Find the solutions of the system of congruences

[Math100]

Homework Statement

Find the solutions of the system of congruences:
$3x+4y\equiv 5\pmod {13}$
$2x+5y\equiv 7\pmod {13}$.

Relevant Equations

None.

Consider the system of congruences:
$3x+4y\equiv 5\pmod {13}$
$2x+5y\equiv 7\pmod {13}$.
Then
\begin{align*}
&3x+4y\equiv 5\pmod {13}\implies 6x+8y\equiv 10\pmod {13}\\
&2x+5y\equiv 7\pmod {13}\implies 6x+15y\equiv 21\pmod {13}.\\
\end{align*}
Observe that $[6x+15y\equiv 21\pmod {13}]-[6x+8y\equiv 10\pmod {13}]$ produces $7y\equiv 11\pmod {13}$.
This means $7y\equiv 11\pmod {13}\implies 14y\equiv 22\pmod {13}\implies y\equiv 9\pmod {13}$.
Thus
\begin{align*}
&3x+4y\equiv 5\pmod {13}\implies 3x+4(9)\equiv 5\pmod {13}\\
&\implies 3x+36\equiv 5\pmod {13}\implies 3x\equiv 8\pmod {13}\\
&\implies 12x\equiv 32\pmod {13}\implies -x\equiv 6\pmod {13}\\
&\implies x\equiv -6\pmod {13}\implies x\equiv 7\pmod {13}.\\
\end{align*}
Therefore, the solutions are $x\equiv 7\pmod {13}$ and $y\equiv 9\pmod {13}$.

 

[fresh_42]

Homework Statement:: Find the solutions of the system of congruences:
$3x+4y\equiv 5\pmod {13}$
$2x+5y\equiv 7\pmod {13}$.
Relevant Equations:: None.

Consider the system of congruences:
$3x+4y\equiv 5\pmod {13}$
$2x+5y\equiv 7\pmod {13}$.
Then
\begin{align*}
&3x+4y\equiv 5\pmod {13}\implies 6x+8y\equiv 10\pmod {13}\\
&2x+5y\equiv 7\pmod {13}\implies 6x+15y\equiv 21\pmod {13}.\\
\end{align*}
Observe that $[6x+15y\equiv 21\pmod {13}]-[6x+8y\equiv 10\pmod {13}]$ produces $7y\equiv 11\pmod {13}$.
This means $7y\equiv 11\pmod {13}\implies 14y\equiv 22\pmod {13}\implies y\equiv 9\pmod {13}$.
Thus
\begin{align*}
&3x+4y\equiv 5\pmod {13}\implies 3x+4(9)\equiv 5\pmod {13}\\
&\implies 3x+36\equiv 5\pmod {13}\implies 3x\equiv 8\pmod {13}\\
&\implies 12x\equiv 32\pmod {13}\implies -x\equiv 6\pmod {13}\\
&\implies x\equiv -6\pmod {13}\implies x\equiv 7\pmod {13}.\\
\end{align*}
Therefore, the solutions are $x\equiv 7\pmod {13}$ and $y\equiv 9\pmod {13}$.

Correct. And see, just as if it was over the rationals. All numbers different from $0$ have an inverse. We can therefore solve the linear equation system as usual. This time you calculated the intersection point of two straights in the plane.

If you draw the straights, they will look a bit weird because they wrap around $13$.