Chapter 7.1, Problem 59E

To determine

Tofind:The intersection point of the system of equations $\{\begin{array}{c}{x}^2+y^2=169\\ x^2-8y=104\end{array}$ using graphical utility and verify the result by checking them in the original system.

Answer to Problem 59E

The intersection point of the system of equations $\{\begin{array}{c}{x}^2+y^2=169\\ x^2-8y=104\end{array}$ are $\left(0,-13\right)$ , $\left(12,5\right)$ and $\left(-12,5\right)$ .

Explanation of Solution

Given information:

The given system of equations is $\{\begin{array}{c}{x}^2+y^2=169\\ x^2-8y=104\end{array}$ .

Calculation:

Draw the graph for the equations $x^2+y^2=169$ and $x^2-8y=104$ .

  

From the above graph, the intersection points are $\left(0,-13\right)$ , $\left(12,5\right)$ and $\left(-12,5\right)$ . Therefore, the solution of the system of equations $\{\begin{array}{c}{x}^2+y^2=169\\ x^2-8y=104\end{array}$ points are $\left(0,-13\right)$ , $\left(12,5\right)$ and $\left(-12,5\right)$ .

Verify the solution by substituting the values of intersection point in the system of equations.

Substitute $0$ for $x$ and $-13$ for $y$ in the first equation.

  $\begin{array}{c}{x}^2+y^2=169\\ {\left(0\right)}^2+{\left(-13\right)}^2=169\\ 169=169\end{array}$

Substitute $0$ for $x$ and $-13$ for $y$ in the second equation.

  $\begin{array}{c}{x}^2-8y=104\\ 0^2-8\left(-13\right)=104\\ 104=104\end{array}$

From the above, it is clear that both the equations are satisfied. So, the ordered pair $\left(0,-13\right)$ is a solution of the system of equations.

Substitute $12$ for $x$ and $5$ for $y$ in the first equation.

  $\begin{array}{c}{x}^2+y^2=169\\ {\left(12\right)}^2+{\left(5\right)}^2=169\\ 169=169\end{array}$

Substitute $12$ for $x$ and $5$ for $y$ in the second equation.

  $\begin{array}{c}{x}^2-8y=104\\ {12}^2-8\left(5\right)=104\\ 104=104\end{array}$

From the above, it is clear that both the equations are satisfied. So, the ordered pair $\left(12,5\right)$ is a solution of the system of equations.

Substitute $-12$ for $x$ and $5$ for $y$ in the first equation.

  $\begin{array}{c}{x}^2+y^2=169\\ {\left(-12\right)}^2+{\left(5\right)}^2=169\\ 169=169\end{array}$

Substitute $-12$ for $x$ and $5$ for $y$ in the second equation.

  $\begin{array}{c}{x}^2-8y=104\\ {\left(-12\right)}^2-8\left(5\right)=104\\ 104=104\end{array}$

From the above, it is clear that both the equations are satisfied. So, the ordered pair $\left(-12,5\right)$ . is a solution of the system of equations.

Therefore, the intersection points of the system of equations $\{\begin{array}{c}{x}^2+y^2=169\\ x^2-8y=104\end{array}$ are $\left(0,-13\right)$ , $\left(12,5\right)$ and $\left(-12,5\right)$ .