4.4: Applications of Linear Systems
 Page ID
 18350
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Learning Objectives
 Set up and solve applications involving relationships between numbers.
 Set up and solve applications involving interest and money.
 Set up and solve mixture problems.
 Set up and solve uniform motion problems (distance problems).
Problems Involving Relationships between Real Numbers
We now have the techniques needed to solve linear systems. For this reason, we are no longer limited to using one variable when setting up equations that model applications. If we translate an application to a mathematical setup using two variables, then we need to form a linear system with two equations.
Example \(\PageIndex{1}\)
The sum of two numbers is \(40\) and their difference is \(8\). Find the numbers.
Solution:
Identify variables:
Let \(x\) represent one of the unknown numbers.
Let \(y\) represent the other unknown number.
Set up equations:
When using two variables, we need to set up two equations. The first key phrase, “the sum of the two numbers is \(40\),” translates as follows:
\(x+y=40\)
And the second key phrase, “the difference is \(8\),” leads us to the second equation:
\(xy=8\)
Therefore, our algebraic setup consists of the following system:
\(\left\{\begin{aligned} x+y&=40 \\ xy&=8 \end{aligned}\right.\)
Solve:
We can solve the resulting system using any method of our choosing. Here we choose to solve by elimination. Adding the equations together eliminates the variable \(y\).
\(\begin{aligned} x\color{red}{+y}&=40 \\ \underline{+\quad x\color{red}{y}}&\underline{=8} \\ 2x&=48 \\ x&=24 \end{aligned}\)
Once we have \(x\), back substitute to find \(y\).
\(\begin{aligned} x+y&=40 \\ \color{OliveGreen}{24}\color{black}{+y}&=40 \\ 24+y\color{Cerulean}{24}&=40\color{Cerulean}{24} \\ y&=16 \end{aligned}\)
Check:
The sum of the two numbers should be \(42\) and their difference \(8\).
\(\begin{aligned} 24+16&=40 \\ 2416&=8 \end{aligned}\)
Answer:
The two numbers are \(24\) and \(16\).
Example \(\PageIndex{2}\)
The sum of \(9\) times a larger number and twice a smaller is \(6\). The difference of \(3\) times the larger and the smaller is \(7\). Find the numbers.
Solution:
Begin by assigning variables to the larger and smaller number.
Let \(x\) represent the larger number.
Let \(y\) represent the smaller number.
The first sentence describes a sum and the second sentence describes a difference.
This leads to the following system:
\(\left\{\begin{aligned} 9x+2y&=6\\3xy&=7 \end{aligned}\right.\)
Solve using the elimination method. Multiply the second equation by \(2\) and add.
\(\left\{\begin{aligned} 9x+2y&=6 \\ 3xy&=7 \end{aligned}\right. \stackrel{\times 2}{\Rightarrow} \left\{\begin{aligned} 9x+2y&=6\\ 6x2y&=14 \end{aligned}\right.\)
\(\begin{aligned} 9x\color{red}{+2y}&=6\\ \underline{+\quad 6x\color{red}{2y}}&\underline{=14} \\ 15x&=20 \\ x&=\frac{20}{15} \\ x&=\frac{4}{3} \end{aligned}\)
Back substitute to find \(y\).
\(\begin{aligned} 3xy&=7 \\ 3\left(\color{OliveGreen}{\frac{4}{3}} \right) \color{black}{y}&=7 \\ 4y&=7 \\ y&=3 \\ y&=3 \end{aligned}\)
Answer:
The larger number is \(\frac{4}{3}\) and the smaller number is \(−3\).
Exercise \(\PageIndex{1}\)
The sum of two numbers is \(3\). When twice the smaller number is subtracted from \(6\) times the larger the result is \(22\). Find the numbers.
 Answer

The two numbers are \(−\frac{1}{2}\) and \(\frac{7}{2}\).
