Question 138195: I also have to set up a three equation problem as well but once again I don't understand it.
If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid number plus 2 times the first number is 3. The first number plus 3 times the second number plus the third number is 18. Find the three numbers.
Thank you for your help.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1.
a+2c-b = 1
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The third number plus 2 times the first number is 3.
c +2a = 3
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The first number plus 3 times the second number plus the third number is 18.
a+3b+c = 18
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Find the three numbers.
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Rearrange the equations:
a - b + 2c = 1
2a + 0 + c = 3
a + 3b + c = 18
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Using Matrix methods I get:
a= 0
b= 5
c= 3
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Cheers,
Stan H.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
First number: x
Second number: y
Third number: z
The sum of the 1st number and 2 times the third: x + 2z
Then subtract the 2nd number x + 2z - y, or x - y + 2z to keep things in order.
The result is 1: x - y + 2z = 1
2 times the 1st number: 2x
Plus the 3rd number: 2x + z
'is' means equals, so 2x + z = 3
The 1st number, x, plus 3 times the 2nd, 3y, plus the 3rd, z
is (=) 18, so:
x + 3y + z = 18
So, here are your three equations.
x - y + 2z = 1
2x + z = 3
x + 3y + z = 18
Looking at the coefficients, I would say that Gauss-Jordan elimination is the
best solution method. But you do it however you like.
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