Answer
The system of equations has a unique solution and the solution is $\left[ \left. \left( x,y \right) \right|\left( \frac{29}{22},-\frac{5}{11} \right) \right]$.
Work Step by Step
Let us consider system of equations:
$\begin{align}
& 2x-3y=4 \\
& 4x+5y=3 \\
\end{align}$
So, for the equations ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$.
If $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$, then the system of equations has a unique solution.
Here, ${{a}_{1}}=2,\text{ }{{a}_{2}}=4,\text{ }{{b}_{1}}=-3,\text{ and }{{b}_{2}}=5$.
Therefore,
$\begin{align}
& \frac{2}{4}\ne \frac{-3}{5} \\
& \frac{1}{2}\ne -\frac{3}{5}
\end{align}$
Therefore, the system of equations has a unique solution.
Multiply the first equation $2x-3y=4$ by -2 to obtain the coefficient of x equal to -4, so that the variable x is eliminated.
$-4x+6y=-8$
And add equations $2x=3y+4\text{ and }4x=3-5y$ to obtain:
$\begin{align}
& 4x+5y-4x+6y=3-8 \\
& 11y=-5 \\
& y=-\frac{5}{11}
\end{align}$
Put the value $y=-\frac{5}{11}$ y in the equation $y=-\frac{5}{11}$ to obtain:
$\begin{align}
& 4x-\frac{25}{11}=3 \\
& 4x=3+\frac{25}{11} \\
& 4x=\frac{58}{11} \\
& x=\frac{29}{22}
\end{align}$
