Learning Objectives
- Determine whether a solution reveals that a system has one, many, or no solutions
- Interpret the solution to a system of equations that represents profits for a business
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different [latex]y[/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[/latex].
Example: Solving an Inconsistent System of Equations
Solve the following system of equations.
[latex]\begin{array}{l}\text{ }x=9 - 2y\hfill \\ x+2y=13\hfill \end{array}[/latex]
Solution
We can approach this problem in two ways. Because one equation is already solved for [latex]x[/latex], the most obvious step is to use substitution.
[latex]\begin{array}{l}x+2y=13\hfill \\ \left(9 - 2y\right)+2y=13\hfill \\ 9+0y=13\hfill \\ 9=13\hfill \end{array}[/latex]
Clearly, this statement is a contradiction because [latex]9\ne 13[/latex]. Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
[latex]\begin{array}{l}\text{ }x=9 - 2y\hfill \\ 2y=-x+9\hfill \\ \text{ }y=-\frac{1}{2}x+\frac{9}{2}\hfill \end{array}[/latex]
We then convert the second equation expressed to slope-intercept form.
[latex]\begin{array}{l}x+2y=13\hfill \\ \text{ }2y=-x+13\hfill \\ \text{ }y=-\frac{1}{2}x+\frac{13}{2}\hfill \end{array}[/latex]
Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect.
[latex]\begin{array}{l}\begin{array}{l}\\ y=-\frac{1}{2}x+\frac{9}{2}\end{array}\hfill \\ y=-\frac{1}{2}x+\frac{13}{2}\hfill \end{array}[/latex]
Analysis of the Solution
Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.
Try It
Solve the following system of equations in two variables.
[latex]\begin{array}{l}2y - 2x=2\\ 2y - 2x=6\end{array}[/latex]
Solution
No solution. It is an inconsistent system.
Expressing the Solution of a System of Dependent Equations Containing Two Variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[/latex].
Example: Finding a Solution to a Dependent System of Linear Equations
Find a solution to the system of equations using the addition method.
[latex]\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}[/latex]
Solution
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating [latex]x[/latex]. If we multiply both sides of the first equation by [latex]-3[/latex], then we will be able to eliminate the [latex]x[/latex] -variable.
[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \left(-3\right)\left(x+3y\right)=\left(-3\right)\left(2\right)\hfill \\ \text{ }-3x - 9y=-6\hfill \end{array}[/latex]
Now add the equations.
[latex]\begin{array} \hfill−3x−9y=−6 \\ \hfill+3x+9y=6 \\ \hfill \text{_____________} \\ \hfill 0=0 \end{array}[/latex]
We can see that there will be an infinite number of solutions that satisfy both equations.
Analysis of the Solution
If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.
[latex]\begin{array}{l}\text{ }x+3y=2\hfill \\ \text{ }3y=-x+2\hfill \\ \text{ }y=-\frac{1}{3}x+\frac{2}{3}\hfill \\ 3x+9y=6\hfill \\ \text{ }9y=-3x+6\hfill \\ \text{ }y=-\frac{3}{9}x+\frac{6}{9}\hfill \\ \text{ }y=-\frac{1}{3}x+\frac{2}{3}\hfill \end{array}[/latex]
Look at the graph below. Notice the results are the same. The general solution to the system is [latex]\left(x, -\frac{1}{3}x+\frac{2}{3}\right)[/latex].
Writing the general solution
In the previous example, we presented an analysis of the solution to the following system of equations:
[latex]\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}[/latex]
After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as [latex]\left(x, -\frac{1}{3}x+\frac{2}{3}\right)[/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot. It tells us that x can be anything, x is x. It also tells us that y is going to depend on x, just like when we write a function rule. In this case, depending on what you put in for x, y will be defined in terms of x as [latex]-\frac{1}{3}x+\frac{2}{3}[/latex].
In other words, there are infinitely many (x,y) pairs that will satisfy this system of equations, and they all fall on the line [latex]f(x)-\frac{1}{3}x+\frac{2}{3}[/latex].
Try It
Solve the following system of equations in two variables.
[latex]\begin{array}{l}\begin{array}{l}\\ \text{ }\text{}\text{}y - 2x=5\end{array}\hfill \\ -3y+6x=-15\hfill \end{array}[/latex]
Solution
The system is dependent so there are infinite solutions of the form [latex]\left(x,2x+5\right)[/latex].
try it
Use Desmos to write three different systems:
- A system of equations with one solution
- A system of equations with no solutions
- A system of equations with infinitely many solutions
Using Systems of Equations to Investigate Profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price. The revenue function is shown in orange in the graph below.
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[/latex] -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
Example: Finding the Break-Even Point and the Profit Function Using Substitution
Given the cost function [latex]C\left(x\right)=0.85x+35,000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point and the profit function.
Solution
Write the system of equations using [latex]y[/latex] to replace function notation.
[latex]\begin{array}{l}\begin{array}{l}\\ y=0.85x+35,000\end{array}\hfill \\ y=1.55x\hfill \end{array}[/latex]
Substitute the expression [latex]0.85x+35,000[/latex] from the first equation into the second equation and solve for [latex]x[/latex].