Interest and Money Problems
In this section, the interest and money problems should seem familiar. The difference is that we will be making use of two variables when setting up the algebraic equations.
Example \(\PageIndex{3}\)
A roll of \(32\) bills contains only $\(5\) bills and $\(10\) bills. If the value of the roll is $\(220\), then how many of each bill are in the roll?
Solution:
Begin by identifying the variables.
Let \(x\) represent the number of $\(5\) bills.
Let \(y\) represent the number of $\(10\) bills.
When using two variables, we need to set up two equations. The first equation is created from the fact that there are \(32\) bills.
\(x+y=32\)
The second equation sums the value of each bill: the total value is $\(220\).
$\(5\cdot x+\)$\(10\cdot y=\)$\(220\)
Present both equations as a system; this is our algebraic setup.
\(\left\{\begin{aligned} x+y&=32 \\ 5x+10y&=220 \end{aligned}\right.\)
Here we choose to solve by elimination, although substitution would work just as well. Eliminate \(x\) by multiplying the first equation by \(−5\).
\(\left\{\begin{aligned} x+y&=32 \\ 5x+10y&=220 \end{aligned}\right. \stackrel{\times (5)}{\Rightarrow} \left\{\begin{aligned} 5x5y&=160 \\ 5x+10y&=220 \end{aligned}\right.\)
Now add the equations together:
\(\begin{aligned} \color{red}{5x}\color{black}{5y}&=160 \\ \underline{+\quad\color{red}{5x}\color{black}{+10y}}&\underline{=220} \\ 5y&=60 \\ \frac{5y}{\color{Cerulean}{5}}&=\frac{60}{\color{Cerulean}{5}} \\ y&=12 \end{aligned}\)
Once we have \(y\), the number of $\(10\) bills, back substitute to find \(x\).
\(\begin{aligned} x+y&=32 \\ x+\color{OliveGreen}{12}&=32 \\ x+12\color{Cerulean}{12}&=32\color{Cerulean}{12} \\ x&=20 \end{aligned}\)
Answer:
There are twenty $\(5\) bills and twelve $\(10\) bills. The check is left to the reader.
Example \(\PageIndex{4}\)
A total of $\(6,300\) was invested in two accounts. Part was invested in a CD at a \(4\frac{1}{2}\)% annual interest rate and part was invested in a money market fund at a \(3\frac{3}{4}\)% annual interest rate. If the total simple interest for one year was $\(267.75\), then how much was invested in each account?
Solution:
Let \(x\) represent the amount invested at \(4\frac{1}{2}\)%\(=4.5\)%\(=0.045\)
Let \(y\) represent the amount invested at \(3\frac{3}{4}\)%\(=3.75\)%\(=0.0375\)
The total amount in both accounts can be expressed as
\(x+y=6,300\)
To set up a second equation, use the fact that the total interest was $\(267.75\). Recall that the interest for one year is the interest rate times the principal \((I=prt=pr⋅1=pr)\). Use this to add the interest in both accounts. Be sure to use the decimal equivalents for the interest rates given as percentages.
\(\begin{array}{ccccc}{\color{Cerulean}{interest\:from\:the\:CD}}&{\color{Cerulean}{+}}&{\color{Cerulean}{interest\:from\:the\:fund}}&{\color{Cerulean}{=}}&{\color{Cerulean}{total\:interest}}\\{0.045x}&{+}&{0.375y}&{=}&{267.75} \end{array}\)
These two equations together form the following linear system:
\(\left\{\begin{aligned} x+y&=6,300 \\ 0.045x+0.0375y&=267.75 \end{aligned}\right.\)
Eliminate \(y\) by multiplying the first equation by \(−0.0375\).
\(\left\{\begin{aligned} x+y&=6,300 \\ 0.045x+0.0375y&=267.75 \end{aligned}\right. \stackrel{\times (0.0375)}{\Rightarrow} \left\{\begin{aligned} 0.0375x0.0375y&=236.25 \\ 0.045x+0.0375y&=267.75 \end{aligned}\right.\)
Next, add the equations together to eliminate the variable \(y\).
