college
posted by Sharnell .
1. Solve.
2x+3y+4z=2
5x2y+3z=0
x5y2z=4
A.
(3, 4, 2)
B.
(2 ,1, 2)
C.
(2, 2, 2)
D.
(1, 1, 2)

The school subject is math, not college.

You could either substitute the 4 sets of values into one equation or start solving:
1. 2x+3y+4z=2
2. 5x2y+3z=0
3. x5y2z=4
Equ 1 x 2
4x+6y+8z=4
Equ 2 x 3
15x6y+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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