Using your third equation, solve for X and substitute it in for X in your other two equations...
X = 3y - 4z - 14
3(3y - 4z - 14) + 2y - z = 8
9y - 12z - 42 + 2y - z = 8
11y - 13z - 42 = 8
11y - 13z = 50
-3(3y - 4z - 14) + 4y + 5z = -14
9y - 12z - 42 + 2y - z = 8
-9y + 12z + 42 + 4y + 5z = -14
-5y + 17z + 42 = -14
Take these two equations and use linear combinations
5(11y - 13z = 50)
11(-5y + 17z = -56)
55y - 65z = 250
+ -55y + 187x = -616
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122z = -366 ----> ( Z = -3 )
Now that we know Z = -3, substitute it back into two of your original equations.
3x + 2y - (-3) = 8
3x + 2y + 3 = 8
-3x + 4y = 1
-3x + 4y + 5(-3) = -14
-3x + 4y - 15 = -143x + 2y = 5
Use these to do Linear Combinations
3x + 2y = 5
+ -3x + 4y = 1
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6y = 6 ------> ( Y = 1 )
Now Take your Original Equation 3x+2y-z=8 Substitute Z and Y to find X
3x + 2(1) - (-3) = 8
3x + 2 + 3 = 8
3x + 5 = 8
3x = 3
x = 1
So Z = -3, Y = 1, and X = 1