Question 370426: I must solve this system if possible, I worked the system and received the answer below...Im wondering if it is correct.
3x+y+z=0
2x-y+z=0
2x+y+z=0
My answer is
z=4, y=-25, x=18
Thank U in advance.
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! I must solve this system if possible, I worked the system and received the answer below...Im wondering if it is correct.
3x+y+z=0
2x-y+z=0
2x+y+z=0
My answer is
z=4, y=-25, x=18
-----------------
Sub your answers into the eqns:
3x+y+z=0
3*18 - 25 + 4 = 33, not 0
--> not correct.
Hint: when all the coefficients are zero,
x = y = z = 0
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
z=4, y=-25, x=18
These do not check in the first equation
3x + y + z = 0
3(18) + (-25) + 4 = 0
54 - 25 + 4 = 0
33 = 0
These do not check in the second equation, either.
2x - y + z = 0
2(18) - (-25) + 4 = 0
36 + 25 + 4 = 0
65 = 0
These do not check in the third equation, either.
2x + y + z = 0
2(18) + (-25) + 4 = 0
36 - 25 + 4 = 0
15 = 0
The only solution is x=0, y=0, z=0
The other tutor is right in one respect, but wrong in another respect.
When the numbers on the right (not coefficients, as he stated) are all
aero, it's true that x = y = z = 0 is a solution. However that is not
necessarily the only solution. It was in your case, but not in all
cases.
For example:
x + y + 3z = 0
x + 2y + 4z = 0
x + 3y + 5z = 0
This one has infinitely many solutions.
x = y = z = 0 is indeed a solution, but also
x=-2, y=-1, and z=1 is a solution and
x=4, y=2, and z=-2 is also a solution.
The general solution is
x = -2k, y = -k, and z = k, where k is any number
Edwin
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