Posted by ashley on Monday, April 2, 2012 at 8:31pm.
systems of linear equations in 3 variables
1/3x2/3y+z=0
1/2x3/4y+z=1/4
2xy+z=1

algebra  Charlie, Monday, April 2, 2012 at 8:46pm
Mult 1st eq by 3; x2y+3z = 0
Mult 2nd eq by 4: 2x  3y + 4z = 1
Add 2nd and 3rd: 4y+5z = 2
Mut 1st by 2: 2x4y + 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: 2xy2 = 1
2(x2y = 6) + (2xy=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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