Question

1. What is the solution to the system?

[1] x +2y+3z=6
[2] y+2z=0
[3] z=2
a. (10, 0, 2)
b. (8, -4, 2)
c. (6, 0, 2)
d. (-8, 4, 2)

Answer 1

x + 2y + 3z = 6.....[1]
y + 2z = 0 ...[2]
z = 2 ...[3]

sub in eq [3] into eq [2]
y + 2z = 0
y + 2(2) = 0
y + 4 = 0
y = -4

now sub in -4 for y and 2 for z back into eq [1]
x + 2y + 3z = 6
x + 2(-4) + 3(2) = 6
x - 8 + 6 = 6
x - 2 = 6
x = 6 + 2
x = 8

solution is : (8,-4,2)