SOLUTION: The feed mill pays a farmer $6930.00 for the 1st delivery, $5475.00 for the 2nd delivery, and $8879.50 for the 3rd delivery. The table shows the number of bushels included in each

Question 775527: The feed mill pays a farmer $6930.00 for the 1st delivery, $5475.00 for the 2nd delivery, and $8879.50 for the 3rd delivery. The table shows the number of bushels included in each delivery. Use the table to write and solve a system of equations to find the price per bushel that the farmer received for each crop.
1st delivery: corn- 900; wheat- 540; soybeans- 360
2nd delivery: corn- 1125; wheat; 150; soybeans- 225
3rd delivery: corn- 860; wheat; 645; soybeans- 645
I know how to do 3x3s and system of equations; but I've been stuck on this problem for over an hour and I can't figure out what variable to get rid of.
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
The feed mill pays a farmer $6930.00 for the 1st delivery, $5475.00 for the 2nd delivery, and $8879.50 for the 3rd delivery. The table shows the number of bushels included in each delivery. Use the table to write and solve a system of equations to find the price per bushel that the farmer received for each crop.
1st delivery: corn- 900; wheat- 540; soybeans- 360
2nd delivery: corn- 1125; wheat; 150; soybeans- 225
3rd delivery: corn- 860; wheat; 645; soybeans- 645
I know how to do 3x3s and system of equations; but I've been stuck on this problem for over an hour and I can't figure out what variable to get rid of.

Let the price per bushel of corn, wheat, and soybeans be C, W, and S, respectively

Then, for 1st, 2nd, and 3rd deliveries, we have:

900C + 540W + 360S = 6,930 ------ eq (i)

1,125C + 150W + 225S = 5,475 ------- eq (ii)

860C + 645W + 645S = 8,879.5 ------- eq (iii)

These are some pretty big numbers that you have to deal with, but for starters I would factor out a 90 in eq (i), so it becomes: 10C + 6W + 4S = 77

I would then multiply eq (i) by 225, and eq (ii) by - 4 to get rid of S, which would leave me with the variables, C & W.

I would then multiply eq (i) by 645 and eq (iii) by - 4 to again get rid of S, and leaves me with the variables, C & W.

You could then solve the 2X2 system for the variables, C (corn) & W (wheat), and then finally, S (soybeans).
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