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4.4: Applications of Linear Systems

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• Page ID 18350

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.

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:

And the second key phrase, “the difference is \(8\),” leads us to the second equation:

Therefore, our algebraic setup consists of the following system:

\(\left\{\begin{aligned} x+y&=40 \\ x-y&=8 \end{aligned}\right.\)

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 \(y\).

\(\begin{aligned} x+y&=40 \\ \color{OliveGreen}{24}\color{black}{+y}&=40 \\ 24+y\color{Cerulean}{-24}&=40\color{Cerulean}{-24} \\ y&=16 \end{aligned}\)

The sum of the two numbers should be \(42\) and their difference \(8\).

\(\begin{aligned} 24+16&=40 \\ 24-16&=8 \end{aligned}\)

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.

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:

\(\left\{\begin{aligned} 9x+2y&=6\\3x-y&=7 \end{aligned}\right.\)

Solve using the elimination method. Multiply the second equation by \(2\) and add.

\(\left\{\begin{aligned} 9x+2y&=6 \\ 3x-y&=7 \end{aligned}\right. \stackrel{\times 2}{\Rightarrow} \left\{\begin{aligned} 9x+2y&=6\\ 6x-2y&=14 \end{aligned}\right.\)

\(\begin{aligned} 9x\color{red}{+2y}&=6\\ \underline{+\quad 6x\color{red}{-2y}}&\underline{=14} \\ 15x&=20 \\ x&=\frac{20}{15} \\ x&=\frac{4}{3} \end{aligned}\)

Back substitute to find \(y\).

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.

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?

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.

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 \(−5\).

Now add the equations together:

Once we have \(y\), the number of $\(10\) bills, back substitute to find \(x\).

\(\begin{aligned} x+y&=32 \\ x+\color{OliveGreen}{12}&=32 \\ x+12\color{Cerulean}{-12}&=32\color{Cerulean}{-12} \\ x&=20 \end{aligned}\)

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?

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 \(−0.0375\).

\(\left\{\begin{aligned} x+y&=6,300 \\ 0.045x+0.0375y&=267.75 \end{aligned}\right. \stackrel{\times (-0.0375)}{\Rightarrow} \left\{\begin{aligned} -0.0375x-0.0375y&=-236.25 \\ 0.045x+0.0375y&=267.75 \end{aligned}\right.\)

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.

\(\begin{aligned} x+y&=6,300 \\ \color{OliveGreen}{4,200}\color{black}{+y}&=6,300 \\ 4,200+y\color{Cerulean}{-4,200}&=6,300\color{Cerulean}{-4,200} \\ y&=2,100 \end{aligned}\)

$\(4,200\) 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 two-day 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?

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:

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?

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,

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.

\(\begin{array}{ccccc}{\color{Cerulean}{salt\:in2\%\:solution}}&{\color{Cerulean}{+}}&{\color{Cerulean}{salt\:in\:5\%\:solution}}&{\color{Cerulean}{=}}&{\color{Cerulean}{salt\:in\:the\:end\:solution}}\\{0.02x}&{+}&{0.05y}&{=}&{0.025(72)}\end{array}\)

\(\left\{\begin{aligned} x+y&=72 \\ 0.02x+0.05y&=0.025(72) \end{aligned}\right. \stackrel{\times (-0.02)}{\Rightarrow} \left\{\begin{aligned} -0.02x-0.02y&=-1.44 \\ 0.02x+0.05y&=1.8 \end{aligned}\right.\)

\(\begin{aligned} \color{red}{-0.02x}\color{black}{-0.02y}&=-1.44 \\ \underline{+\quad\color{red}{0.02x}\color{black}{+0.05y}}&\underline{=1.8} \\ 0.03y&=0.36 \\ \frac{0.03y}{\color{Cerulean}{0.03}}&=\frac{0.36}{\color{Cerulean}{0.03}} \\ y&=12 \end{aligned}\)

\(\begin{aligned} x+y&=72 \\ x+\color{OliveGreen}{12}&=72 \\ x+12\color{Cerulean}{-12}&=72\color{Cerulean}{-12} \\ x&=60 \end{aligned}\)

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?

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.

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.

