Posted by Diane on Sunday, July 5, 2009 at 11:18pm.
take the first equation, and the third equation.
mulitiply the first by 2, adn the third by 3.
that implies that x is zero.
then the first and second
multply the first by 10, the second by 9
-3y=6 or y=-2
solve for z in any equation, check your answers.
Since you don't specify which method, let's use good ol' elimination.
#1x2 ---> 6x+12y-18z=6
#3x3 ---> 33x+12y-18z=6
subtract, by luck both z and y disappear
x = 0
#2x3 ---> 15x-21y+30z=-12
#3x5 ---> 55x+20y-30z=10
70x-y=-2 but x=0, so
y = 2
sub back into #3 to get z = 1
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