Problem with solving systems of linear congruence of two variables

Question

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.

Fractions are well defined for denominators coprime to the modulus. See https://math.stackexchange.com/a/3151617/431789 — David Diaz, Oct 10 '19 at 18:50
@DavidDiaz That information makes this problem much easier, I solved it thanks to this link. Thank you very much. — Crystal McMillian, Oct 10 '19 at 19:03

score 1 · Accepted Answer · answered Oct 10 '19 at 18:59

1

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.

answered Oct 10 '19 at 18:59

May I ask how you went from $24x + 32y$ $\equiv$ $40$ $mod$ $13$ to $-2x + 6y$ $\equiv$ $1$ $mod$ $13$? Specifically the $24x + 32y$ to $-2x + 6y$ part. – Crystal McMillian Oct 10 '19 at 19:06
Write $24x$ as $26x-2x$, $32y$ as $26y+6y$ and $40$ as $39+1$ – Oct 10 '19 at 19:08
@Crystal Mcmilliam,Do you see it now ? – Oct 10 '19 at 19:09
Yes, thank you for the clarification. – Crystal McMillian Oct 10 '19 at 19:21

Bill Dubuque · Answer 2 · 2020-04-10T19:25:18.310

Cramer's Rule works since the determinant $= 7\,$ is invertible, being coprime to the modulus $13$

$$\begin{align} \left[\begin{matrix} 5 &\!\!\! -4 \\ -2 &\! 3 \end{matrix}\right]\ \times\ &\left\{\, \left[\begin{matrix} 3 & 4 \\ 2 & 5 \end{matrix}\right] \left[\begin{matrix} x \\ y \end{matrix}\right] \equiv \left[\begin{matrix} 5 \\ 7 \end{matrix}\right]\,\right\}\pmod{\!13}\\[.8em] \Longrightarrow &\ \ \ \left[\begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix}\right] \left[\begin{matrix} x \\ y \end{matrix}\right] \equiv \left[\begin{matrix} 10 \\ 11 \end{matrix}\right] \end{align}\qquad$$

$$\begin{align}{\rm Therefore}\ \ \ &7x\equiv 10 \iff x\equiv \dfrac{10}7\equiv \dfrac{20}{14}\equiv \dfrac{7}1\\[.5em] &7y\equiv\ 11 \iff y\equiv \dfrac{11}7\equiv \dfrac{22}{14}\equiv \dfrac{9}1\end{align}\!\! $$

We evaluated the modular fractions using Gauss's algorithm. – Bill Dubuque Oct 10 '19 at 21:58 — Bill Dubuque, Oct 10 '19 at 21:58

Problem with solving systems of linear congruence of two variables

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