Subsection Solving Systems by Graphing
A biologist wants to know the average weights of two species of birds in a wildlife preserve. She sets up a feeder whose platform is actually a scale and mounts a camera to monitor the feeder. She waits until the feeder is occupied only by members of the two species she is studying, doves and blue jays. Then she takes a picture, which records the number of each species on the scale and the total weight registered.
From her two best pictures, she obtains the following information. The total weight of three blue jays and six doves is \(48\) ounces, and the total weight of five blue jays and two doves is \(32\) ounces. Using these data, the biologist estimates the average weight of a blue jay and of a dove. She begins by assigning variables to the two unknown quantities:
\begin{align*}
\amp\text{Average weight of a blue jay:} ~~b \\
\amp\text{Average weight of a dove:} ~~d
\end{align*}
Because there are two variables, the biologist must write two equations about the weights of the birds. In each of the two photos,
\begin{equation*}
(\text{weight of blue jays}) + (\text{weight of robins}) = \text{total weight}
\end{equation*}
Thus,
\begin{alignat*}{3}
3b \amp {}+{} \amp 6d {}\amp={} \amp 48\\
5b \amp {}+{} \amp 2d {}\amp={} \amp 32
\end{alignat*}
This pair of equations is an example of a linear system of two equations in two unknowns (or a \(2\times 2\) linear system, for short). A solution to the system is an ordered pair of numbers, \((b, r)\text{,}\) that satisfies both equations in the system.
Recall that every point on the graph of an equation represents a solution to that equation. A solution to both equations corresponds to a point on both graphs. Therefore, a solution to the system is a point where the two graphs intersect.
From the figure above, it appears that the intersection point is \((4, 6)\text{,}\) so we expect that the values \(b=\alert{4}\) and \(d=\blert{6}\) form the solution to the system. We can check by verifying that these values satisfy both equations in the system.
\begin{align*}
3(\alert{4}) + 6(\blert{6})\amp\stackrel{?}{=}48 \amp\amp \blert{\text{True}}\\
5(\alert{4}) + 2(\blert{6})\amp\stackrel{?}{=}32 \amp\amp \blert{\text{True}}
\end{align*}
Both equations are true, so we conclude that the average weight of a blue jay is \(4\) ounces, and the average weight of a robin is \(6\) ounces.
We can obtain graphs for the equations in a system quickly and easily using a calculator.
Example 8.1
Use your calculator to solve the system by graphing.
\begin{align*}
y \amp = 1.7x + 0.4\\
y \amp = 4.1x + 5.2
\end{align*}
Solution
We set the graphing window to
\begin{align*}
\text{Xmin} \amp = 9.4 \amp\amp \text{Xmax} = 9.4\\
\text{Ymin} \amp = 10 \amp\amp \text{Ymax} = 10
\end{align*}
and enter the two equations. We can see in the figure that the two lines intersect in the third quadrant. We use the TRACE
key to find the coordinates of the intersection point, \((2,3)\text{.}\) The solution to the system is \(x=2\text{,}\) \(y=3\text{.}\)
Checkpoint 8.3

Solve the system of equations
\begin{align*}
y \amp = 0.7x + 6.9\\
y \amp =1.2x  6.4
\end{align*}
by graphing. Use the window
\begin{align*}
\text{Xmin} \amp = 9.4 \amp\amp \text{Xmax} = 9.4\\
\text{Ymin} \amp = 10 \amp\amp \text{Ymax} = 10
\end{align*}
Checkpoint 8.6
Solve the system of equations
\begin{gather*}
y = 47x  1930\\
y + 19x = 710
\end{gather*}
by graphing. Use the intersect feature in the window
\begin{align*}
\text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 94\\
\text{Ymin} \amp = 2000 \amp\amp \text{Ymax} = 1000
\end{align*}
In previous algebra courses, you learned two algebraic techniques for solving \(2\times 2\) linear systems: substitution and elimination. See Algebra Skills Refresher Section A.5 if you would like to review these techniques.
Example 8.7
Rani kayaks downstream for \(45\) minutes and travels a distance of \(6000\) meters. On the return journey upstream, she covers only \(4800\) meters in \(45\) minutes. How fast is the current in the river, and how fast would Rani kayak in still water? (Give your answers in meters per minute.)
Solution

