An alternative way of solving this question is as follows:
The system of equations that have been given to us is:
$$(a*x)+(a*y)=c \tag1$$
$$(b*x)+(c*y)=c\tag2$$
Now we consult the first to get those entries whose value is $c$. So we have
$$a+c=c\tag3$$
$$b+c=c\tag4$$
$$c+b=c\tag5$$
Now we need to choose $x,y$ in such a manner that the LHS of the equations in $(1)$ and $(2)$ matches with the LHS of the equations given in $(3),(4),(5)$.
Let us make 3 tables from the right table:
Table 1
$$\begin{array}{|c|c|} \hline a*a&a\\
\hline
b*c&a\\
\hline
\end{array}$$
Table 2
$$\begin{array}{|c|c|} \hline c*c&b\\
\hline
a*b&b\\
\hline b*a&b\\
\hline \end{array}$$
Table 3
$$\begin{array}{|c|c|} \hline c*b&c\\
\hline b*b&c\\
\hline a*c&c\\
\hline c*a&c\\ \hline\end{array}$$
Trying match $(1)$ of the form $(3)$ we see from table 1 and 3, $x=a, y=c$ but this when put into $(2)$ we get $b+b=c$ which is not present in $(3)$ to $(5)$. So $x=a,y=c$ is not possible.
So we shall continue in this manner, first trying to make equation $(1)$ like the identities $(3)$ to $(5)$ and then checking the corresponding values of $x,y$ in equation $(2)$.
Then we shall continue, trying to make equation $(2)$ like the identities $(3)$ to $(5)$ and then checking the corresponding values of $x,y$ in equation $(1)$.