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Hint: We first find the nature of system of linear equations, if the nature of system of equations are intersecting lines then we will have a unique solution, if the nature of system of equations are coincident then we will have infinite number of solutions, if the nature of system of equations are parallel then there will be no solutions.

Complete step-by-step answer:

First, we will find the nature of a pair of linear equations.

If ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ are pair of linear equations in two variables

If $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then lines are intersecting.

If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ then lines are parallel.

If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ then lines are coincident.

Pair of linear equations are $3=2x+y\cdot \cdot \cdot \cdot \cdot (1)$ and $9=4x-y\cdot \cdot \cdot \cdot \cdot (2)$

Now, we will find the nature of linear equations.

$\begin{align}

& {{a}_{1}}=2,{{b}_{1}}=1,{{c}_{1}}=-3 \\

& {{a}_{2}}=4,{{b}_{2}}=-1,{{c}_{2}}=9 \\

\end{align}$$$$$

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{2}{4}=\dfrac{1}{2}$ and $\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{1}{-1}=-1$

Here $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then lines are intersecting.

So, it has a unique solution.

So, now we will solve for x and y.

We will add both equations (1) and (2)

$\Rightarrow 12=6x$

$\Rightarrow x=2$

Now we will substitute the obtained value of $x$ in either equation 1 or in equation 2. Now we will substitute value of $x$ i.e. 2 in equation 1 we will get,

$\begin{align}

& \Rightarrow 3=2\times (2)+y \\

& \Rightarrow y=3-4 \\

& \Rightarrow y=-1 \\

\end{align}$

We obtained the value of $x$ and $y$ by solving equation 1 and equation 2.

The values of $x$ and $y$ are 2 and -1 respectively.

$x=2,y=-1$.

Note: While solving a pair of linear equations in two variables first we will find the nature of a pair of linear equations in two variables and then, we proceed to solve if lines are not parallel.

Complete step-by-step answer:

First, we will find the nature of a pair of linear equations.

If ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ are pair of linear equations in two variables

If $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then lines are intersecting.

If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$ then lines are parallel.

If $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$ then lines are coincident.

Pair of linear equations are $3=2x+y\cdot \cdot \cdot \cdot \cdot (1)$ and $9=4x-y\cdot \cdot \cdot \cdot \cdot (2)$

Now, we will find the nature of linear equations.

$\begin{align}

& {{a}_{1}}=2,{{b}_{1}}=1,{{c}_{1}}=-3 \\

& {{a}_{2}}=4,{{b}_{2}}=-1,{{c}_{2}}=9 \\

\end{align}$$$$$

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{2}{4}=\dfrac{1}{2}$ and $\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{1}{-1}=-1$

Here $\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}$ then lines are intersecting.

So, it has a unique solution.

So, now we will solve for x and y.

We will add both equations (1) and (2)

$\Rightarrow 12=6x$

$\Rightarrow x=2$

Now we will substitute the obtained value of $x$ in either equation 1 or in equation 2. Now we will substitute value of $x$ i.e. 2 in equation 1 we will get,

$\begin{align}

& \Rightarrow 3=2\times (2)+y \\

& \Rightarrow y=3-4 \\

& \Rightarrow y=-1 \\

\end{align}$

We obtained the value of $x$ and $y$ by solving equation 1 and equation 2.

The values of $x$ and $y$ are 2 and -1 respectively.

$x=2,y=-1$.

Note: While solving a pair of linear equations in two variables first we will find the nature of a pair of linear equations in two variables and then, we proceed to solve if lines are not parallel.

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