What dimensional space would the column picture of four equations with two unknown be in? This question was taken from MIT OCW and Introduction to Linear Algebra by Gilbert Strang
For four linear equations in two unknowns $x$ and $y$, the row picture shows four ____. The column picture is in ____ -dimensional space. The equations have no solution unless the vector on the right side is a combination of ____.
I am having trouble understanding everything in this question and just seeing the answer isn't helping.
What I understand so far: 
The row picture would show four lines since each equation could at most be some kind of combination of the other.
What dimension would the column picture be in? The answer saids 4-D, but I don't understand why.
The vector on the right side would have to be a combination of the four columns? I don't understand this either.
I always get confused when there are a different number of equations and variables. I never know what dimensional space I am working with then. I want this broken down to me as if it were being explained to a five year old. 
 A: Here's the general situation:
$$\begin{cases} ax + by = c \\ dx + ey = f \\ gx + hy = i \\ jx + ky = l\end{cases}$$
Notice that this system of equations is equivalent to the matrix equation
$$\pmatrix{a & b \\ d & e \\ g & h \\ j & k}\pmatrix{x \\ y} = \pmatrix{c \\ f \\ i \\ l}$$
Can you see this?
"Row picture" and "column picture" are not standard terms so I can't be sure what Dr. Strang means.  But my guess is just that he wants you to see that each column vector of the transformation matrix $\pmatrix{a & b \\ d & e \\ g & h \\ j & k}$ has $4$ entries and each row vector (for instance the second row vector is $\pmatrix{d & e}$) has $2$ entries.

You also mention that you'd like to know why the RHS of the $A\vec x=\vec b$ equation must be a linear combination of the columns of the matrix $A$.  Can you see that $$\pmatrix{a & b \\ d & e \\ g & h \\ j & k}\pmatrix{x \\ y} = \pmatrix{ax +by \\ dx + ey \\ gx+hy \\ jx+ky} = \pmatrix{ax \\ dx \\ gx \\ jx} + \pmatrix{by \\ ey \\ hy \\ ky} = x\pmatrix{a \\ d \\ g \\ j} + y\pmatrix{b \\ e \\ h \\ k}\ ?$$
Notice that this last expression is just a linear combination of the columns of $A$.  Just plug this back into the $A\vec x=\vec b$ equation to get $$x\pmatrix{a \\ d \\ g \\ j} + y\pmatrix{b \\ e \\ h \\ k} = \pmatrix{c \\ f \\ i \\ l}$$  Now it should be clear that if $A\vec x=\vec b$ has a solution then $\vec b$ must be a linear combination of the columns of $A$.
A: The "column" picture treats the coefficients of a single unknowns in all the equations as a vector.  If the equations were, say
3x- 2y= 8
4x+ 3y= 2
2x- y= 1
5x+ 7y= 3 
Then the "column" picture gives the vectors <3, 4, 2, 5>, <-2, 3, -1, 7>, and <8, 2, 1, 3>.  The equations can be thought of as the vector equation x<3, 4, 2, 5>+ y<-2, 3, -1, 7>= <8, 2, 1, 3>.  Since these are four dimensional vectors, only two vectors cannot span the space.  In general two vectors will span only a two dimensional subspace of a four dimensional space.  There will be x, y that satisfy this equation only if the right side (here, <8, 2, 1, 3>) happens to be in that subspace.
