Im a student studying linear algebra and my professor has said this to me on the fly. I have encountered that as long as the equation is not trivial ($1 = 1$), or equations are not the same ($y = x$, $x = y$), then each equation of a vector space removes 1 dimension of the vector space it is based on. For example the vector space $W$:
$W = \{(x,y,z)\in\mathbb{R}^3:x=y\}$
then a basis for $W$ could be: $\{(1,1,0),(0,0,1)\}$
$\dim(\mathbb{R}^3)=3$
but $\dim(W)=2$
so my question is why does this happen? if it even is true
Good question.
I think it helps to think geometrically. In three dimensions the solutions to an equation like $$ ax + by +cz = 0 $$ form a plane. That has dimension $2 = 3-1$.
If you have another such equation it defines another plane (unless it happens to be proportional to this equation). The intersection of those two planes is a line, so dimension $3-2 =1$.
Three planes (in general position) intersect in a point, with dimension $3-3=0$.
You can't see the pictures in higher dimensions but the geometry works the same way. You use the algebra of linear independence to formalize that geometric intuition.