Asked by Stephy
                Solving Systems of Equations in Three Variables:
3x+3y-z=-3
-2x-3y+2z=3
x-6y+3z=0
PLease show work
            
        3x+3y-z=-3
-2x-3y+2z=3
x-6y+3z=0
PLease show work
Answers
                    Answered by
            Henry
            
    +3x + 3y - z = -3
-2x - 3y + 2z = 3
x - 6y + 3z = 0
Eq1: 3x + 3y - z = -3
Eq2: -2x - 3y + 2z = 3
Sum: x + 0 + z = 0
z = -x
Eq2: -2x - 3y + 2z = 3
Eq3: x - 6y + 3z = 0
Multiply Eq2 by 6 and Eq3 by (-3):
-12x - 18y + 12z = 18
-3x + 18y - 9z = 0
Sum: -15x + 0 + 3z = 18
Replace z with -x:
-15x-3x = 18
-18x = 18
X = -1
Z = -x = 1.
In Eq1, replace X with -1. and z with 1:
3*(-1) + 3y - 1 = -3
-3 + 3y -1 = -3
3y = 1
Y = 1/3
Solution = (X, Y, Z) = (-1, 1/3, 1).
    
-2x - 3y + 2z = 3
x - 6y + 3z = 0
Eq1: 3x + 3y - z = -3
Eq2: -2x - 3y + 2z = 3
Sum: x + 0 + z = 0
z = -x
Eq2: -2x - 3y + 2z = 3
Eq3: x - 6y + 3z = 0
Multiply Eq2 by 6 and Eq3 by (-3):
-12x - 18y + 12z = 18
-3x + 18y - 9z = 0
Sum: -15x + 0 + 3z = 18
Replace z with -x:
-15x-3x = 18
-18x = 18
X = -1
Z = -x = 1.
In Eq1, replace X with -1. and z with 1:
3*(-1) + 3y - 1 = -3
-3 + 3y -1 = -3
3y = 1
Y = 1/3
Solution = (X, Y, Z) = (-1, 1/3, 1).
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