Intermediate Algebra for College Students (7th Edition)

The points $\underline{\left( -14,-20 \right)}$ and $\underline{\left( 2,-4 \right)}$ are a solution of both equation.
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.