Question

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).

Answer 1

(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).