SOLUTION: 2x+3y-z=4 3x+2y+2z=6 4x-y+3z=5 solve in three variables

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Question 44966: 2x+3y-z=4
3x+2y+2z=6
4x-y+3z=5
solve in three variables
Answer by josmiceli(19441) About Me  (Show Source):
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(1) 2x + 3y - z = 4
(2) 3x + 2y + 2z = 6
(3) 4x - y + 3z = 5
multiply (3) by 2
8x - 2y + 6z = 10
(2) 3x + 2y + 2z = 6
add them
11x + 8z = 16
multiply (3) by 3
(3) 12x - 3y + 9z = 15
(1) 2x + 3y - z = 4
add them
14x + 8z = 19
11x + 8z = 16
subtract
14x + 8z = 19
-11x -8z = -16
3x = 3
x = 1
-------------
11x + 8z = 16
11*1 +8z = 16
8z = 5
z = 5/8
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(1) 2x + 3y - z = 4
2*1 + 3y - 5/8 = 4
16 + 24y - 5 = 32
24y = 21
y = 21/24
y = 7/8
-------------
check
(1) 2x + 3y - z = 4
(2) 3x + 2y + 2z = 6
(3) 4x - y + 3z = 5
---------------
(1) 2*1 + 3*(7/8) - 5/8 = 4
(2) 3*1 + 2*(7/8) + 2*(5/8) = 6
(3) 4*1 - 7/8 + 3*(5/8) = 5
----------------
(1) 16/8 + 21/8 - 5/8 = 32/8
32/8 = 32/8
OK
(2) 3*1 + 2*(7/8) + 2*(5/8) = 6
24/8 + 14/8 + 10/8 = 48/8
48/8 = 48/8
OK
(3) 4*1 - 7/8 + 3*(5/8) = 5
32/8 - 7/8 + 15/8 = 40/8
40/8 = 40/8
OK