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Question 472985: Does the system have a solution? If so, could you help me find it?
x+y=9
5x+y=29
x-y=2
How do I graph the function f(x)=[x-2]+3?
How do I solve the system :
x-y+z=1
2x+2y=4
-x+4y+3z=6
Thank you for your time.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! x+y=9
5x+y=29
x-y=2
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You have 3 equations with 2 variables.
Adding the 1st and 3rd you get 2x = 11
x = 5.5
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Using x+y = 9, y = 3.5
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Trying that in the 2nd equation you would get:
5(5.5) + 3.5 = 29
27.5 + 3.5 = 29
31 = 29
Wrong.
The three lines do not meet at a point so there is
no solution for the system.
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How do I graph the function f(x)=[x-2]+3?
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Is that supposed to be absolute (x-2)?
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How do I solve the system :
x-y+z=1
2x+2y+0z=4
-x+4y+3z=6
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Divide thru the 2nd equation by 2
How do I solve the system :
x-y+z=1
x+y+0z=2
-x+4y+3z=6
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Subtract the 1st equation from the 2nd.
Add the 1st equation to the 3rd.
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x-y+z=1
0x+2y-z=1
0x+3y+4z=7
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Divide thru the 2nd equation by 2.
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x-y+z=1
0x+y-(1/2)z=(1/2)
0x+3y+4z=7
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Subtract 3 times the 2nd equation from the 3rd:
x-y+z=1
0x+y-(1/2)z=(1/2)
0x+0y+(11/2)z=11/2
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Solve for "z" in the 3rd equation:
x-y+z=1
0x+y-(1/2)z=(1/2)
0x+0y+z=1
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Use z=1 to solve for "y" in the 2nd equation:
y-(1/2) = 1/2
y = 1
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Use z = 1 and y = 1 to solve for "x" in the 1st equation:
x -1+1 = 1
x = 1
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Solution:
x = y = z = 1
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Cheers,
stan H.
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