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)

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)