\{\begin{array}{rrrl} (\alert{-0.018})\amp(x\amp +y\amp =35)\\ \amp 0.018x \amp + 0.032y\amp = 0.022(35) \end{array}\right.\\
= \left\{\begin{array}{rrl} -0.018x\amp -0.018y\amp =-0.63\\ 0.018x \amp + 0.032y\amp = 0.022(35) \end{array}\right.
\end{align*}
Now we add the resulting equations together.
\begin{equation*}
\begin{aligned} \alert{-0.018x}\amp -0.018y\amp =-0.63\\ \alert{0.018x} \amp + 0.032y\amp = 0.022(35)\\ \hline\\ 0.014y\amp = 0.14\\ y\amp =\frac{0.14}{0.014}\\ y\amp = 10 \end{aligned}
\end{equation*}
Now we use this for a substitution to find the value of \(x\text{.}\)
\begin{equation*}
\begin{aligned} x+y\amp = 35\\ x+\alert{10}\amp =35\\ x\amp =25 \end{aligned}
\end{equation*}
We need \(25\) ounces of the \(1.8\)% saline solution and \(10\) ounces of the \(3.2\)% saline solution.
Example 3.24.
An \(80\)% antifreeze concentrate is to be mixed with water to produce a \(48\)-liter mixture containing \(25\)% antifreeze. How much water and antifreeze concentrate is needed?
Solution.
We begin by identifying two variables: let \(x\) represent the volume in liters of \(80\)% antifreeze concentrate needed and let \(y\) represent the volume in liters of water needed.
The total amount of the mixture needed is \(48\) liters. This leads to the equation
\begin{equation*}
x+y=48.
\end{equation*}
The second equation adds up the amount of antifreeze from each solution in the correct percentages. The amount of antifreeze in the end result is \(25\)% of the \(48\) liters, or \(0.25(48)\text{.}\)
\begin{equation*}
0.80x+0\cdot y=0.25(48)
\end{equation*}
This leads us to the following linear system:
\begin{equation*}
\left\{\begin{array}{rrl} x\amp +y\amp =48\\ 0.80x \amp +\amp = 0.25(48)=12 \end{array}\right.
\end{equation*}
We will use the second equation to find \(x\text{:}\)
\begin{equation*}
\begin{aligned} 0.80x\amp =12\\ x\amp =\frac{12}{0.80}\\ x\amp =15 \end{aligned}
\end{equation*}
Now we use this for a substitution to find the value of \(y\text{.}\)
\begin{equation*}
\begin{aligned} x+y\amp = 48\\ \alert{15}+y\amp =48\\ y\amp =33 \end{aligned}
\end{equation*}
We need to mix \(33\) liters of water with \(15\) liters of antifreeze concentrate.
Subsection 3.2.2 Distance Applications
Distance traveled is equal to the average rate times the time traveled at that rate, \(d=r\cdot t\text{.}\) These uniform motion problems usually have a lot of data, so it helps to first organize that data in a chart and then set up a linear system. In this section, you are encouraged to use two variables.
Example 3.25.
Naomi traveled a total of \(4\) hours and \(875\) miles by car and by plane. Driving to the airport by car, she averaged \(50\) miles per hour. In the air, the plane averaged \(320\) miles per hour. How long did it take her to drive to the airport?
Solution.
We are asked to find the time it takes her to drive to the airport; this indicates that time is the unknown quantity.
We begin by identifying two variables: let \(x\) represent the time (in hours) it took to drive to the airport and let \(y\) represent the time (in hours) spent in the air.
Since the total time traveled was \(4\) hours, we have our first equation
\begin{equation*}
x+y=4.
\end{equation*}
We use the formula \(d=r\cdot t\) to fill in the unknown distances:
\begin{gather*}
\text{Distance traveled in the car: }\, d=r\cdot t=50\cdot x\\
\text{Distance traveled in the air: }\, d=r\cdot t=320\cdot y
\end{gather*}
The total amount of miles traveled is 875, so this leads to the equation
\begin{equation*}
50x+320y=875.
\end{equation*}
This leads us to the following linear system:
\begin{equation*}
\left\{\begin{array}{rrl} x\amp +y\amp =4\\ 50x \amp +320y\amp = 875 \end{array}\right.
\end{equation*}
We will solve using the elimination method.
\begin{align*}
\left\{\begin{array}{rrrl} (\alert{-50})\amp(x\amp +y\amp =4)\\ \amp 50x \amp +320y\amp = 875 \end{array}\right.\\
= \left\{\begin{array}{rrrl} \amp -50x\amp -50y\amp =-200\\ \amp 50x \amp +320y\amp = 875 \end{array}\right.
