1) What is the value of y in the system x+3y-z=-3, 6x=2y-1 and x=z+3 IF x = 5/6 and z = 23/6?

Question

2) Given the systems x+y+z=-14 and 2x+y+6z=16, how many number of solutions to the system are there?

Answer 1

1. strange question

Without looking at your last part of the given x and z values, just solving the system ....
from the 2nd:
2y = 6x+1
y = (6x+1)/2

from the 3rd:
z = x-3

back into the 1st:
x + 3y - z = -3
x + 3(6x+1)/2 - (x-3) = -3
times 2
2x + 18x + 3 - 2x + 6 = -6
18x = -15
x = -15/18 =-5/6

then y = (6(-5/6) + 1)/2 = -2
z = -5/6 - 3 = -23/6

So there is one uniques solution of (-5/6, -2, -23/6)
and the value of y is -2

Now when you give an x value and a z value different from the actual solution, it will depend on which equation you sub into.
There would be 3 different answers for the value of y

2. Each equation represents a plane. Since the 2 planes are not parallel, they will intersect in a line.
There will be an infinite number of points on that line, thus there will be an infinite number of solutions for your system.