Solve the system of equations.

2x - y + z = - 7

x - 3y + 4z = - 19

-x + 4y - 3z = 18.

A. There is one solution (1, - 2, 3).

B. There is one solution (-1, - 2, - 3).

C. There is one solution (1, 2, 3).

D. There is one solution (-1, 2, - 3).

(1) 2x-y+z=-7

(2) x-3y+4z=-19

(3) - x+4y-3z=18

Using the method of elimination:

Adding equation (2) and (3):

(x-3y+4z) + (-x+4y-3z) = (-19) + (18)

x-3y+4z-x+4y-3z=-19+18

y+z=-1

Solving for z:

y+z-y=-1-y

z=-1-y

Multiplying the third equation by 2:

(3) - x+4y-3z=18

2 (-x+4y-3z=18)

-2x+8y-6z=36

Adding with equation (1)

(2x-y+z) + (-2x+8y-6z) = (-7) + (36)

2x-y+z-2x+8y-6z=-7+36

7y-5z=29

Replacing z=-1-y in the equation above:

7y-5 (-1-y) = 29

7y+5+5y=29

12y+5=29

Solving for y. Subtracting 5 both sides of the equation.

12y+5-5=29-5

12y=24

Dividing both side of the equation by 12:

12y/12=24/12

y=2

Replacing y=2 in z=-1-y

z=-1-2→z=-3

Replacing y=2 and z=-3 in the equation (2); and solving for x:

(2) x-3y+4z=-19

x-3 (2) + 4 (-3) = - 19

x-6-12=-19

x-18=-19

Adding 18 both sides of the equation:

x-18+18=-19+18

x=-1

There is one solution: (x, y, z) = (-1,2,-3)

Answer: Option D. There is one solution (-1, 2, - 3).