8.2 5 Please solve. Thanks!

Algebra
Tutor: None Selected Time limit: 1 Day

May 4th, 2015

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. 


May 4th, 2015

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