First, we choose variables for the two unknown quantities.
Rani's speed in still water: 
\(r\) 
Speed of the current: 
\(s\) 

We must write two equations using the variables \(r\) and \(s\text{.}\)
First, we organize the information into a table. When Rani travels downstream, the current in the river helps her, so her effective speed is \(r+s\text{.}\) When she travels upstream she is fighting the current, so her speed is actually \(rs\text{.}\)

Rate 
Time 
Distance 
Downstream 
\(r+s\) 
\(45\) 
\(6000\) 
Upstream 
\(rs\) 
\(45\) 
\(4800\) 
Using the formula \(~~\text{Rate} \times \text{Time} = \text{Distance}~~\text{,}\) we write one equation describing Rani's journey downstream, and a second equation for the journey upstream.
\begin{align*}
(r + s) \cdot 45 \amp = 6000\\
(r  s) \cdot 45 \amp = 4800
\end{align*}
We apply the distributive law to write each equation in standard form.
\begin{align*}
45r + 45s \amp = 6000 \amp\amp(1)\\
45r  45s \amp = 4800 \amp\amp(2)
\end{align*}

To solve the system, we eliminate the variable \(s\) by adding the two equations vertically.
\(45r\) 
\(+\) 
\(45s\) 
\(=\) 
\(6000\) 
\(+45r\) 
\(\) 
\(45s\) 
\(=\) 
\(4800\) 
\(90r\) 
\(\) 
\(\) 
\(=\) 
\(10,800\) 
We now have an equation in one variable only, which we can solve for \(r\text{.}\)
\begin{align*}
90r \amp = 10,800\amp\amp \blert{\text{Divide both sides by 90.}}\\
r \amp = 120
\end{align*}
To solve for \(s\) we substitute \(r = 120\) into any previous equation involving both \(r\) and \(s\text{.}\) We will use Equation (1).
\begin{align*}
45(120) + 45s\amp = 6000 \amp\amp \blert{\text{Simplify the left side.}}\\
5400 + 45s \amp = 6000 \amp\amp \blert{\text{Subtract 5400 from both sides.}}\\
45s \amp = 600 \amp\amp \blert{\text{Divide both sides by 45; reduce.}}\\
s\amp = \frac{40}{3}
\end{align*}
The speed of the current is \(\dfrac{40}{3}\text{,}\) or \(13\dfrac{1}{3} \) meters per minute, and Rani's speed in still water is \(120\) meters per minute.
Checkpoint 8.8
It took Leon \(7\) hours to fly the same distance that Marlene drove in \(21\) hours. Leon flies \(120\) miles per hour faster than Marlene drives. At what speed did each travel?

Choose variables for the unknown quantities, and fill in the table.

Rate 
Time 
Distance 
Leon 
\(\) 
\(\) 
\(\) 
Marlene 
\(\) 
\(\) 
\(\) 
Write one equation about the Leon's and Marlene's speeds.
Write a second equation about distances.
Solve the system and answer the question in the problem.
Answer

