Let D denote Delta, B for Beta, and S for Sigma.

7D + 15B + 5S = 128 (for painting)

3D + 7B + 5S = 88 (for drying)

4D + 5B + 5S = 86(for polishing)

Let's eliminate S first.

7D + 15B + 5S = 128

-1[3D + 7B + 5S = 88]-1

4D +8B =40(equation 1)

-1[4D + 5B + 4S = 86]-1

7D + 15B + 5S =128

3D +10B =42 (equation 2)

Let's now use equation 1 and eqaution 2 to solve for thevalues of D and B.

In solving for D and B, let's eliminate D first.

4D +8B =40 also equals D +2B =10

3(10-2B) +10B =42

30 - 6B +10B = 42

4B = 12

B = 3.

Let's now substitute the value of B.

D+2(3)=10

D= 4.

Now, let's solve for S.

7(4) + 15(3) + 5S = 128

28 + 45 + 5S = 128

5S = 55

S = 11.

Therefore, Epsilon Motor Company produces 4 Deltas, 3 Betas, and 11 Sigmas in a month.

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