A football team has an away game, and the bus breaks down. The coaches decide to drive the players to the game in cars and vans. Four players can ride in each car. Six players can ride in each van. There are 48 players on the team. The equation 4.3 + 6y = 48
models this situation, where x is the number of cars and y is the number of vans

Question:

Find 4 possible solutions in the context of the problem.

The 4 possible solutions in the context of the problem are (0, 8), (3, 6), (6, 3), and (9, 2)

Given the equation that models the situation expressed as:

4x + 6y = 48

We are to find 4 possible solutions for x and y. Note that the values of x and y must be integers.

If there are no cars i.e. at when x = 0

4(0) + 6y = 48

0 + 6y = 48

6y = 48

y = 48/6

y = 8

One of the solution is (0, 8)

If there are 3 cars, i.e. at when x = 0

4(3) + 6y = 48

12 + 6y = 48

6y = 48 - 12

y = 36/6

y = 6

The second solution will be (3, 6)

If there are 6 cars, i.e. at when x = 6

4(6) + 6y = 48

24 + 6y = 48

6y = 48 - 24

y = 24/6

y = 4

The third solution will be (6, 3)

If there are 9 cars, i.e. at when x = 9

4(9) + 6y = 48

36 + 6y = 48

6y = 48 - 36

y = 12/6

y = 2

The fourth solution will be (9, 2)

Hence the 4 possible solutions in the context of the problem are (0, 8), (3, 6), (6, 3), and (9, 2)

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Answer     2x and 7y

Step-by-step explanation: