Equations of planes and lines in 3-space I'm reading Strang's book "Linear Algebra and it's applications" and he writes in the first chapter that an equation involving two variables in still a plane in 3-space.
"The second plane is 4u - 6v = -2. It is drawn vertically, because w can take any value. The coefficient of w is zero, but this, remains a plane in 3-space. (The equation 4u = 3, or even the extreme case u = 0, would still describe a plane.)"
I don't quiete understand why, I always thought that you would need three variables to create a plane. E.g. the intersection of the equations 2u+ v+ w= 5 and 4u - 6v = -2 is a line, presumably because when you solve those two equations you get an equation with two variables, but then why isn't 4u - 6v = -2 a line as well?

A linear equation reduces the dimension of the ambient space by 1.  You can think of it as restricting one variable, as a function of all the others.
Hence, a linear equation in 2 dimensions is a one-dimensional space, or a line.  A linear equation in 3 dimensions is a 2-dimensional space, or a plane.  And so on.

Followup, as requested.  A system of 2 linear equations generally reduces the dimension of the ambient space by 2 -- each of them reduces the dimension by 1.  In a plane, this means a point (0-dimensional).  In 3-space, this means a line (1-dimensional).
The reason it's "generally" is that the two equations may actually be the same, such as $$x+y=1, 2x+2y=2$$
