Answer
The system of equations has a unique solution and the solution is $\left\{ \left( \frac{41}{7},\frac{36}{7} \right) \right\}$.
Work Step by Step
Let us consider the system of equations:
$4x-3y=8$
$2x-5y=-14$
And for 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, ${{a}_{1}}=4,\text{ }{{a}_{2}}=2,\text{ }{{b}_{1}}=-3\text{ and }{{b}_{2}}=-5$.
Therefore,
$\frac{4}{2}\ne \frac{-3}{-5}$
Or,
$2\ne \frac{3}{5}$
Thus the system of equations has a unique solution.
Multiply the first equation $2x-5y=-14$ by 2 to obtain the coefficient of x equal to 4, so that variable x is eliminated, $-4x+10y=28$
Add equation $4x=3y+8\text{ and }2x=-14+5y$:
$\begin{align}
& 4x-3y-4x+10y=8+28 \\
& 7y=36 \\
& y=\frac{36}{7}
\end{align}$
Put the value of $y=\frac{36}{7}$ in equation $2x-5y=-14$ as follows:
$\begin{align}
& 2x-5y=-14 \\
& 2x-5\left( \frac{36}{7} \right)=-14 \\
& 2x-\frac{180}{7}=-14 \\
& 2x=-14+\frac{180}{7}
\end{align}$
So, further simplify it to get:
$\begin{align}
& 2x=-14+\frac{180}{7} \\
& =\frac{-98+180}{7} \\
& =\frac{82}{7}
\end{align}$
It implies that:
$\begin{align}
& x=\frac{82}{7\times 2} \\
& =\frac{82}{14} \\
& =\frac{41}{7}
\end{align}$
It is, $x=\frac{41}{7}$
Now, verify the solution by substituting $x=\frac{41}{7}$ and $y=\frac{36}{7}$ into the equation $4x=3y+8$ as follows:
$\begin{matrix}
4x=3y+8 \\
4\left( \frac{41}{7} \right)\overset{?}{\mathop{=}}\,3\left( \frac{36}{7} \right)+8 \\
\frac{164}{7}\overset{?}{\mathop{=}}\,\frac{108}{7}+8 \\
\frac{164}{7}=\frac{164}{7} \\
\end{matrix}$
It is true.
Now, put $x=\frac{41}{7}$ and $y=\frac{36}{7}$ in the equation $2x-5y=-14$, to verify it as follows:
$\begin{align}
& 2x-5y=-14 \\
& 2\left( \frac{41}{7} \right)-5\left( \frac{36}{7} \right)=-14 \\
& \frac{82}{7}-\frac{180}{7}=-14 \\
& \frac{-98}{7}=-14
\end{align}$
The result comes out to be:
$-14=-14$
So, both equation (I) and (II) are satisfied by the solution, $x=\frac{41}{7}$ and $y=\frac{36}{7}$.
Hence, the system of equations has a unique solution and the solution is $\left\{ \left( \frac{41}{7},\frac{36}{7} \right) \right\}$.
