Let [latex]c[/latex] = the number of children and [latex]a[/latex] = 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 [latex]$25.00[/latex] by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying [latex]$50.00[/latex] by the number of adults, [latex]50a[/latex]. The total revenue is [latex]$70,000[/latex]. 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 [latex]1[/latex]. 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 [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.

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	[latex]d[/latex], distance driven in miles
Outputs	[latex]K(d)[/latex]: cost, in dollars, for renting from Keep on Trucking [latex]M(d)[/latex]: cost, in dollars, for renting from Move It Your Way
Initial Value	Up-front fee: [latex]K(0) = 20[/latex] and [latex]M(0) = 16[/latex]
Rate of Change	[latex]K(d) = $0.59[/latex]/mile and [latex]P(d) = $0.63[/latex]/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)

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 [latex]100[/latex] 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 [latex]100[/latex] miles are driven, that is [latex]d>100[/latex].

We will use the following table to help us solve this mixture problem:

	Amount	Part	Total
Start
Add
Final

We start with [latex]70[/latex]mL of solution, and the unknown amount can be [latex]x[/latex]. The part is the percentages, or concentration of solution [latex]0.5[/latex] for start, [latex]0.8[/latex] for add.

	Amount	Part	Total
Start	[latex]70[/latex]mL	[latex]0.5[/latex]
Add	[latex]x[/latex]	[latex]0.8[/latex]
Final	[latex]70+x[/latex]	[latex]0.6[/latex]

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	[latex]70[/latex]mL	[latex]0.5[/latex]	[latex]35[/latex]
Add	[latex]x[/latex]	[latex]0.8[/latex]	[latex]0.8x[/latex]
Final	[latex]70+x[/latex]	[latex]0.6[/latex]	[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]

[latex]35[/latex]mL of [latex]80\%[/latex] solution must be added to [latex]70[/latex]mL of [latex]50\%[/latex] solution to get a [latex]60\%[/latex] 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.