The athletic department just received a shipment of footballs and basketballs. You don’t know how many there are, and don’t feel like counting. Here’s what you do know: each football costs 12 dollars, each basketball costs 15 dollars, and the credit card receipt says the total for the shipment was 1335 dollars. You also remember the number of basketballs ordered was 5 more than twice the number of footballs.

To avoid counting, set up a system of equations representing the above situation, and solve it algebraically to determine the number of footballs.

B = 2F + 5

12F + 15B = 1335

In the second equation, substitute 2F+5 for B and solve for F.

To solve this problem algebraically, let's set up a system of equations based on the given information.

Let's assume that the number of footballs is represented by the variable "f," and the number of basketballs is represented by the variable "b."

We know that each football costs $12, so the total cost of footballs would be 12f.
Similarly, each basketball costs $15, so the total cost of basketballs would be 15b.

The credit card receipt says the total cost of the shipment was $1335, so we can set up the first equation:

12f + 15b = 1335 ... (Equation 1)

We also know that the number of basketballs ordered was 5 more than twice the number of footballs. This can be expressed as:

b = 2f + 5 ... (Equation 2)

Now we have a system of two equations with two variables. We can solve this system to find the value of "f" (the number of footballs) algebraically.

To do this, we can substitute Equation 2 into Equation 1:

12f + 15(2f + 5) = 1335

Simplify the equation:

12f + 30f + 75 = 1335
42f = 1335 - 75
42f = 1260
f = 1260 / 42
f = 30

Therefore, the number of footballs in the shipment is 30.