In this section, you will...
write and solve systems of linear equations that represent real world applications
SectionApplications 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:
Example110
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*}
\begin{cases}
(\alert{2})\amp (4x\amp +5y\amp =7)\\
(\alert{5})\amp 3 x\amp -2y\amp =11)\\
\end{cases}=
\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 \(-1\text{.}\)
Example111
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 \(7\text{.}\)
Example112
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 (\(I=prt=pr1=pr\)). 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*}
\begin{cases}
x\amp +y\amp =12,800\\
0.03125x \amp + 0.0475y \amp = 465
\end{cases}
\end{equation*}
Eliminate \(x\) by multiplying the first equation by \(-0.03125\text{.}\)
\begin{equation*}
\begin{cases}
(\alert{-0.03125})\amp ( x\amp +y\amp =12,800)\\
\amp 0.03125x \amp + 0.0475y \amp = 465\\
\end{cases}\\=
\begin{cases}
-0.03125x\amp -0.03125y\amp = -400\\
0.03125x \amp + 0.0475y \amp = 465
\end{cases}
\end{equation*}
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 \(3.125\)%.
Example113
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*}
\begin{cases}
n\amp +d\amp =58\\
0.05n \amp + 0.10d\amp = 4.20
\end{cases}
\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.
Example114
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.
SubsectionMixture 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.
Example115
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*}
\begin{cases}
x\amp +y\amp =35\\
0.018x \amp + 0.032y\amp = 0.022(35)
\end{cases}
\end{equation*}
Here we will solve using the elimination method.
\begin{equation*}
\begin{cases}
(\alert{-0.018})\amp(x\amp +y\amp =35)\\
\amp 0.018x \amp + 0.032y\amp = 0.022(35)
\end{cases}\\
=\begin{cases}
-0.018x\amp -0.018y\amp =-0.63\\
0.018x \amp + 0.032y\amp = 0.022(35)
\end{cases}
\end{equation*}
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.
Example116
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*}
\begin{cases}
x\amp +y\amp =48\\
0.80x \amp +\amp = 0.25(48)=12
\end{cases}
\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.
SubsectionDistance 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.
Example117
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*}
\begin{cases}
x\amp +y\amp =4\\
50x \amp +320y\amp = 875
\end{cases}
\end{equation*}
We will solve using the elimination method.
\begin{equation*}
\begin{cases}
(\alert{-50})\amp(x\amp +y\amp =4)\\
\amp 50x \amp +320y\amp = 875
\end{cases}\\
=
\begin{cases}
\amp -50x\amp -50y\amp =-200\\
\amp 50x \amp +320y\amp = 875
\end{cases}
\end{equation*}
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.
Example118
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.