\end{align*}
Now we add the resulting equations together.
\begin{equation*}
\begin{aligned} \alert{-50x}\amp -50y =-200\\ \alert{50x} \amp +320y= 875\\ \hline\\ \amp \, 270 y = 675 \end{aligned}
\end{equation*}
This results in \(y=\frac{675}{270}=\frac{5}{2}\text{.}\) Now we substitute to find the time \(x\) it took to drive to the airport:
\begin{equation*}
\begin{aligned} x+y\amp =4\\ x+\alert{\frac{5}{2}}\amp =4\\ x\amp = \frac{8}{2}-\frac{5}{2}\\ x\amp =\frac{3}{2} \end{aligned}
\end{equation*}
It took Naomi \(1.5\) hours to drive to the airport.
It is not always the case that time is the unknown quantity. Read the problem carefully and identify what you are asked to find. This defines your variables.
Example 3.26.
Flying with the wind, Dale piloted a light aircraft traveling \(240\) miles in \(2\) hours. Dale then turned against the wind and traveled another \(135\) miles in \(1.5\) hours. Find the speed of the airplane and the speed of the wind (you may assume that Dale flew at a constant rate and the wind blew at a constant rate.).
Solution.
We begin by identifying two variables: let \(a\) represent the speed of the airplane in miles per hour and let \(w\) represent the speed of the wind in miles per hour.
With the wind, the airplane’s total speed is \(a+w\text{.}\) Flying against the wind, the total speed is \(a-w\text{.}\)
We use the formula \(d=r\cdot t\) to find the unknown distances. Be careful when grouping the quantities that represent the rate in parentheses:
\begin{gather*}
\text{Flight with wind: }\, d=r\cdot t\implies 240=(a+w)\cdot 2\\
\text{Flight against wind: }\, d=r\cdot t\implies 135=(a-w)\cdot 1.5
\end{gather*}
If we divide both sides of the first equation by \(2\) (or multiply by \(\frac{1}{2}\)) and both sides of the second equation by \(1.5=\frac{3}{2}\) (or multiply by the reciprocal \(\frac{2}{3}\)), then we obtain the following equivalent system:
\begin{equation*}
\begin{cases} \left(\alert{\frac{1}{2}}\right)(240=(a+w)\cdot 2)\\ \left(\alert{\frac{2}{3}}\right)\left(135=(a-w)\cdot\frac{3}{2}\right)\\ \end{cases} = \begin{cases} 120=a+w\\ 90=a-w\\ \end{cases}
\end{equation*}
Here \(w\) is lined up to eliminate.
\begin{equation*}
\begin{aligned} a\alert{+w}\amp =120\\ a \alert{-w}\amp =90\\ \hline\\ 2a\amp = 210 \end{aligned}
\end{equation*}
This results in \(a=\frac{210}{2}=105\text{.}\) Now we substitute to find the speed \(w\) of the wind:
\begin{equation*}
\begin{aligned} a+w\amp = 120\\ \alert{105}+w\amp =120\\ w\amp =15 \end{aligned}
\end{equation*}
Dale piloted his airplane at a speed of \(105\) miles per hour and the speed of the wind is \(15\) miles per hour.
Exercises 3.2.3 Exercises
1. Dimensions of a Room.
This is similar to Problem 1 from your Section 3.2 worksheet.
The length of a rectangular room is \(3\) feet more than the width. The perimeter of the room is \(34\) feet. Find the dimensions of the room.
The width of the room is and the length of the room is . (Be sure to include in your answers)
Hint.
Start by defining variables for the two quantities we’re looking for. Can you find a relationship between the unknown quantities and the given information?
Since the length is \(3\) more than the width, can you write and equation relating the length, the width, and \(3\text{?}\) Can you use the perimeter formula to write an equation relating the length, the width, and the given perimeter?
2. Mystery Numbers.
This is similar to Problem 2 from your Section 3.2 worksheet.
Find two numbers whose sum is \(11\) and whose difference is \(29\text{.}\)
The two numbers are (separate the two numbers with a comma).
Hint.
Define a variable for each of the numbers. What does it mean that their sum is \(11\text{?}\) What does it mean that their difference is \(29\text{?}\)
3. Ticket Sales.
This is similar to Problem 3 from your Section 3.1 worksheet.
Leslie Knope and April Ludgate are selling tickets to a Lil Sebastian memorial Concert. Leslie sells \(6\) standard tickets and \(6\) VIP tickets for a total of $\(138\text{.}\) April sells \(3\) standard tickets and \(7\) VIP tickets for a total of $\(129\text{.}\) What is the cost of each type of ticket?
The standard tickets cost $ each and the VIP tickets cost $ each.
4. Coins.
This is similar to Problem 4 from your Section 3.2 worksheet.
