Example: Solving an Inconsistent System of Equations

Solve the following system of equations.

[latex]\begin{gathered}&x=9 - 2y \\ &x+2y=13 \end{gathered}[/latex]

Answer: 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{align}x+2y&=13 \\ \left(9 - 2y\right)+2y&=13 \\ 9+0y&=13 \\ 9&=13 \end{align}[/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{gathered}x=9 - 2y \\ 2y=-x+9 \\ y=-\frac{1}{2}x+\frac{9}{2} \end{gathered}[/latex]

We then convert the second equation expressed to slope-intercept form.

[latex]\begin{gathered}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}[/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{gathered}y=-\frac{1}{2}x+\frac{9}{2} \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}[/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.

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{gathered}x+3y=2\\ 3x+9y=6\end{gathered}[/latex]

Answer: 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{align}x+3y&=2 \\ \left(-3\right)\left(x+3y\right)&=\left(-3\right)\left(2\right) \\ -3x - 9y&=-6 \end{align}[/latex]

Now add the equations.

[latex]\begin{align} −3x−9y&=−6 \\ +3x+9y&=6 \\ \hline 0&=0 \end{align}[/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{align}\begin{gathered}x+3y=2 \\ 3y=-x+2 \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered} \hspace{2cm} \begin{gathered} 3x+9y=6 \\9y=-3x+6 \\ y=-\frac{3}{9}x+\frac{6}{9} \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered}\end{align}[/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].  

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.

Answer: Write the system of equations using [latex]y[/latex] to replace function notation.

[latex]\begin{align} y&=0.85x+35{,}000 \\ y&=1.55x \end{align}[/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{gathered}0.85x+35{,}000=1.55x\\ 35{,}000=0.7x\\ 50{,}000=x\end{gathered}[/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{align}P\left(x\right)&=1.55x-\left(0.85x+35{,}000\right) \\ &=0.7x - 35{,}000 \end{align}[/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.  

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?

Answer: Let c = the number of children and a = the number of adults in attendance. The total number of people is 2,000. 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{gathered}c+a=2,000\\ 25c+50a=70{,}000\end{gathered}[/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{gathered}c+a=2{,}000\\ a=2{,}000-c\end{gathered}[/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{align} 25c+50\left(2{,}000-c\right)&=70{,}000 \\ 25c+100{,}000 - 50c&=70{,}000 \\ -25c&=-30{,}000 \\ c&=1{,}200 \end{align}[/latex]

Substitute [latex]c=1{,}200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{align}1{,}200+a&=2{,}000 \\ a&=800 \end{align}[/latex]

We find that 1,200 children and 800 adults bought tickets to the circus that day.

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.[footnote]Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/[/footnote] When will Keep on Trucking, Inc. be the better choice for Jamal?

Answer: 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{align}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{align}[/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{align}K\left(d\right)&=M\left(d\right) \\ 0.59d+20&=0.63d+16 \\ 4&=0.04d \\ 100&=d \\ d&=100 \end{align}[/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].

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?

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	[latex]x[/latex]	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	[latex]x[/latex]	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{align}35+0.8x& = 42+0.6x \\ 0.2x&=7 \\ \frac{0.2}{0.2}x&=\frac{7}{0.2} \\ x&=35 \end{align}[/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.