All the solutions for this system 5x+33y = 6 (mod 13) and 7x + 2y = 9 (mod 13)

Question

I want all the solutions for this system. 5x + 3y = 6 (mod 13) and 7x + 2y = 9 (mod 13)... Thanks

Answer 1

Doing arithmetic modulo $\;13\;$ all the time :

$$\begin{align*}&I\;\;\;&5x+33y=6\iff5x+7y=6\\ &II\;\;\;&7x+2y=9\iff7x+2y=9\end{align*}$$

Multiply I by $\;-7\;$ and II by $\;5\;$ :

$$\begin{align*}&I\;\;\;&4x+3y=10\\ &II\;\;\;&9x+10y=6\end{align*}\;\implies I+II\rightarrow\;\;\;\;0=1\;\;\leftarrow\text{contradiction}$$

and thus the equation has no solution.

Another way: solve as over the reals (again, multiplying I by $\;-7\;$ and etc.)

$$\begin{align*}&I\;\;\;&5x+33y=6\\ &II\;\;\;&7x+2y=9\end{align*}\;\;\implies$$

$$\begin{align*}I\;\;\;&-35x-231y=-42\\ II\;\;\;\;\;&\;\;\;\;\;35x+10y=45\end{align*}\;\;\implies-221y=3$$

Yet the above is already $\;0=3\;$ since $\;221=13\cdot17=0\pmod{13}\;$

Answer 2

We can write $$5x+33y=13a+6\ \ \ \ (1)$$ $$7x+2y=13b+9\ \ \ \ (2)$$ where $a,b$ are integers

$(1)\cdot7-(2)\cdot5\implies 221y=7(13a+6)-5(13b+9)=13(7a-5b)-3$

$$\iff3=13(7a-5b-17y)$$ which is impossible as the Right Hand Side is divisible by $13$ unlike the Left one