\(\begin{array}{c|c} {Equation\:1:}&{Equation\:2:}\\{\color{Cerulean}{-10}\color{black}{(x+y)=}\color{Cerulean}{-10}\color{black}{(8)}}&{\color{Cerulean}{100}\:\color{black}{0.50x+0.10y=}\color{Cerulean}{100}\color{black}{(0.32)(8)}}\\{-10x-10y=-80}&{50x+10y=256} \end{array}\)

We obtain the following equivalent system:

Add the equations and then solve for \(x\):

\(\begin{aligned} x+y&=8 \\ \color{OliveGreen}{4.4}\color{black}{+y}&=8 \\ 4.4+y\color{Cerulean}{-4.4}&=8\color{Cerulean}{-4.4} \\ x&=3.6 \end{aligned}\)

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?

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?

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.

\(\left\{\begin{aligned} x+y&=8 &\color{Cerulean}{\leftarrow\:total\:time\:traveled} \\ 6-x+350y&=1,930 &\color{Cerulean}{\leftarrow\:total\:distance\:traveled} \end{aligned}\right.\)

Now back substitute to find the time it took to drive to the airport \(x\):

\(\begin{aligned} x+y&=8 \\ x+\color{OliveGreen}{5}&=8 \\ x&=3 \end{aligned}\)

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.

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.

\(\left\{\begin{aligned} 1,365&=(x+w)\cdot 3 &\color{Cerulean}{\leftarrow\:distance\:traveled\:with\:the\:wind} \\ 870&=(x-w)\cdot 2 &\color{Cerulean}{\leftarrow\:distance\:traveled\:against\:the\:wind} \end{aligned}\right.\)

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:

\(\left\{\begin{aligned} 1,365&=(x+w)\cdot 3 \\ 870&=(x-w)\cdot 2 \end{aligned}\right. \begin{aligned} &\stackrel{\div 3}{\Rightarrow} \\ &\stackrel{\div 2}{\Rightarrow} \end{aligned} \left\{\begin{aligned} 455&=x+w \\ 435&=x-w \end{aligned}\right.\)

\(\begin{aligned} x\color{red}{+w}&=455 \\ \underline{+\quad x\color{red}{-w}}&\underline{=435} \\ 2x&=890 \\ \frac{2x}{\color{Cerulean}{2}}&=\frac{890}{\color{Cerulean}{2}} \\ x&=455 \end{aligned}\)

\(\begin{aligned} x+w&=455 \\ \color{OliveGreen}{455}\color{black}{+w}&=455 \\ w&=10 \end{aligned}\)

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?

The speed of the boat is \(9\) miles per hour and the speed of the current is \(3\) miles per hour.

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• 1. Multiple Choice Edit 1 minute 1 pt A system of linear equations is...  Math Magic One Equation Always more than 2 equations A set of 2 or more equations
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Solve for x and y by graphing the system

Manny’s Music Rental charges a fee of $40 plus $25 per month to rent a saxophone. Sid’s Saxophones charges $25 plus $30 per month to rent the same saxophone.

Write a system of linear equations to represent this situation.

y = 40x + 25

y = 30x + 25

y = 25x + 40

y = 25x + 30

• 9. Multiple Choice Edit 3 minutes 1 pt What does the x represent in this situation? Manny’s Music Rental charges a fee of $40 plus $25 per month to rent a saxophone. Sid’s Saxophones charges $25 plus $30 per month to rent the same saxophone. y = 25x + 40 y = 30x + 25 price of a saxophone rental fee total cost number of months
• 10. Multiple Choice Edit 45 seconds 1 pt The solution of a system of linear equations is... the y-intercept (b) the intersection of the two equations on a graph the second equations' answers the slope (m)

Solve the system by graphing. What does this solution represent?

115; the total cost of renting which is the same for both rental services

(115, 3); the total cost of renting a saxophone will be $115 after 3 months for both services

(3, 115); the total cost of renting saxophone will be $115 after 3 months for both services

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What are systems of equations?

A system of equations is a set of one or more equations involving a number of variables..

The solutions to systems of equations are the variable mappings such that all component equations are satisfied—in other words, the locations at which all of these equations intersect. To solve a system is to find all such common solutions or points of intersection.

Systems of linear equations are a common and applicable subset of systems of equations. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. The system is said to be inconsistent otherwise, having no solutions. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent).

More general systems involving nonlinear functions are possible as well. These possess more complicated solution sets involving one, zero, infinite or any number of solutions, but work similarly to linear systems in that their solutions are the points satisfying all equations involved. Going further, more general systems of constraints are possible, such as ones that involve inequalities or have requirements that certain variables be integers.

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