- #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} ##.