- #1
Math100
- 779
- 220
- 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} ##.
## 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} ##.