Chapter 9.7, Problem 2OE

To determine

To calculate: To find the number of real solutions of system of equations

Answer to Problem 2OE

Number of real solutions is $0$

Explanation of Solution

Given information: System of equations $x^2+y^2=4;x+y=8$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}{x}^2+y^2=4----(1)\\ x+y=8-----(2)\end{array}$

From equation (2)

  $\begin{array}{l}x+y=8\\ y=8-x\end{array}$

Substituting the value in (1):

  $\begin{array}{l}{x}^2+y^2=4\\ x^2+{\left(8-x\right)}^2=4\\ x^2+64+x^2-16x=4\\ 2x^2-16x+60=0\\ x^2-8x+30=0\\ x^2-8x+30=0\end{array}$

Here, both solutions of $x$ are imaginary

Thus, no real solutions of system of equations

Conclusion:

Hence, number of real solutions is $0$