Asked by ashley
                systems of linear equations in 3 variables
1/3x-2/3y+z=0
1/2x-3/4y+z=1/4
-2x-y+z=1
            
        1/3x-2/3y+z=0
1/2x-3/4y+z=1/4
-2x-y+z=1
Answers
                    Answered by
            Charlie
            
    Mult 1st eq by 3; x-2y+3z = 0
Mult 2nd eq by 4: 2x - 3y + 4z = 1
Add 2nd and 3rd: -4y+5z = 2
Mut 1st by 2: 2x-4y + 6z = 0
Add that answer and 3rd eq: -5y + 7z = 1
5(-4y + 5z = 2) + -4(-5y + 7z = 1)
-20y + 25z = 10 + 20y - 28z = -4
-3z =6
z = -2
Sub into the 1st eq: x - 2y -6 =0
Sub into the 3rd eq: -2x-y-2 = 1
2(x-2y = 6) + (-2x-y=3)
2x - 4y = 12 + -2x -y = 3
-5y =15
y = -3
sub into the 3rd eq
-2x -(-3) +(-2) = 1
-2x +1 = 1
x = 0
(0,-3,-2)
    
Mult 2nd eq by 4: 2x - 3y + 4z = 1
Add 2nd and 3rd: -4y+5z = 2
Mut 1st by 2: 2x-4y + 6z = 0
Add that answer and 3rd eq: -5y + 7z = 1
5(-4y + 5z = 2) + -4(-5y + 7z = 1)
-20y + 25z = 10 + 20y - 28z = -4
-3z =6
z = -2
Sub into the 1st eq: x - 2y -6 =0
Sub into the 3rd eq: -2x-y-2 = 1
2(x-2y = 6) + (-2x-y=3)
2x - 4y = 12 + -2x -y = 3
-5y =15
y = -3
sub into the 3rd eq
-2x -(-3) +(-2) = 1
-2x +1 = 1
x = 0
(0,-3,-2)
                    Answered by
            bobpursley
            
    It might be easier if you 
a) multiplied the first equation by 3
b) the second equation by 4
    
a) multiplied the first equation by 3
b) the second equation by 4
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