Exemplo 11 de Algebra and Trigonometry, 11.1 Systems of Linear Equations: Two Variables

Writing and Solving a System of Equations in Two Variables

Problema e solução em inglês, como no livro original.

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?

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Solução
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.
$c + a = 2,000$
The revenue from all children can be found by multiplying $$ 25.00$ by the number of children, $25 c .$ The revenue from all adults can be found by multiplying $$ 50.00$ by the number of adults, $50 a .$ The total revenue is $$ 70,000.$ We can use this to write an equation for the revenue.
$25 c + 50 a = 70,000$
We now have a system of linear equations in two variables.

$\begin{matrix} c + a = 2,000 \\ 25 c + 50 a = 70,000 \end{matrix}$
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 $a .$
$\begin{matrix} c + a = 2,000 \\ a = 2,000 - c \end{matrix}$
Substitute the expression $2,000 - c$ in the second equation for $a$ and solve for $c .$
$\begin{array}{l} 25 c + 50 (2,000 - c) = 70,000 \\ 25 c + 100,000 - 50 c = 70,000 \\ - 25 c = −30,000 \\ c = 1,200 \end{array}$
Substitute $c = 1,200$ into the first equation to solve for $a .$
$\begin{array}{l} 1,200 + a = 2,000 \\ a = 800 \end{array}$
We find that $1,200$ children and $800$ adults bought tickets to the circus that day.

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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?

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