Chapter 7.1, Problem 66E

To determine

Tofind:Thesolutions of the system of equations $\{\begin{array}{c}{x}^2+y^2=4\\ x-y=5\end{array}$ algebraically and graphically.

Answer to Problem 66E

There is no real point of intersection.

Explanation of Solution

Given information:

The given system of equations is $\{\begin{array}{c}{x}^2+y^2=4\\ x-y=5\end{array}$ .

Calculation:

Draw the graph for the equations $x^2+y^2=4$ and $x-y=5$ .

  

From the above graph, there is no real point of intersection.

Determine the solution of the system of equations algebraically.

By the revised equations of $\left(1\right)$ and $\left(2\right)$ .

  $\begin{array}{c}\pm \sqrt{4-x^2}=x-5\\ 4-x^2={\left(x-5\right)}^2\\ 2x^2-10x+21=0\\ x=\frac{10\pm \sqrt{68i}}{4}\end{array}$

Hence, there is no real point of intersection.

Therefore, there is no real point of intersection.