Find the solutions of ## 7x+3y\equiv 6\pmod {11} ##.

[Math100]

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 $y\equiv 3\pmod {11}$.

 

[SammyS]

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 $y\equiv 3\pmod {11}$.

Or you could observe that:
$\displaystyle 7x+3y+4x+2y\equiv 6+9 \pmod {11}$

so that $\displaystyle 11x+5y\equiv 15 \pmod {11}$

which reduces to $\displaystyle 0x+5y\equiv 4 \pmod {11}$ .

After a bit of work $\displaystyle y\equiv 3 \pmod {11}$ .

 

[fresh_42]

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 $y\equiv 3\pmod {11}$.

Correct.

I have observed that the coefficients of $x$ add up to $11$ so I simply added both equations and got $5y\equiv 4\pmod{11}.$ I got from $5\cdot 9 \equiv 4\cdot 3\equiv 1\pmod{11}$ the inverses $5^{-1}\equiv 9\pmod{11}$ and $4^{-1}\equiv 3\pmod{11}.$ Thus $y\equiv 4\cdot 9\equiv 3\pmod{11}$ and $x\equiv 4^{-1} (9-2y)\equiv 3\cdot (9-2\cdot 3)\equiv 9\pmod{11}.$