## Precalculus (6th Edition) Blitzer

The solutions of the stated algebraic equations are $\left( 3,2 \right),\left( 3,-2 \right),\left( -3,2 \right)$ and $\left( -3,-2 \right)$ .
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.