Posted by **ashley** on Monday, April 2, 2012 at 8:31pm.

systems of linear equations in 3 variables

1/3x-2/3y+z=0

1/2x-3/4y+z=1/4

-2x-y+z=1

- algebra -
**Charlie**, Monday, April 2, 2012 at 8:46pm
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)

- algebra -
**bobpursley**, Monday, April 2, 2012 at 8:51pm
It might be easier if you

a) multiplied the first equation by 3

b) the second equation by 4

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