How can two equations with three variables be in R2, not in R3? I am currently taking 18-06sc Linear Algebra from MIT's Opencourseware. One of the things that has baffled me is how a 2x3 matrix is in R2. In my mind the underlying equations in this matrix would be:
2x + 3y + 7z = 4
1x + 2y + 3z = 10
Since each one of these equations has three variables, graphing it would create a plane in R3. This seems to contradict the fact that the matrix form of these two equations would be in R2. please help me understand this. 

A $2 \times 3$ matrix represents a linear transformation from $\Bbb R^3$  to $\Bbb R^2$.
For your example, $M = \begin{bmatrix}2&3&7 \\ 1&2&3\end{bmatrix}$ takes a vector ${\mathbf v} = \begin{bmatrix} x \\y \\z\end{bmatrix} \in \Bbb R^3$ and sends it to the vector $$M{\bf v} = \begin{bmatrix}2&3&7 \\ 1&2&3\end{bmatrix}\begin{bmatrix} x \\y \\z\end{bmatrix} = \begin{bmatrix}2x + 3y + 7z \\ x + 2y + 3z\end{bmatrix} \in \Bbb R^2.$$
You're right that geometrically, each single equation determines a two-dimensional plane,
\begin{align*}
P_1 &= \{(x, y, z) \in \Bbb R^3 : 2x + 3y + 7z = 4\} \quad {\rm and} \\
P_2 &= \{(x, y, z) \in \Bbb R^3 : x + 2y + 3z = 10\}, 
\end{align*}
each of which is a subspace of $\Bbb R^3$.
To solve both equations simultaneously amounts to finding all solutions of $$M{\bf v} = \begin{bmatrix}4 \\ 10\end{bmatrix},$$
which corresponds geometrically to finding the intersection $P_1 \cap P_2$ of those two planes, which will be a line in $\Bbb R^3$. This will be the (one-dimensional) subspace of $\Bbb R^3$ that gets sent to the point $\begin{bmatrix}4 \\ 10\end{bmatrix}$ by the matrix $M$.
