college algebra
posted by Diane .
Solve for
3x + 6Y  9z = 3
5x  7y + 10z = 4
11x +4y  6z = 2

take the first equation, and the third equation.
mulitiply the first by 2, and the third by 3.
6x+12y18z=6
33x+12y18z=6
that implies that x is zero.
then the first and second
6y9z=3
7y+10z=4
multply the first by 10, the second by 9
60y90z=30
63y+90z=36
add,
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+12y18z=6
#3x3 > 33x+12y18z=6
subtract, by luck both z and y disappear
x = 0
#2x3 > 15x21y+30z=12
#3x5 > 55x+20y30z=10
add
70xy=2 but x=0, so
y = 2
sub back into #3 to get z = 1
x=0
y=2
z=1 
2x^3(2x^2+4x+3)