Answer
The solutions of the stated algebraic equations are $\left( 3,2 \right),\left( 3,-2 \right),\left( -3,2 \right)$ and $\left( -3,-2 \right)$ .
Work Step by Step
Let us consider that the equations are as follows:
${{x}^{2}}-4{{y}^{2}}=-7$ (I)
$3{{x}^{2}}+{{y}^{2}}=31$ (II)
Now proceed with the steps as follows:
Step 1:
When necessary, multiply either equation or both equations by the appropriate number so that the sum of the coefficients of ${{x}^{2}}$ or the sum of the coefficients of ${{y}^{2}}$ is zero.
Thus, multiply equation (II) by $4$ so that the sum of the coefficients of ${{y}^{2}}$ becomes zero.
$\begin{align}
& {{x}^{2}}-4{{y}^{2}}+\left( 12{{x}^{2}}+4{{y}^{2}} \right)=-7+4(31) \\
& {{x}^{2}}-4{{y}^{2}}+\left( 12{{x}^{2}}+4{{y}^{2}} \right)=-7+124 \\
& 13{{x}^{2}}=117 \\
& {{x}^{2}}=9
\end{align}$
$\begin{align}
& {{x}^{2}}-9=0 \\
& (x-3)(x+3)=0
\end{align}$
It gives:
$\begin{align}
& x-3=0 \\
& x=3
\end{align}$
Or
$\begin{align}
& x+3=0 \\
& x=-3
\end{align}$
Therefore, $x=\pm 3$ are the two solutions of both equations.
Step 2:
Substitute the values of $x$ in equation (I) to find out the value of $y$ as given below,
When $x=+3\,or\,x=-3$
$\begin{align}
& {{x}^{2}}-4{{y}^{2}}=-7 \\
& {{\left( \pm 3 \right)}^{2}}-4{{y}^{2}}=-7 \\
& 9+7=4{{y}^{2}} \\
& 16=4{{y}^{2}}
\end{align}$
${{y}^{2}}=4$
$\begin{align}
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{y}^{2}}-4=0 \\
& (y-2)(y+2)=0 \\
& y=\pm 2
\end{align}$
Therefore, $y=\pm 2$ are the two solutions of this equation.
Step 3:
Verify the values of $x$ and $y$ in both of the equations:
Now start by taking the pair $\left( \pm 3,\pm 2 \right)$, and put $\,x=-3\text{or}-3$ and $y=+2\text{or}-2$.
$\begin{align}
& {{x}^{2}}-4{{y}^{2}}=-7 \\
& {{\left( 3 \right)}^{2}}-4\times {{\left( 2 \right)}^{2}}=-7 \\
& 9-16=-7 \\
& -7=-7
\end{align}$
True.
$\begin{align}
& {{x}^{2}}-4{{y}^{2}}=-7 \\
& {{\left( 3 \right)}^{2}}-4\times {{\left( -2 \right)}^{2}}=-7 \\
& 9-16=-7 \\
& -7=-7
\end{align}$
True.
$\begin{align}
& {{x}^{2}}-4{{y}^{2}}=-7 \\
& {{\left( -3 \right)}^{2}}-4\times {{\left( -2 \right)}^{2}}=-7 \\
& 9-16=-7 \\
& -7=-7
\end{align}$
True.
$\begin{align}
& {{x}^{2}}-4{{y}^{2}}=-7 \\
& {{\left( -3 \right)}^{2}}-4\times {{\left( 2 \right)}^{2}}=-7 \\
& 9-16=-7 \\
& -7=-7
\end{align}$
True.
And,
$\begin{align}
& 3{{x}^{2}}+{{y}^{2}}=31 \\
& 3\times {{\left( 3 \right)}^{2}}+{{\left( 2 \right)}^{2}}=31 \\
& 27+4=31 \\
& 31=31
\end{align}$
True.
$\begin{align}
& 3{{x}^{2}}+{{y}^{2}}=31 \\
& 3\times {{\left( 3 \right)}^{2}}+{{\left( -2 \right)}^{2}}=31 \\
& 27+4=31 \\
& 31=31
\end{align}$
True.
$\begin{align}
& 3{{x}^{2}}+{{y}^{2}}=31 \\
& 3\times {{\left( -3 \right)}^{2}}+{{\left( -2 \right)}^{2}}=31 \\
& 27+4=31 \\
& 31=31
\end{align}$
True.
$\begin{align}
& 3{{x}^{2}}+{{y}^{2}}=31 \\
& 3\times {{\left( -3 \right)}^{2}}+{{\left( 2 \right)}^{2}}=31 \\
& 27+4=31 \\
& 31=31
\end{align}$
True.
