Using the ticket prices, write an equation for the total money collected. Each student ticket adds $4 and each adult ticket adds $6 to the total of $400.

You now have two equations with two variables. Consider using elimination: can you combine the equations so that one variable cancels out?

Once you find a value for the number of adult tickets, verify it by checking both the total number of tickets (80) and the total money ($400).

In Desmos, use $x$ for the number of adult tickets and $y$ for the number of student tickets. Type these two equations as separate lines:

• x + y = 80
• 4y + 6x = 400 Desmos will graph both lines on the coordinate plane.

Look for the point where the two lines intersect. The $x$-coordinate of that intersection represents the number of adult tickets, and the $y$-coordinate represents the number of student tickets.

Take the $x$-value from the intersection and plug it into the expression 6x + 4(80 - x) in Desmos to verify that it gives an output of 400. The $x$-value that makes this true is the number of adult tickets sold.

Let $s$ be the number of student tickets and $a$ be the number of adult tickets.

From the problem:

• Total tickets: $s + a = 80$.

Student tickets cost $4 each, and adult tickets cost $6 each.

So the money equation is:

This represents $4$ dollars for each of the $s$ student tickets plus $6$ dollars for each of the $a$ adult tickets, totaling $400$ dollars.

To make the numbers smaller, divide every term in the money equation by $2$:

Now you have a simpler system:

• $s + a = 80$
• $2s + 3a = 200$.

Use the first equation to help eliminate $s$.

Multiply the first equation by $2$:

Now subtract this new equation from the simplified money equation:

We found that $a = 40$, so the club sold 40 adult tickets.

Therefore, the correct answer is 40 (choice B).