3x + 6Y - 9z = 3
5x - 7y + 10z = -4
11x +4y - 6z = 2
college algebra - bobpursley, Sunday, July 5, 2009 at 11:32pm
take the first equation, and the third equation.
mulitiply the first by 2, and 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.
college algebra - Reiny, Sunday, July 5, 2009 at 11:38pm
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
college algebra - Anonymous, Thursday, July 9, 2009 at 8:18pm