\(\begin{aligned} 0.0375x\color{red}{0.0375y}&=236.25 \\ \underline{+\quad 0.045x\color{red}{+0.0375y}}&\underline{=267.75} \\ 0.0075x&=31.5 \\ \frac{0.0075x}{\color{Cerulean}{0.0075}}&=\frac{31.5}{\color{Cerulean}{0.0075}} \\ x&=4,200 \end{aligned}\)
Back substitute.
\(\begin{aligned} x+y&=6,300 \\ \color{OliveGreen}{4,200}\color{black}{+y}&=6,300 \\ 4,200+y\color{Cerulean}{4,200}&=6,300\color{Cerulean}{4,200} \\ y&=2,100 \end{aligned}\)
Answer:
$\(4,200\) was invested at \(4\frac{1}{2}\)% and $\(2,100\) was invested at \(3\frac{3}{4}\)%
At this point, we should be able to solve these types of problems in two ways: with one variable and now with two variables. Setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear.
Exercise \(\PageIndex{2}\)
On the first day of a twoday meeting, \(10\) coffees and \(10\) doughnuts were purchased for a total of $\(20.00\). Since nobody drank the coffee and all the doughnuts were eaten, the next day only \(2\) coffees and \(14\) doughnuts were purchased for a total of $\(13.00\). How much did each coffee and each doughnut cost?
 Answer

Coffee: $\(1.25\); doughnut: $\(0.75\)
Mixture Problems
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 
Table 4.4.1
In other words, we multiply the percentage times the total to get the amount of each part of the mixture.
Example \(\PageIndex{5}\)
A \(2\)% saline solution is to be combined and mixed with a \(5\)% saline solution to produce \(72\) ounces of a \(2.5\)% saline solution. How much of each is needed?
Solution:
Let \(x\) represent the amount of \(2\)% saline solution needed.
Let \(y\) represent the amount of \(5\)% saline solution needed.
The total amount of saline solution needed is \(72\) ounces. This leads to one equation,
\(x+y=72\)
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.
\(\begin{array}{ccccc}{\color{Cerulean}{salt\:in2\%\:solution}}&{\color{Cerulean}{+}}&{\color{Cerulean}{salt\:in\:5\%\:solution}}&{\color{Cerulean}{=}}&{\color{Cerulean}{salt\:in\:the\:end\:solution}}\\{0.02x}&{+}&{0.05y}&{=}&{0.025(72)}\end{array}\)
Solve.
\(\left\{\begin{aligned} x+y&=72 \\ 0.02x+0.05y&=0.025(72) \end{aligned}\right. \stackrel{\times (0.02)}{\Rightarrow} \left\{\begin{aligned} 0.02x0.02y&=1.44 \\ 0.02x+0.05y&=1.8 \end{aligned}\right.\)
\(\begin{aligned} \color{red}{0.02x}\color{black}{0.02y}&=1.44 \\ \underline{+\quad\color{red}{0.02x}\color{black}{+0.05y}}&\underline{=1.8} \\ 0.03y&=0.36 \\ \frac{0.03y}{\color{Cerulean}{0.03}}&=\frac{0.36}{\color{Cerulean}{0.03}} \\ y&=12 \end{aligned}\)
Back substitute.
\(\begin{aligned} x+y&=72 \\ x+\color{OliveGreen}{12}&=72 \\ x+12\color{Cerulean}{12}&=72\color{Cerulean}{12} \\ x&=60 \end{aligned}\)
Answer:
We need \(60\) ounces of the \(2\)% saline solution and \(12\) ounces of the \(5\)% saline solution.
Example \(\PageIndex{6}\)
A \(50\)% alcohol solution is to be mixed with a \(10\)% alcohol solution to create an \(8\)ounce mixture of a \(32\)% alcohol solution. How much of each is needed?
