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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