## Precalculus (6th Edition) Blitzer

The solutions of the equations ${{x}^{2}}+{{y}^{2}}=5$ and ${{x}^{2}}+{{\left( y-8 \right)}^{2}}=41$ are $\left( \sqrt{\frac{31}{16}},\frac{7}{4} \right)$ and$\left( -\sqrt{\frac{31}{16}},\frac{7}{4} \right)$ .
Let us consider that the equations are as follows: ${{x}^{2}}+{{y}^{2}}=5$ (I) ${{x}^{2}}+{{\left( y-8 \right)}^{2}}=41$ (II) Now, demonstrate the following steps as follows: Step 1: By eliminating the variable ${{x}^{2}}$, multiply −1 in the above equation (I) by operating the addition method in equations (I) and (II) and simplify as given below: \begin{align} & -{{x}^{2}}-{{y}^{2}}+\left( {{x}^{2}}+{{\left( y-8 \right)}^{2}} \right)=-5+41 \\ & {{\left( y-8 \right)}^{2}}-{{y}^{2}}=36 \\ & {{y}^{2}}+64-16y-{{y}^{2}}=36 \\ & 28=16y \end{align} Simplify further, $y=\frac{7}{4}$ Therefore, $y=\frac{7}{4}$ is the simplified solution of both the equations. Step 2: Substitute the value of $y$ in the equation (I) to find out the value of $x$ and solve as given below: For $y=\frac{7}{4}$, \begin{align} & {{x}^{2}}+{{y}^{2}}=5 \\ & {{x}^{2}}+{{\left( \frac{7}{4} \right)}^{2}}=5 \\ & {{x}^{2}}+\frac{49}{16}=5 \\ & {{x}^{2}}=5-\frac{49}{16} \end{align} Simplify further, \begin{align} & {{x}^{2}}=\frac{80-49}{16} \\ & {{x}^{2}}=\frac{31}{16} \end{align} So, $x=\pm \sqrt{\frac{31}{16}}$ is the simplified solution of this equation. Step 3: And verify the values of $x$ and $y$ in both of the equations. Now start by taking the pair $\left( \sqrt{\frac{31}{16}},\frac{7}{4} \right)$, and put $x=\pm \sqrt{\frac{31}{16}}$ and $y=\frac{7}{4}$. \begin{align} & {{x}^{2}}+{{y}^{2}}=5 \\ & {{\left( \sqrt{\frac{31}{16}} \right)}^{2}}+{{\left( \frac{7}{4} \right)}^{2}}=5 \\ & \frac{31}{16}+\frac{49}{16}=5 \\ & \frac{80}{16}=5 \end{align} Simplify further, $5=5$ Thus, the values satisfy the equation. And, \begin{align} & {{x}^{2}}+{{\left( y-8 \right)}^{2}}=41 \\ & {{\left( \sqrt{\frac{31}{16}} \right)}^{2}}+{{\left( \frac{7}{4}-8 \right)}^{2}}=41 \\ & \frac{31}{16}+\frac{625}{16}=41 \\ & \frac{656}{16}=41 \end{align} Simplify further, $41=41$ Thus, the values satisfy the equation. Now, start the check by $\left( -\sqrt{\frac{31}{16}},\frac{7}{4} \right)$, and put $x=-\sqrt{\frac{31}{16}}$ and $y=\sqrt{\frac{7}{4}}$. \begin{align} & {{x}^{2}}+{{y}^{2}}=5 \\ & {{\left( -\sqrt{\frac{31}{16}} \right)}^{2}}+{{\left( \frac{7}{4} \right)}^{2}}=5 \\ & \frac{31}{16}+\frac{49}{16}=5 \\ & \frac{80}{16}=5 \end{align} Simplify further, $5=5$ Thus, the values satisfy the equation. And, \begin{align} & {{x}^{2}}+{{\left( y-8 \right)}^{2}}=41 \\ & {{\left( -\sqrt{\frac{31}{16}} \right)}^{2}}+{{\left( \frac{7}{4}-8 \right)}^{2}}=41 \\ & \frac{31}{16}+\frac{625}{16}=41 \\ & \frac{656}{16}=41 \end{align} Simplify further, $41=41$ Thus, the values satisfy the equation. Hence, the points $\left( \sqrt{\frac{31}{16}},\frac{7}{4} \right)$ and $\left( -\sqrt{\frac{31}{16}},\frac{7}{4} \right)$ are the solutions of the system.