Is it possible to solve for $3$ or more unknowns with only $1$ equation Example: 
$$x+3y+7z=221$$
or
$$2x+5y+z+9i=383$$
 A: If the only thing you have is the constraint $$x + 3y+ 7z = 221$$ then you know that $x = 221-3y-7z$ and so all solutions to your problem can be characterized as
$$
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
= \begin{pmatrix} 221-3y-7z \\ y \\ z \end{pmatrix}
= \begin{pmatrix} 221 \\ 0 \\ 0 \end{pmatrix}
+ y\begin{pmatrix} -3 \\ 1 \\ 0 \end{pmatrix}
+ z\begin{pmatrix} -7 \\ 0 \\ 1 \end{pmatrix}
$$
A: Sure, but you won't get a unique solution. For instance, $$x=221, y=z=0$$ is a solution of the first equation and so is $$x=-1,y=74,z=0$$
In fact, there will be infinite solutions.
A: Yes, but not uniquely. That means you can choose values for two of the variables and solve for the third one. If you like to visualize this equation, think of an infinitely large plane in 3D space. The orientation of the plane to your axes are given by the coefficients 1,3 and 7. The constant term 221 tells you how far away your plane is from the origin. Any solution to the equation will be a point on that plane. So, there are infinitely many solutions.
Imagine you have another equation, with coefficients that are not a multiple of the first equation. This means  that the orientation of the plane to that second equation will be different from the first one. The intersection of those planes will be a line. Your set of solutions shrunk from being points on an infinite plane to just points on an infinite line.
Finally imagine you had another equation, that has coefficients that aren't a sum of multiples of the first two equations. This means the equation is linearly independent from the others. Visualize it as another plane in 3D space and look at the intersection. That now will be a single point. That point is your unique solution.
If you have more equations they must be some sum of multiples of the other three equations.