[latex]\begin{array}{c}0.85x+35,000=1.55x\\ 35,000=0.7x\\ 50,000=x\end{array}[/latex]
Then, we substitute [latex]x=50,000[/latex] into either the cost function or the revenue function.
[latex]1.55\left(50,000\right)=77,500[/latex]
The break-even point is [latex]\left(50,000,77,500\right)[/latex].
The profit function is found using the formula [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex].
[latex]\begin{array}{l}P\left(x\right)=1.55x-\left(0.85x+35,000\right)\hfill \\ \text{ }=0.7x - 35,000\hfill \end{array}[/latex]
The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex].
Analysis of the Solution
The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.
We see from the graph below that the profit function has a negative value until [latex]x=50,000[/latex], when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.
Writing a System of Linear Equations Given a Situation
It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.
How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.
- Identify the input and output of each linear model.
- Identify the slope and y-intercept of each linear model.
- Find the solution by setting the two linear functions equal to another and solving for x, or find the point of intersection on a graph.
Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.
Example: Writing and Solving a System of Equations in Two Variables
The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?
Solution
Let c = the number of children and a = the number of adults in attendance.
The total number of people is [latex]2,000[/latex]. We can use this to write an equation for the number of people at the circus that day.
[latex]c+a=2,000[/latex]
The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.
[latex]25c+50a=70,000[/latex]
We now have a system of linear equations in two variables.
[latex]\begin{array}{c}c+a=2,000\\ 25c+50a=70,000\end{array}[/latex]
In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].
[latex]\begin{array}{c}c+a=2,000\\ a=2,000-c\end{array}[/latex]
Substitute the expression [latex]2,000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].
[latex]\begin{array}{l} 25c+50\left(2,000-c\right)=70,000\hfill \\ 25c+100,000 - 50c=70,000\hfill \\ \text{ }-25c=-30,000\hfill \\ \text{ }c=1,200\hfill \end{array}[/latex]
Substitute [latex]c=1,200[/latex] into the first equation to solve for [latex]a[/latex].
[latex]\begin{array}{l}1,200+a=2,000\hfill \\ \text{ }\text{}a=800\hfill \end{array}[/latex]
We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.
Try It
Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?
Sometimes, a system can inform a decision. In our next example, we help answer the question, “Which truck rental company will give the best value?”
Example: Building a System of Linear Models to Choose a Truck Rental Company
Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile. When will Keep on Trucking, Inc. be the better choice for Jamal?
Solution
The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.
| Input |
d, distance driven in miles |
| Outputs |
K(d): cost, in dollars, for renting from Keep on TruckingM(d) cost, in dollars, for renting from Move It Your Way |
| Initial Value |
Up-front fee: K(0) = 20 and M(0) = 16 |
| Rate of Change |
K(d) = $0.59/mile and P(d) = $0.63/mile |
A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations
[latex]\begin{array}{l}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{array}[/latex]
Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\left(d\right)<M\left(d\right)[/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\left(d\right)[/latex] function is smaller.
These graphs are sketched above, with K(d) in blue.
To find the intersection, we set the equations equal and solve:
[latex]\begin{array}{l}K\left(d\right)=M\left(d\right)\hfill \\ 0.59d+20=0.63d+16\hfill \\ 4=0.04d\hfill \\ 100=d\hfill \\ d=100\hfill \end{array}[/latex]
This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\left(d\right)[/latex] is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[/latex].
The applications for systems seems almost endless, but we will just show one more. In the next example, we determine the amount 80% methane solution to add to a 50% solution to give a final solution of 60%.
Example: Solve a Chemical Mixture Problem
A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?
Show Answer
We will use the following table to help us solve this mixture problem:
|
Amount |
Part |
Total |
| Start |
|
|
|
| Add |
|
|
|
| Final |
|
|
|
We start with 70 mL of solution, and the unknown amount can be x. The part is the percentages, or concentration of solution 0.5 for start, 0.8 for add.
|
Amount |
Part |
Total |
| Start |
70mL |
0.5 |
|
| Add |
x |
0.8 |
|
| Final |
[latex]70+x[/latex] |
0.6 |
|
Add amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.
|
Amount |
Part |
Total |
| Start |
70mL |
0.5 |
35 |
| Add |
x |
0.8 |
[latex]0.8x[/latex] |
| Final |
[latex]70+x[/latex] |
0.6 |
[latex]42+0.6x[/latex] |
Multiply amount by part to get total. be sure to distribute on the last row:[latex](70 + x)0.6[/latex].
If we add the start and add entries in the Total column, we get the final equation that represents the total amount and it’s concentration.
[latex]\begin{array}{c}35+0.8x = 42+0.6x\\0.2x=7\\\frac{0.2}{0.2}x=\frac{7}{0.2}\\x=35\end{array}[/latex]
35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane.
The same process can be used if the starting and final amount have a price attached to them, rather than a percentage.
