Posted by **Sharnell** on Tuesday, January 20, 2009 at 10:16am.

1. Solve.

2x+3y+4z=2

5x-2y+3z=0

x-5y-2z=-4

A.

(3, 4, 2)

B.

(-2 ,1, 2)

C.

(2, 2, -2)

D.

(1, 1, -2)

- college -- Math -
**Ms. Sue**, Tuesday, January 20, 2009 at 10:22am
The school subject is __math__, not college.

- college -
**Dr Russ**, Tuesday, January 20, 2009 at 10:59am
You could either substitute the 4 sets of values into one equation or start solving:

1. 2x+3y+4z=2

2. 5x-2y+3z=0

3. x-5y-2z=-4

Equ 1 x 2

4x+6y+8z=4

Equ 2 x 3

15x-6y+9z=0

add 19x+17z=4

from by inspection z must be opposite sign to x and x equals z so B) and C) are possible answers.

You could either continue solving the equations or substitute the two sets in say equation 1.

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