Let d represent the number of Deltas produced
Let b represent the number of Betas produced
Let s represent the number of Sigmas produced
Each Delta requires 9 hours of painting. Each Beta requires 33 hours of painting and each Sigma requires 4 hours of painting. If the company has 190 hours of painting, then the equation representing this situation is
9d + 33b + 4s = 190
Each Delta requires 2 hours of drying. Each Beta requires 8 hours of drying and each Sigma requires 3 hours of drying. If the company has 66 hours of drying, then the equation representing this situation is
2d + 8b + 3s = 66
Each Delta requires 5 hours of polishing. Each Beta requires 6 hours of polishing and each Sigma requires 2 hours of polishing. If the company has 54 hours of polishing, then the equation representing this situation is
5d + 6b + 2s = 54
From the third equation, if we divide both sides by 2, we have
2.5d + 3b + s = 27
s = 27 - 2.5d - 3b
We would substitute s = 27 - 2.5d - 3b into the first and second equations. By substituting s = 27 - 2.5d - 3b into the first equation, we have
9d + 33b + 4(27 - 2.5d - 3b) = 190
9d + 33b + 108 - 10d - 12b = 190
33b - 12b + 9d - 10d = 190 - 108
21b - d = 82 equation 4
By substituting s = 27 - 2.5d - 3b into the second equation, we have
2d + 8b + 3(27 - 2.5d - 3b) = 66
2d + 8b + 81 - 7.5d - 9b = 66
2d - 7.5d + 8b - 9b = 66 - 81
- 5.5d - b = - 15
b = - 5.5d + 15
We would substitute b = - 5.5d + 15 into equation 4. We have
21(- 5.5d + 15) - d = 82
- 115.5d + 315 - d = 82
- 115.5d - d = 82 - 315
- 116.5d = - 233
d = - 233/- 116.5
d = 2
We would substitute d = 2 into b = - 5.5d + 15. We have
b = - 5.5 * 2 + 15 = - 11 + 15
b = 4
We would substitute b = 4 and d = 2 into s = 27 - 2.5d - 3b. We have
s = 27 - 2.5 * 2 - 3 * 4 = 27 - 5 - 12
s = 10
Therefore, the Epsilon motors produce 2 Deltas, 4 Betas and 10 Sigmas in a month