Rate 
Time 
Distance 
Leon 
\(x\) 
\(7\) 
\(7x\) 
Marlene 
\(y\) 
\(21\) 
\(21y\) 
\(x=y+ 120\) or \(x  y =120\)
\(7x = 21y\)
\((180, 60)\text{.}\) Leon flies at \(180\) mph; Marlene drives at \(60\) mph.
Subsection Inconsistent and Dependent Systems
Because two straight lines do not always intersect at a single point, a \(2\times 2\) system of linear equations does not always have a unique solution. In fact, there are three possibilities, as illustrated below.
The graphs may be the same line, as shown in figure (a).
The graphs may be parallel but distinct lines, as shown in figure (b).
The graphs may intersect in one and only one point, as shown in figure (c).
Example 8.9
Solve the system
\begin{gather*}
y = x + 5\\
2x + 2y = 3
\end{gather*}
Solution
We use the calculator to graph both equations on the same axes, as shown below. First, we rewrite the second equation in slopeintercept form by solving for \(y\text{.}\)
\begin{align*}
2x + 2y \amp = 3\amp\amp \blert{\text{Substract } 2x \text{ from both sides.}}\\
2y \amp = 2x + 3\amp\amp \blert{\text{Divide both sides by 2.}}\\
y \amp = x + 1.5
\end{align*}
Next, we enter the equations as
\begin{align*}
Y_1 \amp= X + 5\\
Y_2 \amp = X + 1.5
\end{align*}
The lines do not intersect within the viewing window; they appear to be parallel. If we look again at the equations of the lines, we recognize that both have slope \(1\) but different \(y\)intercepts, so they are parallel. Because parallel lines never meet, there is no solution to the system.
A system with no solutions, such as the system in Example 8.9, is called inconsistent. A \(2\times 2\) system of linear equations is inconsistent when the two equations correspond to parallel lines. This situation occurs when the lines have the same slope but different \(y\)intercepts.
Checkpoint 8.10
Without graphing, show that the following system is inconsistent:
\begin{gather*}
3y = \frac{3}{2}x  1\\
2x  4y = 3
\end{gather*}
AnswerBoth lines have slope \(\frac{1}{2}\text{.}\)
A linear system with infinitely many solutions is called dependent. A \(2\times 2\) system is dependent when the two equations actually describe the same line. This situation occurs when the two lines have the same slope and the same \(y\)intercept.
Example 8.11
Solve the system
\begin{gather*}
x = \frac{2}{3}y + 3\\
3x  2y = 9
\end{gather*}
Solution
We begin by putting each equation in slopeintercept form.
\begin{align*}
x \amp = \frac{2}{3}y + 3\amp\amp \blert{\text{Subtract 3.}} \\
x 3\amp = \frac{2}{3}y \amp\amp \blert{\text{Multiply by } \frac{3}{2}.}\\
\frac{3}{2}x\frac{9}{2}\amp = y
\end{align*}
For the second equation,
\begin{align*}
3x  2y \amp = 9 \amp\amp \blert{\text{Subtract } 3x.}\\
2y \amp = 3x + 9 \amp\amp \blert{\text{Divide by } 2.}\\
y \amp = \frac{3}{2}x  \frac{9}{2}
\end{align*}
The two equations are actually different forms of the same equation. Because they are equivalent, they share the same line as a graph, as shown at right. Every point on the first line is also a point on the second line, so every solution to the first equation is also a solution of the second equation. Thus, the system has infinitely many solutions.
Here is a summary of the three cases for a \(2\times 2\) system of linear equations.
Solutions of \(2\times 2\) Linear Systems
Dependent system. All the solutions of one equation are also solutions to the second equation and hence are solutions of the system. The graphs of the two equations are the same line. A dependent system has infinitely many solutions.
Inconsistent system. The graphs of the equations are parallel lines and hence do not intersect. An inconsistent system has no solutions.
Consistent and independent system. The graphs of the two lines intersect in exactly one point. The system has exactly one solution.
Checkpoint 8.12

Graph the system
\begin{gather*}
y = 3x + 6\\
6x + 2y = 15
\end{gather*}
by hand, using either the intercept method or the slopeintercept method.
Identify the system as dependent, inconsistent, or consistent and independent.
It is not always easy to tell from the equations themselves whether there is one solution, no solution, or infinitely many solutions. However, the method of elimination will reveal which of the three cases applies.
Example 8.13
Solve the system
\begin{align*}
2x \amp= 2  3y\\
6y \amp= 7  4x
\end{align*}
Solution
First, we rewrite the system in standard form as
\begin{align*}
2x + 3y \amp= 2 \amp\amp (1)\\
4x + 6y \amp = 7 \amp\amp (2)
\end{align*}
We multiply Equation (1) by \(2\) and add the result to Equation (2) to obtain
\(4x\) 
\(\) 
\(6y\) 
\(=\) 
\(4\) 
\(4x\) 
\(+\) 
\(6y\) 
\(=\) 
\(7\) 
\(0x\) 
\(+\) 
\(0y\) 
\(=\) 
\(3\) 
This equation has no solutions. The system is inconsistent. (Notice that both lines have slope \(\dfrac{2}{3}\text{,}\) but they have different \(y\)intercepts, so their graphs are parallel.)
We generalize the results from Example 8.13 as follows.
Inconsistent and Dependent Systems

If an equation of the form
\begin{equation*}
0x + 0y = k \hphantom{blank}(k\ne 0)
\end{equation*}
is obtained as a linear combination of the equations in a system, the system is inconsistent.