Solution:
Let \(x\) represent the amount of \(50\)% alcohol solution needed.
Let \(y\) represent the amount of \(10\)% alcohol solution needed.
The total amount of the mixture must be \(8\) ounces.
\(x+y=8\)
The second equation adds up the amount of alcohol from each solution in the correct percentages. The amount of alcohol in the end result is \(32\)% of \(8\) ounces, or \(0.032(8)\).
\(\begin{array}{ccccc}{\color{Cerulean}{alcohol\:in\:50\%\:solution}}&{\color{Cerulean}{+}}&{\color{Cerulean}{alcohol\:in\:10\%\:solution}}&{\color{Cerulean}{=}}&{\color{Cerulean}{alcohol\:in\:the\:end\:solution}}\\{0.50x}&{+}&{0.10y}&{=}&{0.32(8)}\end{array}\)
Now we can form a system of two linear equations and two variables as follows:
\(\left\{\begin{aligned} x+y&=8 \\ 0.50x+0.10y&=0.32(8) \end{aligned}\right.\)
In this example, multiply the second equation by \(100\) to eliminate the decimals. In addition, multiply the first equation by \(−10\) to line up the variable \(y\) to eliminate.
\(\begin{array}{cc} {Equation\:1:}&{Equation\:2:}\\{\color{Cerulean}{10}\color{black}{(x+y)=}\color{Cerulean}{10}\color{black}{(8)}}&{\color{Cerulean}{100}\:\color{black}{0.50x+0.10y=}\color{Cerulean}{100}\color{black}{(0.32)(8)}}\\{10x10y=80}&{50x+10y=256} \end{array}\)
We obtain the following equivalent system:
\(\left\{\begin{aligned} x+y&=8 \\ 0.50x+0.10y&=0.32(8) \end{aligned}\right. \begin{aligned} &\stackrel{\times (10)}{\Rightarrow} \\ &\stackrel{\times 100}{\Rightarrow} \end{aligned} \left\{\begin{aligned} 10x10y&=80 \\ 50x+10y&=256 \end{aligned}\right.\)
Add the equations and then solve for \(x\):
\(\begin{aligned} 10x\color{red}{10y}&=80 \\ \underline{+\quad 50x\color{red}{+10y}}&\underline{=256} \\ 40x&=176 \\ \frac{40x}{\color{Cerulean}{40}}&=\frac{176}{\color{Cerulean}{40}} \\ x&=4.4 \end{aligned}\)
Back substitute.
\(\begin{aligned} x+y&=8 \\ \color{OliveGreen}{4.4}\color{black}{+y}&=8 \\ 4.4+y\color{Cerulean}{4.4}&=8\color{Cerulean}{4.4} \\ x&=3.6 \end{aligned}\)
Answer:
To obtain \(8\) ounces of a \(32\)% alcohol mixture we need to mix \(4.4\) ounces of the \(50\)% alcohol solution and \(3.6\) ounces of the \(10\)% solution.
Exercise \(\PageIndex{3}\)
A \(70\)% antifreeze concentrate is to be mixed with water to produce a \(5\)gallon mixture containing \(28\)% antifreeze. How much water and antifreeze concentrate is needed?
 Answer

We need to mix \(3\) gallons of water with \(2\) gallons of antifreeze concentrate.
Uniform Motion Problems (Distance Problems)
Recall that the distance traveled is equal to the average rate times the time traveled at that rate, \(D=r⋅t\).
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 \(\PageIndex{7}\)
An executive traveled a total of \(8\) hours and \(1,930\) miles by car and by plane. Driving to the airport by car, she averaged \(60\) miles per hour. In the air, the plane averaged \(350\) 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.
Let \(x\) represent the time it took to drive to the airport.
Let \(y\) represent the time spent in the air.
