Answer: To solve for c and v, we need to use the equation:
4c + 6v = 80
We have two variables and one equation, so we need another equation to solve for c and v. We can use the fact that the number of students transported by the drivers is 80. Since each car can seat 4 students and each van can seat 6 students, we can write:
4c + 6v = 80
c + v = number of drivers
Since we want to minimize the number of drivers, we want to minimize the value of c + v. We can use substitution to solve for c and v in terms of one variable. Solving for c in terms of v from the first equation, we get:
4c + 6v = 80
4c = 80 - 6v
c = (80 - 6v)/4
Substituting this expression for c into the second equation, we get:
c + v = number of drivers
(80 - 6v)/4 + v = number of drivers
Multiplying both sides by 4 to clear the fraction, we get:
80 - 6v + 4v = 4(number of drivers)
80 - 2v = 4(number of drivers)
20 - 0.5v = number of drivers
We want to minimize the value of c + v, so we want to minimize the value of v. Since v must be a whole number, we want to find the smallest integer value of v that satisfies the equation. We can plug in integer values of v and find the corresponding value of c, and check if the sum of c and v is less than or equal to the current minimum. Starting with v = 1, we get:
v = 1 -> c = 19.5 -> c + v = 20.5
v = 2 -> c = 18.0 -> c + v = 20.0
v = 3 -> c = 16.5 -> c + v = 19.5
v = 4 -> c = 15.0 -> c + v = 19.0
v = 5 -> c = 13.5 -> c + v = 18.5
v = 6 -> c = 12.0 -> c + v = 18.0
The smallest integer value of v that satisfies the equation is v = 6, which gives c = 12. Therefore, we need 12 cars and 6 vans to transport 80 students.
Explanation: