Problem with solving systems of linear congruence of two variables The problem I am given is finding a solution to the following pair of equations:
$3x + 4y$ $\equiv$ $5$ $mod$ $13$
$2x + 5y$ $\equiv$ $7$ $mod$ $13$
By reading the methodology here: How do you solve linear congruences with two variables.
Since the modulos in my problem are also the same, I am able to use substitution, but this leaves fractions in my expressions.
I then used matrix multiplication as follows:
$\begin{bmatrix} 3 & 4\\ 2 & 5\end{bmatrix}$
$\begin{bmatrix}\ x\\ y\end{bmatrix}$
= $\begin{bmatrix}\ 5\\ 7\end{bmatrix}$ 
And solving this also gives me a matrix with fractions, so I am a bit lost on how to solve this problem.
Any help is appreciated. Thank you very much.
 A: We have $3x + 4y \equiv 5\pmod{13}$.
Since $(8,13)=1$ we can multiply $8$ on both sides of the above congruence. This yields , $$24x+32y\equiv 40\pmod{13}$$ $$\implies -2x+6y\equiv 1\pmod {13}$$ 
Now adding the above equation with $2x + 5y \equiv7 \pmod{13}$ we get $$11y\equiv 8\pmod{13}$$.
The above congruence has a solution $y=9$. Substituting this in any one of the above equations will give you $x=7$.
So $x=7$ and $y=9$  satisfies the given system of equations.
A: Cramer's Rule works since the determinant $= 7\,$ is invertible, being coprime to the modulus $13$
$$\begin{align} \left[\begin{matrix} 5 &\!\!\! -4 \\  -2 &\! 3 \end{matrix}\right]\ \times\ 
&\left\{\, \left[\begin{matrix} 3 & 4 \\  2 & 5 \end{matrix}\right]
 \left[\begin{matrix} x \\ y \end{matrix}\right] 
\equiv \left[\begin{matrix} 5 \\  7 \end{matrix}\right]\,\right\}\pmod{\!13}\\[.8em]
\Longrightarrow &\ \ \  \left[\begin{matrix} 7 & 0 \\  0 & 7 \end{matrix}\right] \left[\begin{matrix} x \\ y \end{matrix}\right] \equiv \left[\begin{matrix} 10 \\  11 \end{matrix}\right]
\end{align}\qquad$$
$$\begin{align}{\rm Therefore}\ \ \  &7x\equiv 10 \iff x\equiv \dfrac{10}7\equiv \dfrac{20}{14}\equiv \dfrac{7}1\\[.5em]
 &7y\equiv\ 11 \iff y\equiv \dfrac{11}7\equiv \dfrac{22}{14}\equiv \dfrac{9}1\end{align}\!\! $$