Use the formula \(D=r⋅t\) to fill in the unknown distances.
\(\color{Cerulean}{Distance\:traveled\:in\:the\:car:}\quad\color{black}{D=r\cdot t=60\cdot x}\)
\(\color{Cerulean}{Distance\:traveled\:in\:the\:air:}\quad\color{black}{D=r\cdot t=350\cdot y}\)
The distance column and the time column of the chart help us to set up the following linear system.
\(\left\{\begin{aligned} x+y&=8 &\color{Cerulean}{\leftarrow\:total\:time\:traveled} \\ 6x+350y&=1,930 &\color{Cerulean}{\leftarrow\:total\:distance\:traveled} \end{aligned}\right.\)
Solve.
\(\left\{\begin{aligned} x+y&=8 \\ 60x+350y&=1,930 \end{aligned}\right. \stackrel{\times (60)}{\Rightarrow} \left\{\begin{aligned} 60x60y&=480 \\ 60x+350y&=1,930 \end{aligned}\right.\)
\(\begin{aligned} \color{red}{60x}\color{black}{60y}&=480 \\ \underline{+\quad\color{red}{60x}\color{black}{+350y}}&\underline{=1,930} \\ 290y&=1450 \\ \frac{290y}{\color{Cerulean}{290}}&=\frac{1450}{\color{Cerulean}{290}} \\ y&=5 \end{aligned}\)
Now back substitute to find the time it took to drive to the airport \(x\):
\(\begin{aligned} x+y&=8 \\ x+\color{OliveGreen}{5}&=8 \\ x&=3 \end{aligned}\)
Answer:
It took her \(3\) 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 \(\PageIndex{8}\)
Flying with the wind, an airplane traveled \(1,365\) miles in \(3\) hours. The plane then turned against the wind and traveled another \(870\) miles in \(2\) hours. Find the speed of the airplane and the speed of the wind.
Solution:
There is no obvious relationship between the speed of the plane and the speed of the wind. For this reason, use two variables as follows:
Let \(x\) represent the speed of the airplane.
Let \(w\) represent the speed of the wind.
Use the following chart to organize the data:
With the wind, the airplane’s total speed is \(x+w\). Flying against the wind, the total speed is \(x−w\).
Use the rows of the chart along with the formula \(D=r⋅t\) to construct a linear system that models this problem. Take care to group the quantities that represent the rate in parentheses.
\(\left\{\begin{aligned} 1,365&=(x+w)\cdot 3 &\color{Cerulean}{\leftarrow\:distance\:traveled\:with\:the\:wind} \\ 870&=(xw)\cdot 2 &\color{Cerulean}{\leftarrow\:distance\:traveled\:against\:the\:wind} \end{aligned}\right.\)
If we divide both sides of the first equation by \(3\) and both sides of the second equation by \(2\), then we obtain the following equivalent system:
\(\left\{\begin{aligned} 1,365&=(x+w)\cdot 3 \\ 870&=(xw)\cdot 2 \end{aligned}\right. \begin{aligned} &\stackrel{\div 3}{\Rightarrow} \\ &\stackrel{\div 2}{\Rightarrow} \end{aligned} \left\{\begin{aligned} 455&=x+w \\ 435&=xw \end{aligned}\right.\)
\(\begin{aligned} x\color{red}{+w}&=455 \\ \underline{+\quad x\color{red}{w}}&\underline{=435} \\ 2x&=890 \\ \frac{2x}{\color{Cerulean}{2}}&=\frac{890}{\color{Cerulean}{2}} \\ x&=455 \end{aligned}\)
Back substitute.
\(\begin{aligned} x+w&=455 \\ \color{OliveGreen}{455}\color{black}{+w}&=455 \\ w&=10 \end{aligned}\)
Answer:
The speed of the airplane is \(445\) miles per hour and the speed of the wind is \(10\) miles per hour.
