Answer
The solution set of the equations is $\left\{ \left( -2,1 \right),\left( -1,2 \right) \right\}$.
Work Step by Step
Consider the pair of equations, $\left\{ \begin{align}
& x-y=-3 \\
& {{x}^{2}}+{{y}^{2}}=5
\end{align} \right.$
Consider $x-y=-3$ and solve for x in terms of y.
$x=y-3$
Now, substitute the value of x in the equation ${{x}^{2}}+{{y}^{2}}=5$.
${{\left( y-3 \right)}^{2}}+{{y}^{2}}=5$
Factorize the equation using mathematical operations as follows:
$\begin{align}
& {{y}^{2}}+9-6y+{{y}^{2}}=5 \\
& 2{{y}^{2}}-6y+4=0 \\
& {{y}^{2}}-3y+2=0 \\
& {{y}^{2}}-2y-y+2=0
\end{align}$
Solve this further as,
$\left( y-2 \right)\left( y-1 \right)=0$
Put the factors $\left( y-2 \right),\left( y-1 \right)$ equal to zero and solve for y:
$\begin{align}
& \left( y-2 \right)=0 \\
& y=2
\end{align}$
Or
$\begin{align}
& \left( y-1 \right)=0 \\
& y=1
\end{align}$
Substitute the value of $y=2,y=1$ in the equation $x=y-3$ to obtain the values of x.
For $y=2$
$\begin{align}
& x=2-3 \\
& =-1
\end{align}$
For $y=1$
$\begin{align}
& x=1-3 \\
& =-2
\end{align}$
From here, the solution set is $\left( -2,1 \right)$ and $\left( -1,2 \right)$
Now substitute the values of x and y obtained in the equations $\left\{ \begin{align}
& x-y=-3 \\
& {{x}^{2}}+{{y}^{2}}=5
\end{align} \right.$ to check if they satisfy the equation.
Substitute $\left( -2,1 \right)$ in $x-y=-3$
$\begin{align}
-2-\left( 1 \right)\overset{?}{\mathop{=}}\,-3 & \\
-3=-3 & \\
\end{align}$
Which is true.
So, $\left( -2,1 \right)$ satisfy the equation $x-y=-3$
Substitute $\left( -2,1 \right)$ in ${{x}^{2}}+{{y}^{2}}=5$
$\begin{align}
{{\left( -2 \right)}^{2}}+{{1}^{2}}\overset{?}{\mathop{=}}\,5 & \\
4+1\overset{?}{\mathop{=}}\,5 & \\
5=5 & \\
\end{align}$
Which is true.
So, $\left( -2,1 \right)$ satisfy the equation ${{x}^{2}}+{{y}^{2}}=5$
Substitute $\left( -1,2 \right)$ in $x-y=-3$
$\begin{align}
-1-2\overset{?}{\mathop{=}}\,-3 & \\
-3=-3 & \\
\end{align}$
Which is true.
So, $\left( -1,2 \right)$ satisfy the equation $x-y=-3$
Substitute $\left( -1,2 \right)$ in ${{x}^{2}}+{{y}^{2}}=5$
$\begin{align}
{{\left( -1 \right)}^{2}}+{{2}^{2}}\overset{?}{\mathop{=}}\,5 & \\
1+4\overset{?}{\mathop{=}}\,5 & \\
5=5 & \\
\end{align}$
Which is true.
So, $\left( -1,2 \right)$ satisfy the equation ${{x}^{2}}+{{y}^{2}}=5$
Therefore, the solution set of the system of linear equations is $\left\{ \left( -2,1 \right),\left( -1,2 \right) \right\}$.
