What you have discovered is you really only have "one equation" (the second equation is just the first one multiplied by -4).

So your one equation is:

x_{1}+ (1/2)x_{2}= 3/2.

To "solve" for this, we can only decide what x_{1}should be after we have picked x_{2}(or vice-versa, but since we usually row-reduce from "the bottom up", we designate the non-pivot rows as the "free variables").

So let x_{2}= s. We then get:

x_{1}+ s/2 = 3/2, so:

x_{1}= (3 - s)/2.

Thus "a" solution is ((3 - s)/2,s) (which solution we get will depend on what we choose for s). We can write this as:

(x_{1},x_{2}) = (3/2,0) + s(-1/2,1).

Let's look at what happens with "each term" in our system:

2(3/2) + 0 = 3

-8(3/2) - 4(0) = -12

so (3/2,0) is a PARTICULAR solution (not the ONLY one, but one that works).

Now the other term:

2(-s/2) + s = -s + s = 0.

-8(-s/2) - 4s = 4s - 4s = 0, no matter WHAT number s is.

In other words, any multiple (by s) of (-1/2,1) is a solution to the system:

2x_{1}+ x_{2}= 0

-8x_{1}- 4x_{2}= 0

(called the associated HOMOGENEOUS system).

So if we call the original matrix A, we see that:

A((3 - s)/2,s) = A(3/2,0) + A(s(-1/2,1)) = A(3/2,0) + sA(-1/2,1) = (2(2/2) + 0,-8(3/2) - 4(0)) + s(2(-1/2) + 1, -8(-1/2) - 4) = (3,-12) + s(0,0) = (3,-12) + (0,0) = (3,-12).