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x–3y+3z = 10
2x+ y–3z = –3
x–2y+ z = 6
Add the first two equations above
term by term and the x's will
cancel:
x–3y+3z = 10
2x+ y–3z = –3
-------------
3x-2y = 7
That has z eliminated. So we
eliminate z from the 2nd and
3rd equations at the top:
x–3y+3z = 10
x–2y+ z = 6
We need to multiply x-2y+z=6
through by -3 to make the
z's cancel when we add them
term by term:
x–3y+3z = 10
-3x+6y-3z = -18
---------------
-2x+3y = -8
So we have a system of only
two equations in two unknowns:
3x-2y = 7
-2x+3y = -8
To make the y's cancel multiply
the first one by 3 and the second
one through by 2
9x-6y = 21
-4x+6y = -16
------------
5x = 5
x = 1
Substitute x = 1 in
3x-2y = 7
3(1)-2y = 7
3-2y = 7
-2y = 4
y = -2
Substitute x = 1 and y = -2 in one
of the original equations to find
z.
x–3y+3z = 10
(1)-3(-2)+3z = 10
1 + 6 + 3z = 10
7 + 3z = 10
3z = 3
z = 1
Solution (x,y,z) = (1,-2,1)
Edwin
