How to solve congruence system

Question

What is the step how can I solve following system of congruences (that is one system):

• $7x-8y≡5 \pmod {11}$
• $2x+5y≡9 \pmod {11}$

Answer 1

Hint:

$5\cdot(1)+8\cdot(2)$

$$(35+16)x\equiv5\cdot5+9\cdot8\pmod{11}$$

$$-4x\equiv3-5\pmod{11}\iff2x\equiv1\iff x\equiv2^{-1}\equiv6$$

Answer 2

Adding the first equation with twice the second one we obtain

$$7x-8y+2(2x+5y)≡5+2(9) \implies 11x+2y\equiv23 \implies2y\equiv 1 \mod 11$$

$$\implies 6\cdot 2y\equiv 6\cdot 1 \implies 12y\equiv6\implies y\equiv 6 \mod 11$$

Then from the second equation

$$2x+5y≡9 \implies 6\cdot 2x+6\cdot 5y \equiv 6\cdot 9 \implies12x+4\equiv 10 \implies x\equiv 6 \mod 11$$

Answer 3

You can solve with gaussian elimination of matrix over $\mathbb{F}_{11}$. Noting that $-8\equiv 3\pmod{11}$ we get the following matrix:

$$ \left[ \begin{array}{cc|c} 2 & 5 & 9 \\ 7 & 3 & 5 \end{array} \right] \underset{R_{2}:2R_{1}+R_{2}}{\longrightarrow} \left[ \begin{array}{cc|c} 2 & 5 & 9 \\ 0 & 2 & 1 \end{array} \right] \underset{R_{1}:3R_{2}+R_{1}}{\longrightarrow} \left[ \begin{array}{cc|c} 2 & 0 & 1 \\ 0 & 2 & 1 \end{array} \right] $$

Now we want to divid the second by $2$, for that we need to know what is the multiplicative inverse of $2$ in $\mathbb{F}_{11}$, that is were looking for a positive integer $n\in\{n\in\mathbb{N}\mid n<11\}$ such that $2n\equiv 1 \pmod{11}$, easily enough we know it's $6$ so the solution is: $$ \left[ \begin{array}{cc|c} 1 & 0 & 6 \\ 0 & 1 & 6 \end{array} \right] $$ Or just $x=6,\,y=6$