Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 24

The solutions of the stated algebraic equations are$\left( \sqrt{5},\sqrt{2} \right),\left( \sqrt{5},-\sqrt{2} \right),\left( -\sqrt{5},\sqrt{2} \right)$ and$\left( -\sqrt{5},-\sqrt{2} \right)$.

Let us consider the equations are as given below: $16{{x}^{2}}-4{{y}^{2}}-72=0$ (I) ${{x}^{2}}-{{y}^{2}}-3=0$ (II) Now, demonstrate the following steps as given below: Step 1: If necessary, multiply either of the equations or both equations by a significant number so that the sum of the coefficients of ${{x}^{2}}$ or the sum of the coefficients of ${{y}^{2}}$ is zero. Thus, multiply the equation (II) by $4$ so that the addition of the coefficients of ${{y}^{2}}$ becomes zero. $\begin{align} & 16{{x}^{2}}-4{{y}^{2}}-72+\left( -4{{x}^{2}}+4{{y}^{2}}+12 \right)=0+0 \\ & 12{{x}^{2}}-60=0 \\ & 12{{x}^{2}}=60 \\ & {{x}^{2}}=5 \end{align}$ This gives: $\begin{align} & {{x}^{2}}-5=0 \\ & (x-\sqrt{5})(x+\sqrt{5})=0 \\ & x\pm \sqrt{5}=0 \\ \end{align}$ So, $x=\pm \sqrt{5}$ are the two solutions of both equations. Step 2: Substitute the value of $x$ in equation (I) to find out the value of $y$ as follows: For$x=+\sqrt{5}\,\text{or}\,-\sqrt{5}$: $\begin{align} & 16{{x}^{2}}-4{{y}^{2}}-72=0 \\ & 16{{\left( \pm \sqrt{5} \right)}^{2}}-4{{y}^{2}}-72=0 \\ & 16\times 5-4{{y}^{2}}-72=0 \\ & 80-4{{y}^{2}}-72=0 \end{align}$ It gives: $\begin{align} & 8-4{{y}^{2}}=0 \\ & 8=4{{y}^{2}} \\ & 2={{y}^{2}} \end{align}$ Therefore, $\begin{align} & {{y}^{2}}-2=0 \\ & (y-\sqrt{2})(y+\sqrt{2})=0 \\ & y\pm \sqrt{2}=0 \end{align}$ So, $y=\pm \sqrt{2}$ are the two solutions of this equation. Step 3: And verify the values of $x$ and $y$ in both of the equations: Now start by taking the pair $\left( \pm \sqrt{5},\pm \sqrt{2} \right)$, and put $x=+\sqrt{5}\text{or}-\sqrt{5}$ and $y=+\sqrt{2}\text{or}-\sqrt{2}$ . And, $\begin{align} & {{x}^{2}}-{{y}^{2}}-3=0 \\ & {{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}-3=0 \\ & 5-2-3=0 \\ & 0=0 \end{align}$ Is true. $\begin{align} & {{x}^{2}}-{{y}^{2}}-3=0 \\ & {{\left( \sqrt{5} \right)}^{2}}-{{\left( -\sqrt{2} \right)}^{2}}-3=0 \\ & 5-2-3=0 \\ & 0=0 \end{align}$ Is true $\begin{align} & {{x}^{2}}-{{y}^{2}}-3=0 \\ & {{\left( -\sqrt{5} \right)}^{2}}-{{\left( -\sqrt{2} \right)}^{2}}-3=0 \\ & 5-2-3=0 \\ & 0=0 \end{align}$ Is true $\begin{align} & {{x}^{2}}-{{y}^{2}}-3=0 \\ & {{\left( -\sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}-3=0 \\ & 5-2-3=0 \\ & 0=0 \end{align}$ Is true Hence, the points $\left( \sqrt{5},\sqrt{2} \right),\left( \sqrt{5},-\sqrt{2} \right),\left( -\sqrt{5},\sqrt{2} \right)$ and $\left( -\sqrt{5},-\sqrt{2} \right)$ are the required solutions of the system.