take the first equation, and the third equation.
mulitiply the first by 2, and the third by 3.
6x+12y-18z=6
33x+12y-18z=6
that implies that x is zero.
then the first and second
6y-9z=3
-7y+10z=-4
multply the first by 10, the second by 9
60y-90z=30
-63y+90z=-36
add,
-3y=6 or y=-2
solve for z in any equation, check your answers.
Solve for
3x + 6Y - 9z = 3
5x - 7y + 10z = -4
11x +4y - 6z = 2
3 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
add
70x-y=-2 but x=0, so
y = 2
sub back into #3 to get z = 1
x=0
y=2
z=1
#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
add
70x-y=-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)
1 answer