Answer
The points $\underline{\left( -14,-20 \right)}$ and $\underline{\left( 2,-4 \right)}$ are a solution of both equation.
Work Step by Step
The given set of equations is\[\begin{align}
& 2{{x}^{2}}-{{y}^{2}}=-8 \\
& x-y=6
\end{align}\].
Calculation:
The provided equations are:
\[2{{x}^{2}}-{{y}^{2}}=-8\] (I)
\[x-y=6\] (II)
Simplify as follows:
Step 1:
Solve one of the equations for one variable in terms of the other, solve equation (II) and find the value of$y$in the terms of$x$,
\[x-y=6\]
\[y=x-6\] (III)
Step 2:
Put the value of $y$ in equation (I) simplify as follows,
\[\begin{align}
& 2{{x}^{2}}-{{y}^{2}}=-8 \\
& 2{{x}^{2}}-{{\left( x-6 \right)}^{2}}=-8 \\
& 2{{x}^{2}}-\left( {{x}^{2}}+36-12x \right)=-8 \\
& {{x}^{2}}-36+12x=-8
\end{align}\]
Simplify further,
\[\begin{align}
& {{x}^{2}}-36+12x+8=0 \\
& {{x}^{2}}+12x-28=0
\end{align}\]
Apply quadratic formula in above equation,
${{x}_{1,2}}=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So,
\[\begin{align}
& {{x}_{1,2}}=\frac{-12\pm \sqrt{{{\left( 12 \right)}^{2}}-4\times 1\times -28}}{2} \\
& {{x}_{1,2}}=\frac{-12\pm \sqrt{256}}{2} \\
& {{x}_{1,2}}=\frac{-12\pm 16}{2} \\
& {{x}_{1,2}}=-6\pm 8
\end{align}\]
Simplify further,
\[\begin{align}
& {{x}_{1,2}}=-6+8,-6-8 \\
& {{x}_{1,2}}=2,-14 \\
\end{align}\]
So \[{{x}_{1,2}}=2,-14\] are solutions of this equation.
Step 3:
Put the value of $x$ in equation (III) to find the value of $y$ and simplify as follows,
When $x=2$,
\[\begin{align}
& y=\left( 2 \right)-6 \\
& y=-4 \\
\end{align}\]
And when $x=-14$,
\[\begin{align}
& y=\left( -14 \right)-6 \\
& y=-20 \\
\end{align}\]
Step 4:
Check the proposed solutions in both of the equations, now begin by checking\[\left( 2,-4 \right)\], and replace $x$ with \[2\] and $y$ with\[-4\].
\[\begin{align}
& x-y=6 \\
& 2-\left( -4 \right)=6 \\
& 6=6
\end{align}\]
$\text{True}$.
And,
\[\begin{align}
& 2{{x}^{2}}-{{y}^{2}}=-8 \\
& 2{{\left( 2 \right)}^{2}}-{{\left( -4 \right)}^{2}}=-8 \\
& 8-16=-8 \\
& -8=-8
\end{align}\]
$\text{True}$.
Now begin by checking\[\left( -14,-20 \right)\], and replace $x$ with \[-14\] and $y$ with \[-20\].
\[\begin{align}
& x-y=6 \\
& -14-\left( -20 \right)=6 \\
& -14+20=6 \\
& 6=6
\end{align}\]
Simplify further,
\[3=3\]
$\text{True}$
And,
\[\begin{align}
& 2{{x}^{2}}-{{y}^{2}}=-8 \\
& 2{{\left( -14 \right)}^{2}}-{{\left( -20 \right)}^{2}}=-8 \\
& 392-400=-8 \\
& -8=-8
\end{align}\]
Simplify further,
\[1=1\]
$\text{True}$.
Hence, the points \[\left( 2,-4 \right)\] and \[\left( -14,-20 \right)\] are a solution of the system.