If an equation of the form
\begin{equation*}
0x + 0y = 0
\end{equation*}
is obtained as a linear combination of the equations in a system, the system is dependent.
Checkpoint 8.14 illustrates a dependent system.
Checkpoint 8.14

Use the method of elimination to solve the system
\begin{align*}
3x  4 \amp = y\\
2y + 8 \amp = 6x
\end{align*}
Verify that both equations have the same graph.
Answer
Elimination results in \(0x + 0y = 0\text{.}\)
Both equations have slopeintercept form \(y = 3x  4\text{.}\)
Subsection Applications
Many practical problems involve two or more unknown quantities.
Example 8.15
A cup of rolled oats provides 11 grams of protein. A cup of rolled wheat flakes provides 8.5 grams of protein. Francine wants to combine oats and wheat to make a cereal with 10 grams of protein per cup. How much of each grain will she need in one cup of her mixture?
Solution
Fraction of a cup of oats needed: 
\(x\) 
Fraction of a cup of wheat needed: 
\(y\) 

Because we have two variables, we must find two equations that describe the problem. It may be helpful to organize the information into a table.

Cups 
Grams of protein per cup 
Grams of protein 
Oats 
\(x\) 
\(11\) 
\(11x\) 
Wheat 
\(y\) 
\(8.5\) 
\(8.5y\) 
Mixture 
\(1\) 
— 
\(10\) 
The wheat and oats together will make one cup of mixture, so the first equation is
\begin{equation*}
x + y = 1
\end{equation*}
The \(10\) grams of protein must come form the protein in the oats plus the protein in the wheat. This gives us a second equation:
\begin{equation*}
11x + 8.5y = 10
\end{equation*}
We now have a system of equations.

We will solve the system by graphing. First, solve each equation for \(y\) in terms of \(x\) to get
\begin{align*}
y \amp= x + 1\\
y \amp = (10  11x)/8.5
\end{align*}
Although we could simplify the second equation, the calculator can graph both equations as they are. We know that \(x\) and \(y\) represent fractions of one cup, so we set the window (as shown below) with
\begin{align*}
\text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 0.94\\
\text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 1
\end{align*}
The lines intersect at \((0.6, 0.4)\text{,}\) which we can verify by substituting these values into the original two equations of our system.
Francine needs \(0.6\) cup of oats and \(0.4\) cup of wheat.
Checkpoint 8.16
Etienne plans to open a coffee house, and he has $\(7520\) to spend on furniture. A table costs $\(460\text{,}\) and a chair costs $\(120\text{.}\) Etienne will buy four chairs for each table. How many tables can he buy?
Let \(x\) represent the number of tables Etienne should buy, and let \(y\) represent the number of chairs. Write an equation about the cost of the furniture.
Write a second equation about the number of tables and chairs.
Graph both equations, solve the system, and answer the question in the problem. (Find the intercepts of the graphs to help you choose a window.)
Answer
\(460x + 120y = 7520\)
\(y=4x\)
\(x = 8, y = 32\text{.}\) Etienne should buy \(8\) tables and \(4\) chairs.
Being unable to read exact coordinates from a graph is not always a disadvantage. In many situations, fractional values of the unknowns are not acceptable.
Example 8.17
The mathematics department has $\(40,000\) to set up a new computer lab. The department will need one printer for every four terminals it purchases. If a printer costs $\(560\) and a terminal costs $\(1520\text{,}\) how many of each should the department buy?
Solution
Number of printers: 
\(p\) 
Number of terminals 
\(t\) 

