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Question 1197204: a grain dealer sold to one customer 9th bushels of wheat 3 of corn and 5 of rye for $40.90 to another 3 of wheat 5 of corn and 9 of rye for $47.30 and to a third 5 of wheat 9 of corn and 3 of rye for $46.10 what was the price per bushel of corn
Found 2 solutions by math_tutor2020, josgarithmetic:
Answer by math_tutor2020(3817)   (Show Source):
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Answer: $3.10 per bushel for the price of corn
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Explanation:
Given information
9 bushels of wheat 3 of corn and 5 of rye for $40.90
3 bushels of wheat 5 of corn and 9 of rye for $47.30
5 bushels of wheat 9 of corn and 3 of rye for $46.10
x = price for 1 bushel of wheat
y = price for 1 bushel of corn
z = price for 1 bushel of rye
The three facts in the given information allow us to form these three equations in this order
9x+3y+5z = 40.90
3x+5y+9z = 47.30
5x+9y+3z = 46.10
The matrix form looks like this
Each coefficient makes up the 3x3 matrix on the left hand side; the right hand side values are along the far right column. Ultimately we have a 3x4 matrix.
I'll represent this matrix as a table having 3 rows and 4 columns like thisWe'll do matrix operations to get this matrix into row echelon form (abbreviated as REF).
Let's swap row1 (denoted R1) with row3 (aka R3)
The notation is R1 <--> R3
| 5 | 9 | 3 | 46.1 | R1 <--> R3 | | 3 | 5 | 9 | 47.3 | | | 9 | 3 | 5 | 40.9 | |
To get R1 to have 1 as the first item, we'll multiply everything by 1/5
| 1 | 1.8 | 0.6 | 9.22 | (1/5)*R1 --> R1 | | 3 | 5 | 9 | 47.3 | | | 9 | 3 | 5 | 40.9 | |
The goal is to zero out everything below this '1' in the top left corner.
Compute the operation R2 - 3*R1 which says
"triple an item in R1, then subtract from R2". We subtract straight down. This is why I think the grid lines or boxes from a table are handy to keep things lined up. Use of scratch paper is recommended. A more stronger recommendation is to use a spreadsheet if possible.
The result of R2-3*R1 replaces R2. This is what the notation R2-3R1 --> R2 is referring to.
| 1 | 1.8 | 0.6 | 9.22 | | | 0 | 0.4 | -7.2 | -19.64 | R2 - 3*R1 --> R2 | | 9 | 3 | 5 | 40.9 | |
Now compute R3 - 9*R1 --> R3
Multiply 9 times an item in R1, subtract from R3. The result replaces the value in R3
| 1 | 1.8 | 0.6 | 9.22 | | | 0 | 0.4 | -7.2 | -19.64 | | | 0 | -13.2 | -0.4 | -42.08 | R3 - 9*R1 --> R3 |
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The first nonzero value in R2 is the 0.4 in the second entry.
Multiply this and all of R2 by 1/0.4 so that the first nonzero entry turns into a 1. This is the pivot for this row.
| 1 | 1.8 | 0.6 | 9.22 | | | 0 | 1 | -18 | -49.1 | (1/0.4)*R2 --> R2 | | 0 | -13.2 | -0.4 | -42.08 | |
Now compute R3 + 13.2*R2 to have it replace R3
| 1 | 1.8 | 0.6 | 9.22 | | | 0 | 1 | -18 | -49.1 | | | 0 | 0 | -238 | -690.2 | R3 + 13.2*R2 --> R3 |
The last step to get it to REF is to multiply every item in R3 by (-1/238)
| 1 | 1.8 | 0.6 | 9.22 | | | 0 | 1 | -18 | -49.1 | | | 0 | 0 | 1 | 2.9 | (-1/238)*R3 --> R3 |
The matrix is now in row echelon form
We have gone from this augmented matrix
to this matrix in row echelon form
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The bottom row has
0, 0, 1, 2.90
which translates back to this equation
0x + 0y + 1z = 2.90
or in short
z = 2.90
The price of one bushel of rye is $2.90
In other words, rye costs $2.90 per bushel.
Then use back substitution to find y
The second row has
0, 1, -18, -49.1
which gives this equation
0x+1y-18z = -49.1
aka
y - 18z = -49.1
We'll use the recently found z value to find y
y - 18z = -49.1
y - 18(2.90) = -49.1
y - 52.2 = -49.1
y = -49.1 + 52.2
y = 3.10
The price of a bushel of corn is $3.10
Use further back substitution to find x
The first row is
1, 1.8, 0.6, 9.22
giving us
1x + 1.8y + 0.6z = 9.22
Plug in y = 3.10 and z = 2.90 and solve for x.
1x + 1.8y + 0.6z = 9.22
1x + 1.8*3.10 + 0.6*2.90 = 9.22
1x + 5.58 + 1.74 = 9.22
x + 7.32 = 9.22
x = 9.22 - 7.32
x = 1.90
The price of wheat is $1.90 per bushel
Summary:
x = 1.90
y = 3.10
z = 2.90
which represents the per bushel cost of wheat, corn, and rye in that order.
All costs are in dollars.
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You can use computer software or a graphing calculator to verify the answers.
The linear algebra toolkit is a very valuable resource for situations like this. Let me know if you have any questions using it.
Tools like WolframAlpha and GeoGebra are useful as well.
Or you can plug each x,y,z value into the original equations mentioned.
If we plugged x = 1.90, y = 3.10, and z = 2.90 into the first equation, then,
9x+3y+5z = 40.90
9*1.90+3*3.10+5*2.90 = 40.90
17.10+9.30+14.50 = 40.90
40.90 = 40.90
We get the same thing on both sides, so this is a true equation.
This confirms those x,y,z values work for the first equation.
I'll let you check the other equations. You should get the same number on both sides after replacing those x,y,z values with the numbers mentioned.
Once you discover the three equations are true for those x,y,z values, you will have fully confirmed the solution.
Answer by josgarithmetic(39621) (Show Source):
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