Exercise \(\PageIndex{4}\)
A boat traveled \(24\) miles downstream in \(2\) hours. The return trip, which was against the current, took twice as long. What are the speeds of the boat and of the current?
 Answer

The speed of the boat is \(9\) miles per hour and the speed of the current is \(3\) miles per hour.
Key Takeaways
 Use two variables as a means to simplify the algebraic setup of applications where the relationship between unknowns is unclear.
 Carefully read the problem several times. If two variables are used, then remember that you need to set up two linear equations in order to solve the problem.
 Be sure to answer the question in sentence form and include the correct units for the answer.
Exercise \(\PageIndex{5}\) Applications Involving Numbers
Set up a linear system and solve.
 The sum of two integers is \(54\) and their difference is \(10\). Find the integers.
 The sum of two integers is \(50\) and their difference is \(24\). Find the integers.
 The sum of two positive integers is \(32\). When the smaller integer is subtracted from twice the larger, the result is \(40\). Find the two integers.
 The sum of two positive integers is \(48\). When twice the smaller integer is subtracted from the larger, the result is \(12\). Find the two integers.
 The sum of two integers is \(74\). The larger is \(26\) more than twice the smaller. Find the two integers.
 The sum of two integers is \(45\). The larger is \(3\) less than three times the smaller. Find the two integers.
 The sum of two numbers is zero. When \(4\) times the smaller number is added to \(8\) times the larger, the result is \(1\). Find the two numbers.
 The sum of a larger number and \(4\) times a smaller number is \(5\). When \(8\) times the smaller is subtracted from twice the larger, the result is \(−2\). Find the numbers.
 The sum of \(12\) times the larger number and \(11\) times the smaller is \(−36\). The difference of \(12\) times the larger and \(7\) times the smaller is \(36\). Find the numbers.
 The sum of \(4\) times the larger number and \(3\) times the smaller is \(7\). The difference of \(8\) times the larger and \(6\) times the smaller is \(10\). Find the numbers.
 Answer

1. The integers are \(22\) and \(32\).
3. The integers are \(8\) and \(24\).
5. The integers are \(16\) and \(58\).
7. The two numbers are \(−\frac{1}{4}\) and \(\frac{1}{4}\).
9. The smaller number is \(−4\) and the larger is \(\frac{2}{3}\).
Exercise \(\PageIndex{6}\) Interest and Money Problems
Set up a linear system and solve.
 A $\(7,000\) principal is invested in two accounts, one earning \(3\)% interest and another earning \(7\)% interest. If the total interest for the year is $\(262\), then how much is invested in each account?
 Mary has her total savings of $\(12,500\) in two different CD accounts. One CD earns \(4.4\)% interest and another earns \(3.2\)% interest. If her total interest for the year is $\(463\), then how much does she have in each CD account?
 Sally’s $\(1,800\) savings is in two accounts. One account earns \(6\)% annual interest and the other earns \(3\)%. Her total interest for the year is $\(93\). How much does she have in each account?
 Joe has two savings accounts totaling $\(4,500\). One account earns \(3\frac{3}{4}\)% annual interest and the other earns \(2\frac{5}{8}\)%. If his total interest for the year is $\(141.75\), then how much is in each account?
 Millicent has $\(10,000\) invested in two accounts. For the year, she earns $\(535\) more in interest from her \(7\)% mutual fund account than she does from her \(4\)% CD. How much does she have in each account?
 A small business has $\(85,000\) invested in two accounts. If the account earning \(3\)% annual interest earns $\(825\) more in interest than the account earning \(4.5\)% annual interest, then how much is invested in each account?
 Jerry earned a total of $\(284\) in simple interest from two separate accounts. In an account earning \(6\)% interest, Jerry invested $\(1,000\) more than twice the amount he invested in an account earning \(4\)%. How much did he invest in each account?
 James earned a total of $\(68.25\) in simple interest from two separate accounts. In an account earning \(2.6\)% interest, James invested onehalf as much as he did in the other account that earned \(5.2\)%. How much did he invest in each account?