Since the math department needs four times as many terminals as printers,
\begin{equation*}
t=4p
\end{equation*}
The total cost of the printers will be \(560p\) dollars, and the total cost of the terminals will be \(1520t\) dollars, so we have
\begin{equation*}
560p + 1520t = 40,000
\end{equation*}

We solve the second equation for \(t\) to get
\begin{equation*}
t=(40,000  560p)/1520
\end{equation*}
Now we graph the equations
\begin{align*}
Y_1 \amp =4X\\
Y_2 \amp =(40000  560X)/1520
\end{align*}
on the same set of axes. The second graph is not visible in the standard graphing window, but with a little experimentation we can find an appropriate window setting. The WINDOW values used for the figure below are
\begin{align*}
\text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 9.4\\
\text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 30
\end{align*}
The lines intersect at approximately \((6, 24)\text{.}\) These values satisfy the first equation, but not the second.
\begin{align*}
560(\alert{6})+1520(\blert{24})\amp\stackrel{?}{=}40,000 \\
39,840 \amp\ne 40,000
\end{align*}
The exact solution to the system is \(\left(\dfrac{500}{83},\dfrac{2000}{83}\right)\text{.}\) But this solution is not of practical use, since the math department cannot purchase fractions of printers or terminals. The department can purchase \(6\) printers and \(24\) terminals (with some money left over).
Checkpoint 8.18
The manager for Books for Cooks plans to spend $\(300\) stocking a new diet cookbook. The paperback version costs her $\(5\text{,}\) and the hardback costs $\(10\text{.}\) She finds that she will sell three times as many paperbacks as hardbacks. How many of each should she buy?
Let \(x\) represent the number of hardbacks and \(y\) the number of paperbacks she should buy. Write an equation about the cost of the books.
Write a second equation about the number of each type of book.
Graph both equations and solve the system. (Find the intercepts of the graphs to help you choose a window.) Answer the question in the problem.
Answer
\(10x + 5y = 300\)
\(y=3x\)
\(x = 12, ~y = 36\text{.}\) The manager should buy \(12\) hardbacks and \(36\) paperbacks.
Subsection An Application from Economics
The owner of a retail business must try to balance the demand for his product from consumers with the supply he can obtain from manufacturers. Supply and demand both vary with the price of the product: Consumers usually buy fewer items if the price increases, but manufacturers will be willing to supply more units of the product if its price increases.
The demand function gives the number of units of the product that consumers will buy in terms of the price per unit. The supply function gives the number of units that the producer will supply in terms of the price per unit. The price at which the supply and demand are equal is called the equilibrium price. This is the price at which the consumer and the producer agree to do business.
Example 8.19
A woolens mill can produce \(400x\) yards of fine suit fabric if it can charge \(x\) dollars per yard. The mill's clients in the garment industry will buy \(6000  100x\) yards of wool fabric at a price of \(x\) dollars per yard. Find the equilibrium price and the amount of fabric that will change hands at that price.
Solution
Price per yard: 
\(x\) 
Number of yards: 
\(y\) 

The supply equation tells us how many yards of fabric the mill will produce for a price of \(x\) dollars per yard.
\begin{equation*}
y=400x
\end{equation*}
The demand equation tells us how many yards of fabric the garment industry will buy at a price of \(x\) dollars per yard.
\begin{equation*}
y = 6000  100x
\end{equation*}

We graph the two equations on the same set of axes, as shown below. We set the window values to
\begin{align*}
\text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 94\\
\text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 6200
\end{align*}
and use the TRACE
or the intersect command to locate the solution. The graphs intersect at the point \((12, 4800)\text{.}\)
The equilibrium price is $\(12\) per yard, and the mill will sell \(4800\) yards of fabric at that price.
Checkpoint 8.20
Sanaz can afford to produce \(35x\) pairs of handpainted sunglasses if she can sell them at \(x\) dollars per pair, and the market will buy \(1700  15x\) at \(x\) dollars a pair.
Write the supply and demand equations for the sunglasses.
Find the equilibrium price and the number of sunglasses Sanaz will produce and sell at that price.
Answer
\(S = 35x, ~D = 1700  15x\)
$\(34\text{,}\) \(~1190\) sunglasses