Answer
The solution set is\[\left\{ \left( \frac{1}{2a},\frac{1}{b} \right),a\ne 0,b\ne 0 \right\}\].
Work Step by Step
Multiply \[5\]on both sides to the equation \[4ax+by=3\]to get: \[20ax+5by=15\].
Subtract the second given equation to the above obtained equation from both RHS and LHS as follows:
\[\underline{\begin{align}
& 20ax+5by=15 \\
& -6ax-5by=-8
\end{align}}\]
\[\begin{align}
& 14ax\ \ \ \ \ \ \ \ \ \ =7 \\
& x=\frac{1}{2a},\ \ \ \ \ \ a\ne 0
\end{align}\]
Put \[x=\frac{1}{2a}\]in\[4ax+by=3\] to get:
\[\begin{align}
& 4a\left( \frac{1}{2a} \right)+by=3 \\
& 2+by=3 \\
& by=1 \\
& y=\frac{1}{b},\ \ \ \ \ b\ne 0
\end{align}\]
To verify put\[y=\frac{1}{b}\]and \[x=\frac{1}{2a}\] in\[6ax+5by=8\]as follows:
\[\begin{align}
& 6a\left( \frac{1}{2a} \right)+5b\left( \frac{1}{b} \right)=8 \\
& 3+5=8 \\
& 8=8
\end{align}\]
RHS\[=\]LHS
Hence verified.
Hence, the solution set is\[\left\{ \left( \frac{1}{2a},\frac{1}{b} \right),a\ne 0,b\ne 0 \right\}\].
