5. A volleyball league organizer collected \( \$ 2,040 \) for both divisions of volleyball teams. The Blue division costs \( \$ 160 \) per team and the Red division costs \( \$ 180 \) per team. How many teams will play in each division?

Let \( x \) be the number of teams in the Blue division and \( y \) be the number of teams in the Red division. The equation representing the total collection can be written as: \[ 160x + 180y = 2040 \] To simplify this equation, we can divide everything by 20: \[ 8x + 9y = 102 \] Now, let's look for integer solutions for \( x \) and \( y \). By rearranging the equation, we get: \[ 9y = 102 - 8x \] To find valid values of \( x \) and \( y \), \( 102 - 8x \) must be divisible by 9. We can try some values for \( x \): - If \( x = 0 \): \( y = 11.33 \) (not valid) - If \( x = 1 \): \( y = 10.56 \) (not valid) - If \( x = 2 \): \( y = 9.78 \) (not valid) - If \( x = 3 \): \( y = 9 \) (valid) - If \( x = 4 \): \( y = 8.22 \) (not valid) - If \( x = 5 \): \( y = 7.44 \) (not valid) - If \( x = 6 \): \( y = 6.67 \) (not valid) - If \( x = 7 \): \( y = 5.89 \) (not valid) - If \( x = 8 \): \( y = 5.11 \) (not valid) - If \( x = 9 \): \( y = 4.33 \) (not valid) - If \( x = 10 \): \( y = 3.56 \) (not valid) - If \( x = 11 \): \( y = 2.78 \) (not valid) - If \( x = 12 \): \( y = 2 \) (valid) - If \( x = 13 \): \( y = 1.22 \) (not valid) - If \( x = 14 \): \( y = 0.44 \) (not valid) The valid integer solutions are \( (3, 9) \) and \( (12, 2) \). Thus, there are 3 teams in the Blue division and 9 teams in the Red division, or 12 teams in the Blue division and 2 teams in the Red division.