Suppose one wants to find a basis for the vector subspace of $\mathbb{R}^4$ consisting of the points for which $x+y+z+t=0=x+2y+3z+4t$.
Now this is doable in a "brute force way" by plugging in $x=-(y+z+t)$ into the second equation, then solving for $y$ and then one can pick the two basis vectors by letting the first be the solution vector evaluated at $(z=1,t=0)$ and the second at $(z=0,t=1)$.
However, it feels to me that this whole method should be streamlined by linear algebra somehow. Is there a quick way to see this with a matrix?
More generally if one had $g(x,y,...,z)=h(x,y,...,z)=\cdots=j(x,y,...,z)=0$, how can one find the basis for this subspace?
EDIT: The naive approach would seem to be to consider the matrix with 1 1 1 1 as the top row and 1 2 3 4 as the bottom and row reduce. But this gives that the subspace is two dimensional, which makes sense in this case (coincidentally), but not in general if one had, say, 11 variables. I'm stumped.