Really basic Linear Algebra question

you solve the linear system in two unknowns and two equations

x + 4y = 12 + z
3x + 8y = 4 + 2z

solution is
x = -20
y = (t + 32)/4
z = t

in 3D coordinate system xyz, given equations
x + 4y - z = 12
3x + 8y - 2z = 4

represent two planes
and the infinite solutions

x = -20
y = (t + 32)/4
z = t, for any real t

can be written as (-20, (t + 32)/4, t) for any real t
and represent the intersection line of the two planes

Firstly, multiply both sides of the first equation by 2

You get 2x + 8y - 2z = 24

Then, subtract the second equation by the first one. You get...

3x - 2x + 8y - 8y + 2z - 2z = 4 - 24
x = -20

After you know x, you can use your method to find y and z :)

Either y is dependent on z or z is dependent on y, but you can't have both.

The statements y=0.25z+8 and z=4y-32 are exactly the same, just rearranged.

You have three variable but only two equations, so there won't be a single solution.

I got x = +20, while y is a function of z.