What conditions must the constants satisfy so that each of these systems has a solution? I'm attempting to teach myself linear algebra using this book http://joshua.smcvt.edu/linearalgebra/book.pdf
One of the exercises is:
What conditions must the constants, the b’s, satisfy so that each of
these systems has a solution? Hint. Apply Gauss’s Method and see what happens
to the right side.

x - 3y = b1
3x + y = b2
x + 7y = b3
2x + 4y = b4

The answer given is 
Gauss’s Method shows that this system is consistent if and only if both
b3 = -2b1 + b2 and b4 = -b1 + b2.

I've applied Gauss elimination and gotten
1x - 3y = b1
10y = -3b1 + b2
0y  = 2b1 - b2 + b3
0y  = b1 - b2 + b4

But I still don't understand how the answer is deduced..
Someone help?
Thanks!
 A: You can just solve it like a system of equations (Hint: use elimination), or you can turn it into a matrix and solve it like this:
Step one:
$$ \left[
    \begin{array}{cc|c}
      1&-3&b_{1}\\
      3&1&b_{2}\\
      1&7&b_{3}\\
      2&4&b_{4}
    \end{array}
\right] $$
Leave R1 intact. 
Replace R2 with: R2 - 3(R1).
Replace R3 with: R3 - R1.
Replace R4 with: R4 - 2(R1).
$$ \left[
    \begin{array}{cc|c}
      1&-3&b_{1}\\
      0&10&b_{2} - 3b_{1}\\
      0&10&b_{3} - b_{1}\\
      0&10&b_{4} - 2b_{1}
    \end{array}
\right] $$
Step two:
Leave R1 intact. 
Leave R2 intact.
Replace R3 with: R3 - R2.
Replace R4 with: R4 - R2.
$$ \left[
    \begin{array}{cc|c}
      1&-3&b_{1}\\
      0&10&b_{2} - 3b_{1}\\
      0&0&b_{3} - b_{1} - (b_{2} - 3b_{1})\\
      0&0&b_{4} - 2b_{1} - (b_{2} - 3b_{1})
    \end{array}
\right] $$
Step three (answer): Now that you've got an echelon form of the matrix, you can figure out what your solutions might be. Since the question wants you to find out what must be true of the b's so that you can have at least one solution, and the same rules of math apply, your b's must be something that makes the matrix true. Your equations are:
$$
        \begin{matrix}
        x & -3y & = b_{1} \\
        0x & +10y & = b_{2} - 3b_{1} \\
        0x & +0y & = b_{3} - b_{1} - (b_{2} - 3b_{1})\\
        0x & +0y & = b_{4} - 2b_{1} - (b_{2} - 3b_{1}) \\
        \end{matrix}
$$
Realize that zero must equal zero! Since 0x = 0 and 0y = 0, the right hand side of those last two equations has to equal zero.
Step four (answer): Set the right hand side of the last two equations equal to zero and solve. What you get (left hand side in terms of one thing, right hand side in terms of another) is the answer the book gave you.
0 = b3 - b1 - (b2 - 3b1)
0 = b4 + b1 - b2

Gauss’s Method shows that this system is consistent if and only if both
  b3 = -2b1 + b2 and b4 = -b1 + b2.