 A cash register contains $\(10\) bills and $\(20\) bills with a total value of $\(340\). If there are \(23\) bills total, then how many of each does the register contain?
 John was able to purchase a pizza for $\(10.80\) with quarters and dimes. If he uses \(60\) coins to buy the pizza, then how many of each did he have?
 Dennis mowed his neighbor’s lawn for a jar of dimes and nickels. Upon completing the job, he counted the coins and found that there were \(4\) less than twice as many dimes as there were nickels. The total value of all the coins is $\(6.60\). How many of each coin did he have?
 Two families bought tickets for the big football game. One family ordered \(2\) adult tickets and \(3\) children’s tickets for a total of $\(26.00\). Another family ordered \(3\) adult tickets and \(4\) children’s tickets for a total of $\(37.00\). How much did each adult ticket cost?
 Two friends found shirts and shorts on sale at a flea market. One bought \(5\) shirts and \(3\) shorts for a total of $\(51.00\). The other bought \(3\) shirts and \(7\) shorts for a total of $\(80.00\). How much was each shirt and each pair of shorts?
 On Monday Joe bought \(10\) cups of coffee and \(5\) doughnuts for his office at a cost of $\(16.50\). It turns out that the doughnuts were more popular than the coffee. Therefore, on Tuesday he bought \(5\) cups of coffee and \(10\) doughnuts for a total of $\(14.25\). How much was each cup of coffee?
 Answer

1. $\(5,700\) at \(3\)% and $\(1,300\) at \(7\)%
3. $\(1,300\) at \(6\)% and $\(500\) at \(3\)%
5. $\(8,500\) at \(7\)% and $\(1,500\) at \(4\)%
7. $\(1,400\) at \(4\)% and $\(3,800\) at \(6\)%
9. \(12\) tens and \(11\) twenties
11. \(52\) dimes and \(28\) nickels
13. Shirts: $\(4.50\); shorts: $\(9.50\)
Exercise \(\PageIndex{7}\) Mixture Problems
Set up a linear system and solve.
 A \(15\)% acid solution is to be mixed with a \(25\)% acid solution to produce \(12\) gallons of a \(20\)% acid solution. How much of each is needed?
 One alcohol solution contains \(12\)% alcohol and another contains \(26\)% alcohol. How much of each should be mixed together to obtain \(5\) gallons of a \(14.8\)% alcohol solution?
 A nurse wishes to obtain \(40\) ounces of a \(1.2\)% saline solution. How much of a \(1\)% saline solution must she mix with a \(2\)% saline solution to achieve the desired result?
 A customer ordered \(20\) pounds of fertilizer that contains \(15\)% nitrogen. To fill the customer’s order, how much of the stock \(30\)% nitrogen fertilizer must be mixed with the \(10\)% nitrogen fertilizer?
 A customer ordered \(2\) pounds of a mixed peanut product containing \(15\)% cashews. The inventory consists of only two mixes containing \(10\)% and \(30\)% cashews. How much of each type must be mixed to fill the order?
 How many pounds of pure peanuts must be combined with a \(20\)% peanut mix to produce \(10\) pounds of a \(32\)% peanut mix?
 How much cleaning fluid with \(20\)% alcohol content, must be mixed with water to obtain a \(24\)ounce mixture with \(10\)% alcohol content?
 A chemist wishes to create a \(32\)ounce solution with \(12\)% acid content. He uses two types of stock solutions, one with \(30\)% acid content and another with \(10\)% acid content. How much of each does he need?
 A concentrated cleaning solution that contains \(50\)% ammonia is mixed with another solution containing \(10\)% ammonia. How much of each is mixed to obtain \(8\) ounces of a \(32\)% ammonia cleaning formula?
