SOLUTION: x+y+z=12 x-y+z=6 x-y-z=o. So x,y,and,z=?

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Question 1082665: x+y+z=12
x-y+z=6
x-y-z=o.
So x,y,and,z=?
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
x + y + z = 12   (1)
x - y + z =  6   (2)
x - y - z =  o.  (3)


1)  Subtract  eq(2) from eq(1) (both sides).  You will get  2y = 12 - 6 --->  y = 3.


2)  Add eq(1) and eq(2).   You will get  2x = 12 + 0  --->  x = 6.


3)  Then from EITHER equation  z = 3.

Solved.

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Comment from student: 2x+2y-z=10 x+y+z=11 2x+y-z=7 So x,y,z=?
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My response: Is this instead of "Thanks" ?



Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
x+y+z=12
x-y+z=6
x-y-z=o.
So x,y,and,z=?
x + y + z = 12 ----- eq (i)

x - y + z = 6 ------ eq (ii)

x - y - z = 0 ------ eq (iii)

1) ADD eqs (i) & (iii) to ELIMINATE y and z and find the value of x

2) Subtract eq (ii) from eq (i), or vice-versa to ELIMINATE x and z and find the value of y

3) Subtract eq (iii) from eq (ii) or vice-versa to ELIMINATE x and y and determine the value of z. 

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