Brent has $\(42.95\) in nickels and quarters. He has \(23\) more quarters than nickels. Determine how many coins of each type Brent has.
He has nickels and quarters.
5. Sides of Shapes.
This is similar to Problem 5 from your Section 3.2 worksheet.
The side of a square is \(4\) cm longer than the side of an equilateral triangle (a triangle whose sides are all the same length). The perimeter of the square is \(27\) cm longer than the perimeter of the triangle. Find the length of the sides of the triangle and the length of the sides of the square.
The sides of the triangle are and the sides of the square are (be sure to include )
Hint.
If \(s\) is the length of each side of a square, then the perimeter of the square is \(4s\text{.}\) If \(t\) is the length of each side of an equilateral triangle, then the perimeter of the triangle is \(3t\text{.}\)
6. Jellybeans and Chocolate.
This is similar to Problem 6 from your Section 3.2 worksheet.
Bridget wants to purchase a mix of candy. She decides to make a mix of jelly beans worth $\(20\) per kg and chocolate worth $\(23\) per kg. Together, she gets \(87\) kg of the mixture worth $\(21\) per kg. How many kg of each type of candy did she buy?
She purchased of jellybeans and of chocolate (be sure to include )
Hint.
How much much was spent on jellybeans? On chocolate? In total?
7. Seeds.
This is similar to Problem 7 from your Section 3.2 worksheet.
The Garden Center ordered \(12\) ounces of marigold seed and \(7\) ounces of carnation seed paying $\(267.36\text{.}\) On a different day, they ordered another \(14\) ounces of marigold seed and \(14\) ounces of carnation seed, paying $\(372.12\text{.}\) Find the price per ounce for each type of seed.
The marigold seeds cost dollars per ounce and the carnation seeds cost dollars per ounce.
Hint.
Remember to start by defining a variable for each of the unknowns we’re looking for. Using these variables, how much was spend on marigolds on the first trip? How much was spend on carnation seeds on the first trip?
8. Taxi Fare.
This is similar to Problem 8 from your Section 3.2 worksheet.
A taxi charges a flat rate plus a certain charge per mile. A trip of \(4\) miles costs $\(12.30\text{,}\) while a trip of \(8\) miles costs $\(22.30\text{.}\) Find the flat rate and the charge per mile.
The taxi charges a flat rate of $ and charges $ per mile.
9. Antifreeze.
This is similar to Problem 9 from your Section 3.2 worksheet.
Your car radiator can hold \(20\) liters of fluid. Many automotive stores sell jugs of pure antifreeze, but some also sell jugs of antifreeze that have been premixed with water. Suppose you buy a jug of \(20\)% antifreeze mixture as well as a jug of pure antifreeze. How many liters of each jug must you use to make enough of a \(78\)% mixture to fill your radiator?
You need of the \(20\)% antifreeze and of the pure antifreeze. Be sure to include .
10. Ice Cream.
This is similar to Problem 10 from your Section 3.2 worksheet.
Tony the Ice Cream Maker is making multiple catches of two different flavors of ice cream today. The vanilla ice cream uses \(2\) pounds of sugar and \(4\) gallons of milk per batch. The chocolate ice cream uses \(5\) pounds of sugar and \(6\) gallons of milk per batch. At the end of the day, Tony had used a total of \(69\) pounds of sugar and \(102\) gallons of milk to make the two types of ice cream. How many batches of each type of ice cream did Tony make?
Tony made batches of vanilla ice cream and batches of chocolate ice cream.
11. Rowing in a Canoe.
This is similar to Problem 11 from your Section 3.2 worksheet.
Maddy and Carl row their canoe \(60\) miles downstream in \(4\) hours. After a picnic, they row their canoe back upstream. After \(3\) hours of rowing upstream, they only travel \(15\) miles. Assuming that Maddy and Carl canoe at a constant rate and that the river’s current is constant, find the speed at which Maddy and Carl row in still water.
Maddy and Carl row in still water. Be sure to include .
Hint.
Going downstream means that the speed of the current is adding to Maddy and Carl’s speed, while doing upstream means that it’s working against them (and therefore subtracting from their speed).
Recall the distance formula, \(d=rt\text{.}\)
12. Meat and Cheese.
This is similar to Problem 12 from your Section 3.2 worksheet.
Michelle is picking up sandwich ingredients at the grocery store. She goes to the deli counter and finds roast beef for $\(10.00\) per pound and cheddar cheese for $\(10.25\) per pound. She buys some of each and walks away with \(4.5\) pounds of meat and cheese costing a total of $\(45.50\text{.}\) How many pounds of each ingredient did Michelle buy?
Michelle bought pounds of roast beef and pounds of cheddar cheese.
Hint.
How much money did Michelle spend on beef? How much did she spend on cheese? How much did she spend in total?
