A lawn care store makes a profit of $89.00 for every push mower sold and $236.00 for every riding lawn mower sold. The owner would like to make at least $621.00 a day from the sales of push and riding lawn mowers. Let x = the number of push mowers sold. Let y = the number of riding mowers sold. Which of the following is a list of 3 possible solutions to the graph?(1 point)
Responses
(3,1), (4,2), (1,1)
(3,1), (4,2), (1,1)
(2,1), (2,2), (2,3)
(2,1), (2,2), (2,3)
(3,2), (2,3), (1,4)
(3,2), (2,3), (1,4)
(1,2), (2,2), (3,1)
To solve this problem, we need to find combinations of x and y that satisfy the equation 89x + 236y ≥ 621.
Option 1: (3,1), (4,2), (1,1)
89(3) + 236(1) = 297 + 236 = 533 < 621
89(4) + 236(2) = 356 + 472 = 828 > 621
89(1) + 236(1) = 89 + 236 = 325 < 621
Option 2: (2,1), (2,2), (2,3)
89(2) + 236(1) = 178 + 236 = 414 > 621
89(2) + 236(2) = 178 + 472 = 650 > 621
89(2) + 236(3) = 178 + 708 = 886 > 621
Option 3: (3,2), (2,3), (1,4)
89(3) + 236(2) = 267 + 472 = 739 > 621
89(2) + 236(3) = 178 + 708 = 886 > 621
89(1) + 236(4) = 89 + 944 = 1033 > 621
Option 4: (1,2), (2,2), (3,1)
89(1) + 236(2) = 89 + 472 = 561 < 621
89(2) + 236(2) = 178 + 472 = 650 > 621
89(3) + 236(1) = 267 + 236 = 503 < 621
Out of the given options, option 2: (2,1), (2,2), (2,3) is the only one where all combinations satisfy the equation. Therefore, the correct answer is option 2: (2,1), (2,2), (2,3).