Find the solutions of the following system of congruences

[Math100]

Homework Statement

Find the solutions of the following system of congruences:
$5x+3y\equiv 1\pmod {7}$
$3x+2y\equiv 4\pmod {7}$.

Relevant Equations

None.

Consider the following system of congruences:
$5x+3y\equiv 1\pmod {7}$
$3x+2y\equiv 4\pmod {7}$.
Then
\begin{align*}
&5x+3y\equiv 1\pmod {7}\implies 15x+9y\equiv 3\pmod {7}\\
&3x+2y\equiv 4\pmod {7}\implies 15x+10y\equiv 20\pmod {7}.\\
\end{align*}
Observe that $[15x+10y\equiv 20\pmod {7}]-[15x+9y\equiv 3\pmod {7}]$ produces $y\equiv 17\pmod {7}$.
This means $y\equiv 17\pmod {7}\implies y\equiv 3\pmod {7}$.
Since $3y\equiv 9\pmod {7}\implies 3y\equiv 1-5x\pmod {7}$,
it follows that $1-5x\equiv 9\equiv 2\pmod {7}\implies -5x\equiv 1\pmod {7}$.
Thus
\begin{align*}
&-5x\equiv 1\pmod {7}\implies -15x\equiv 3\pmod {7}\implies -x\equiv 3\pmod {7}\\
&\implies x\equiv -3\pmod {7}\implies x\equiv 4\pmod {7}.\\
\end{align*}
Therefore, the solutions are $x\equiv 4\pmod {7}; y\equiv 3\pmod {7}$.

 

[fresh_42]

Correct.

I only think that you could save a few steps. From $y \equiv 3\pmod{7}$ we get $3x\equiv 4-6\equiv 5\pmod{7}$ and so $x\equiv 5^2\equiv 4\pmod{7}.$

 

[Math100]

Correct.

I only think that you could save a few steps. From $y \equiv 3\pmod{7}$ we get $3x\equiv 4-6\equiv 5\pmod{7}$ and so $x\equiv 5^2\equiv 4\pmod{7}.$

This is way better.