Question

what is the solution to the system of equations?
-4x+4y-2z=-8
-3x-y+4z=0
2x-2y+3z=-4

possibles are
(-3,-7,-4)
(1,-7,-2)
(3,5,-4)
(-3,0,5)

Answer 1

i can solve this, but after an hour because i still have my class.

Answer 2

okay

Answer 3

(-3,-7,-4)
12-28+8=-8
9-7=2

Answer 4

(1,-7,-2)
-4-14+4=-14

Answer 5

(3,5,-4)

-15+20+8=13

Answer 6

(-3,0,5)
12+0-10=2

Answer 7

first we will are going to find equations with two variables,
(-3x-y+4z=0)*4
-12x-4y+16z=0
-4x+4y-2z=-8, add this equations.
you will get: -16x+14z=-8 (eqn. 4)

(-3x-y+4z=0)*-2
6x+2y-8z=0
2x-2y+3z=-4, add
you will get: 8x-5z=-4 (eqn. 5)

add eqn 4 and 5
-16x+14z=-8
(8x-5z=-4)*2 -----> 16x-10z=-8

you will get 4z=-16

z=-4

-4x+4y-2(-4)=-8
-4x+4y+8=-8
-4x+4y=-8-8
-4x+4y=-16 (eqn. 6)

z=-4

-3x-y+4(-4)=0
-3x-y-16=0
-3x-y=16 (eqn. 7)

add eqn 6 and 7.

-4x+4y=-16
(-3x-y=16)*4-----> -12x-4y=64

-16x=48
x=-3

then, x=-3 and z=-4

-4x+4y-2z=-8
-4(-3)+4y-2(-4)=-8
12+4y+8=-8
4y=-8-8-12
4y=-28
y=-7

So, x=-3, y=-7 and z=-4