Let us suppose we have following system
$${5x+7y}\equiv 3\pmod {9}.$$
$${6x+5y}\equiv 4\pmod {9}.$$
I need to solve this task using elimination, first let us eliminate $y$ term, for this let us multiply first equation by $5$ and second one by $7$, we will get
$${25x+35y}\equiv 6\pmod {9}.$$
$${42x+35y}\equiv 1\pmod {9}.$$
Now subtract, we will get $${17x}\equiv {-5}\pmod {9},$$
which is the same as $${17x}\equiv {4}\pmod {9}.$$
I found multiplicative inverse of $17$ and $9$ which is equal to $8$, so I got
\begin{align*} {17\cdot 8 x}&\equiv {4\cdot 8}\pmod {9}\\ {x}&\equiv {32}\pmod {9}, \end{align*}
and the last one reduces to
$${x}\equiv {5}\pmod {9}.$$
Now let us insert this equation to the first one
\begin{align*} {5\cdot 5+7y}&\equiv 3\pmod {9}\\ {7y}&\equiv {-22} \pmod {9}, \end{align*}
which is the same as $${y}\equiv {-88} \pmod {9}$$
or
$${y}\equiv {2} \pmod {9}.$$
So, my final solution is
$${x}\equiv {5}\pmod {9},$$
$${y}\equiv {2} \pmod {9}.$$
Am I right? Thanks in advance.
