Solving Linear Equations with Modulo I'm having trouble approaching the following problem: 
Solve the following system of equation -
$$3x+5y\equiv 14\mod 17$$
$$7x+3y\equiv 6\mod 17$$
Multiplying the top by $3$ and the bottom by $5$, I get $x = -3/5$ and $y = 51/15$, but I'm unsure how to apply (mod $17$) with $2$ variables
 A: So what you do is elimination of variables, and then deal with it differently:
start by eliminating $y$: multiply the first equation by three, second equation by five, then you get:
$$
9x+15y \equiv 42 (\equiv 8) \mod 17 ; 35x+15y \equiv 30 (\equiv 13) \mod 17 
$$
Now, subtract the first equation from the second, and you get:
$$
26x \equiv 5 \mod 17 
$$
This has to be solved, but we can do this through the reverse-Euclid algorithm: we know that there exist integers $x,y$ such that $17x+26y=1$ by Bezout's theorem. So the question is what are these $x$ and $y$?
To do this, note that:
$$
26 = (17 \times 1)+9 ; 17 = (9 \times 1) + 8; 9 = (8 \times 1)+ 1
$$
So we cleverly reverse this procedure:
$$
1 = 9-(8 \times 1) = (26-17  \times 1) - (17-9 \times 1) = (26-17)- (17-(26-17)) = 26-17-17+26-17 = 2 \times 26-3 \times 17
$$
Therefore, $x=2$ satisfies the equation $26x \equiv 1 \mod 17$. Multiply by $5$ and we see that $x=2 \times 5=10$ satisfies $26x \equiv 5 \mod 17$.
Hence, $x=10$ is a solution. Substituting this in the second equation, we see that $70+3y \equiv 6 \mod 17$, which simplifies to $3y \equiv -64 \mod 17$, or $3y \equiv 4 \mod 17$. You can check again via reverse Euclid:
$$
17 = (3 \times 5)+2 ; 3 = (2 \times 1)+1
$$
Therefore
$$
1 = 3-(2 \times 1) = 3 - (17 - 3 \times 5) = 3 \times 6-17 \times 1
$$
So  $y=6$ satisfies $3y \equiv 1 \mod 17$, so multiply by four to get $3y \equiv 4 \mod 17$ (for $y=24$, or $y=7$ when reducing mod $17$).
So the answer is $x=10,y=7 \mod 17$. Let's check this:
$$
3x+5y = 30+35 = 65 \equiv 14 \pmod{17} \\
7x+3y = 70+21 = 91 \equiv 6 \pmod{17}
$$
Hence the equation is solved.
A: A little more automatic: obtain the reduced row echelon form of the augmented matrix in the field $\mathbf Z/17\mathbf Z$.
\begin{align}
&\begin{bmatrix}3&5 &\color{red}{14}\\7&3&\color{red}{6}\end{bmatrix}=\begin{bmatrix}3&5 &\color{red}{-3}\\7&3&\color{red}{6}\end{bmatrix}\stackrel{R_1\leftarrow 6R_1}{-\!\!\!-\!\!\!-\!\!\!\rightsquigarrow}\begin{bmatrix}1&-4 &\color{red}{-1}\\7&3&\color{red}{6}\end{bmatrix}\stackrel{R_2\leftarrow R_2-7R_1}{-\!-\!\!\!\rightsquigarrow}\begin{bmatrix}1&-4 &\color{red}{-1}\\0&-3&\color{red}{-4}\end{bmatrix}\\
\stackrel{R_2\leftarrow -6R_2}{-\!\!\!-\!\!\!-\!\!\!\rightsquigarrow}&\begin{bmatrix}1&-4 &\color{red}{-1}\\0&1&\color{red}{7}\end{bmatrix}\stackrel{R_1\leftarrow R_1-4R_2}{-\!\!\!-\!\!\!-\!\!\!\rightsquigarrow}
\begin{bmatrix}1&0 &\color{red}{-7}\\0&1&\color{red}{7}\end{bmatrix}
\end{align}
The solution is $\;\begin{bmatrix}\color{red}{-7}\\\color{red}{7}\end{bmatrix}=\begin{bmatrix}\color{red}{10}\\\color{red}{7}\end{bmatrix}$.
