Answer
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$\left( -\sqrt{\frac{31}{16}},\frac{7}{4} \right)$ .
Work Step by Step
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.