 A \(50\)% fruit juice concentrate can be purchased wholesale. Best taste is achieved when water is mixed with the concentrate in such a way as to obtain a \(12\)% fruit juice mixture. How much water and concentrate is needed to make a \(50\)ounce fruit juice drink?
 A \(75\)% antifreeze concentrate is to be mixed with water to obtain \(6\) gallons of a \(25\)% antifreeze solution. How much water is needed?
 Pure sugar is to be mixed with a fruit salad containing \(10\)% sugar to produce \(48\) ounces of a salad containing \(16\)% sugar. How much pure sugar is required?
 Answer

1. \(6\) gallons of each
3. \(32\) ounces of the \(1\)% saline solution and \(8\) ounces of the \(2\)% saline solution
5. \(1.5\) pounds of the \(10\)% cashew mix and \(0.5\) pounds of the \(30\)% cashew mix
7. \(12\) ounces of cleaning fluid
9. \(4.4\) ounces of the \(50\)% ammonia solution and \(3.6\) ounces of the \(10\)% ammonia solution
11. \(4\) gallons
Exercise \(\PageIndex{8}\) Uniform Motion Problems
Set up a linear system and solve.
 An airplane averaged \(460\) miles per hour on a trip with the wind behind it and \(345\) miles per hour on the return trip against the wind. If the total round trip took \(7\) hours, then how long did the airplane spend on each leg of the trip?
 The two legs of a \(330\)mile trip took \(5\) hours. The average speed for the first leg of the trip was \(70\) miles per hour and the average speed for the second leg of the trip was \(60\) miles per hour. How long did each leg of the trip take?
 An executive traveled \(1,200\) miles, part by helicopter and part by private jet. The jet averaged \(320\) miles per hour while the helicopter averaged \(80\) miles per hour. If the total trip took \(4\frac{1}{2}\) hours, then how long did she spend in the private jet?
 Joe took two buses on the \(463\)mile trip from San Jose to San Diego. The first bus averaged \(50\) miles per hour and the second bus was able to average \(64\) miles per hour. If the total trip took \(8\) hours, then how long was spent in each bus?
 Billy canoed downstream to the general store at an average rate of \(9\) miles per hour. His average rate canoeing back upstream was \(4\) miles per hour. If the total trip took \(6\frac{1}{2}\) hours, then how long did it take Billy to get back on the return trip?
 Two brothers drove the \(2,793\) miles from Los Angeles to New York. One of the brothers, driving in the day, was able to average \(70\) miles per hour, and the other, driving at night, was able to average \(53\) miles per hour. If the total trip took \(45\) hours, then how many hours did each brother drive?
 A boat traveled \(24\) miles downstream in \(2\) hours. The return trip took twice as long. What was the speed of the boat and the current?
 A helicopter flying with the wind can travel \(525\) miles in \(5\) hours. On the return trip, against the wind, it will take \(7\) hours. What are the speeds of the helicopter and of the wind?
 A boat can travel \(42\) miles with the current downstream in \(3\) hours. Returning upstream against the current, the boat can only travel \(33\) miles in \(3\) hours. Find the speed of the current.
 A light aircraft flying with the wind can travel \(180\) miles in \(1\frac{1}{2}\) hours. The aircraft can fly the same distance against the wind in \(2\) hours. Find the speed of the wind.
 Answer

1. The airplane flew \(3\) hours with the wind and \(4\) hours against the wind.
3. \(3.5\) hours
5. \(4.5\) hours
7. Boat: \(9\) miles per hour; current: \(3\) miles per hour
9. \(1.5\) miles per hour
Exercise \(\PageIndex{9}\) Discussion Board
 Compose a number or money problem that can be solved with a system of equations of your own and share it on the discussion board.
 Compose a mixture problem that can be solved with a system of equations of your own and share it on the discussion board.
 Compose a uniform motion problem that can be solved with a system of equations of your own and share it on the discussion board
 Answer

1. Answers may vary
3. Answers may vary