Why is it that when there are fewer equations than unknowns we have infinite solutions in a system of linear equations? I have an example where x+2y=0, the solutions can be (-2,1), (2,-1),etc to makes sense of the question, but still I am not convinced. I think I need help with solidifying my understanding.
Another question is, does this theorem only applies to homogenous equations and not to non-homogenous equations. So if I have for example,
2x+7y-3z=9 and
3x-5y+9z=10
I don't necessarily have infinitely many solutions? when I can't even solve for one of the variables?
 A: Let us look at a homogenous system with $n \in \mathbb{N}$ unknowns.
A homogenous system always has at least one solution, this is the zero solution
$(x_1, x_2, \dots, x_n) = (0,0,\dots,0)$
So the only question is: do we have other solutions?
Since you have $n$ unknowns the rank of the matrix A of the system is at most $n$. So it doesn't matter how many equations you have: at most n of them can be meaningful i.e. linearly independent.
OK. So then there's this theorem which says:
The homogenous system has a non-zero solution if and only if $rank(A) < n$.
Another theorem says:
The solutions of the homogenous system form a vector space $U$. The dimension of this space is $n-rank(A)$.
In plain words, if you have e.g. $7$ unknowns but $r(A)=5$, then you have infinitely many solutions which depend on $7-5=2$ free parameters.
This solves all questions about the homogenous systems.
There's some more theory for non-homogenous systems which solves all questions which can arise there.
Rouché–Capelli Theorem
For non-homogenous systems, it is possible that you have e.g. 3 equations with 5 unknowns but still you have no solutions. That happens when the equations contradict each other. This happens when $r(A|b) > r(A)$.
Here $A|b$ is the augmented matrix.
So in the case of non-homogenous systems, what you say is not true. If you have less equations than unknowns, you don't necessarily have infinitely many solutions. But you can claim that you either have no solutions or infinitely many.
Example:
$5x + 6y + 1z + 1t - 2v = 12$
$5x + 6y + 1z + 1t - 2v = 10$
$1x - 1y + 0z + 0t + 0v = 1$
Here the first two equations obviously contradict each other.
A few months ago I had these same questions myself. You just need to go through a few chapters of linear algebra and it will all become clear.
A: To build some intuition, I suggest you think about systems of linear equations involving three variables $x,y,z$. Each linear equation represents a plane in 3D space. So an $(x,y,z)$ point that satisfies all of a given collection of equations must lie on the intersection of the corresponding collection of planes.
Suppose there are two equations. Then we have two planes. The planes might be parallel, but distinct. That means there are no intersection points, so the equations have no solution. In general, the two planes will intersect in some line, and every point on that line is a solution, so we have an infinite number of solutions. Or, the two planes might even be coincident, in which case we have an even larger infinity of solutions (roughly speaking).
Now go through the same sort of reasoning with three equations (so three planes). Think about the different ways that three planes can intersect.
What about four equations. If you have four equations and only three unknowns, then someone might have told you that there would be no solution. But clearly the four planes might intersect, if you’re lucky.
Homogeneous equations just correspond to planes that pass through the origin, so it’s not hard to include those in your thought experiment.
Once you’ve grasped all of this, you can translate it all into the language of abstract linear algebra (rank and dimensions and so on).
