- #1

Math100

- 773

- 219

- Homework Statement
- Find the solutions of the following system of congruences:

## 7x+3y\equiv 6\pmod {11} ##

## 4x+2y\equiv 9\pmod {11} ##.

- Relevant Equations
- None.

Consider the following system of congruences:

## 7x+3y\equiv 6\pmod {11} ##

## 4x+2y\equiv 9\pmod {11} ##.

Then ## gcd(7\cdot 2-3\cdot 4, 11)=gcd(2, 11)=1 ##.

This means that ## \exists ## a unique solution.

Observe that

\begin{align*}

&2(7x+3y)-3(4x+2y)\equiv [2(6)-3(9)]\pmod {11}\\

&\implies 2x\equiv -15\pmod {11}\\

&\implies 2x\equiv 7\pmod {11}\\

&\implies 12x\equiv 42\pmod {11}\\

&\implies x\equiv 9\pmod {11}.\\

\end{align*}

Thus

\begin{align*}

&4(7x+3y)-7(4x+2y)]\equiv [4(6)-7(9)]\pmod {11}\implies -2y\equiv -39\pmod {11}\\

&\implies 2y\equiv 39\pmod {11}\implies 12y\equiv 234\pmod {11}\\

&\implies y\equiv 3\pmod {11}.\\

\end{align*}

Therefore, the solutions are ## x\equiv 9\pmod {11} ## and ## y\equiv 3\pmod {11} ##.

## 7x+3y\equiv 6\pmod {11} ##

## 4x+2y\equiv 9\pmod {11} ##.

Then ## gcd(7\cdot 2-3\cdot 4, 11)=gcd(2, 11)=1 ##.

This means that ## \exists ## a unique solution.

Observe that

\begin{align*}

&2(7x+3y)-3(4x+2y)\equiv [2(6)-3(9)]\pmod {11}\\

&\implies 2x\equiv -15\pmod {11}\\

&\implies 2x\equiv 7\pmod {11}\\

&\implies 12x\equiv 42\pmod {11}\\

&\implies x\equiv 9\pmod {11}.\\

\end{align*}

Thus

\begin{align*}

&4(7x+3y)-7(4x+2y)]\equiv [4(6)-7(9)]\pmod {11}\implies -2y\equiv -39\pmod {11}\\

&\implies 2y\equiv 39\pmod {11}\implies 12y\equiv 234\pmod {11}\\

&\implies y\equiv 3\pmod {11}.\\

\end{align*}

Therefore, the solutions are ## x\equiv 9\pmod {11} ## and ## y\equiv 3\pmod {11} ##.