7.1 Systems of linear equations: two variables  (Page 6/20)

Write the system of equations using $y$ to replace function notation.

Substitute the expression $0.85x+\mathrm{35,000}$ from the first equation into the second equation and solve for $x.$

Then, we substitute $x=\mathrm{50,000}$ into either the cost function or the revenue function.

The break-even point is $\left(\mathrm{50,000},\mathrm{77,500}\right).$

The profit function is found using the formula $P(x)=R(x)-C(x).$

The profit function is $P(x)=0.7x\mathrm{-35,000.}$

Let c = the number of children and a = the number of adults in attendance.

The total number of people is $\mathrm{2,000.}$ We can use this to write an equation for the number of people at the circus that day.

The revenue from all children can be found by multiplying $\text{\$}25.00$ by the number of children, $25c.$ The revenue from all adults can be found by multiplying $\text{\$}50.00$ by the number of adults, $50a.$ The total revenue is $\text{\$}\mathrm{70,000.}$ We can use this to write an equation for the revenue.

We now have a system of linear equations in two variables.

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either $c$ or $a.$ We will solve for $a.$

Substitute the expression $\mathrm{2,000}-c$ in the second equation for $a$ and solve for $c.$

Substitute $c=\mathrm{1,200}$ into the first equation to solve for $a.$

We find that $\mathrm{1,200}$ children and $800$ adults bought tickets to the circus that day.