Posted by **Diane** on Sunday, July 5, 2009 at 11:18pm.

Solve for

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.

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.

- 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

add

70x-y=-2 but x=0, so

y = 2

sub back into #3 to get z = 1

x=0

y=2

z=1

- college algebra -
**Anonymous**, Thursday, July 9, 2009 at 8:18pm
2x^3(2x^2+4x+3)

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