Answer
The solution set is\[\left\{ \left( \frac{1}{a},3 \right),a\ne 0 \right\}\].
Work Step by Step
Multiply \[5\]on both sides to the equation \[ax+7y=22\]to get: \[5ax+35y=110\].
Subtract the second given equation to the above obtained equation from both RHS and LHS as follows:
\[\underline{\begin{align}
& 5ax+35y=110 \\
& -5ax-4y=-17
\end{align}}\]
\[\begin{align}
& \ \ \ \ \ \ \ \ \ \ 31y=93 \\
& y=3
\end{align}\]
Put \[y=3\]in\[ax+7y=22\] to get:
\[\begin{align}
& ax+7\left( 3 \right)=22 \\
& ax+21=22 \\
& ax=1 \\
& x=\frac{1}{a},\ \ \ \ \ \ \ \ \ \ a\ne 0
\end{align}\]
To verify put\[y=3\]and \[x=\frac{1}{a}\] in\[5ax+4y=17\] as follows:
\[\begin{align}
& 5a\left( \frac{1}{a} \right)+4\left( 3 \right)=17 \\
& 5+12=17 \\
& 17=17
\end{align}\]
Since RHS\[=\]LHS
Hence verified.
Hence, the solution set is, \[\left\{ \left( \frac{1}{a},3 \right),a\ne 0 \right\}\].
