Adult tickets cost $6 and children’s tickets cost $4. Mrs. LeCompte says that she paid $30 for tickets, for both adults and children. How many of each ticket did she buy?
6a+4c = 30
30-4c must be a multiple of 6, so
c can be 0,3,6,...
then a will be 5,3,1,...
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To find out how many adult and children's tickets Mrs. LeCompte bought, we can use a simple algebraic approach. Let's assume Mrs. LeCompte bought x adult tickets and y children's tickets.
Given that adult tickets cost $6 each, the total cost of adult tickets is 6x dollars.
Given that children's tickets cost $4 each, the total cost of children's tickets is 4y dollars.
According to the information in the question, Mrs. LeCompte paid a total of $30 for both adult and children's tickets. So, we can write the equation:
6x + 4y = 30
Now, we need to find the values of x and y that satisfy this equation. There are multiple ways to solve this, including substitution or elimination, but for this case, let's use a method called "guess and check."
We can start by guessing values of x and y that satisfy the equation and check if they work.
Let's start by assuming x = 1 (1 adult ticket) and see what value of y satisfies the equation:
6(1) + 4y = 30
6 + 4y = 30
4y = 30 - 6
4y = 24
y = 24/4
y = 6
So, if Mrs. LeCompte bought 1 adult ticket and 6 children's tickets, the total cost would be:
6(1) + 4(6) = 6 + 24 = 30
Since the total cost matches and the values of x = 1 and y = 6 satisfy the equation, this is a valid solution.
Therefore, Mrs. LeCompte bought 1 adult ticket and 6 children's tickets.