Chapter 9.8, Problem 20WE

To determine

To calculate: To find the real solutions

Answer to Problem 20WE

Solution is $\left(-2,-3\right)$

Explanation of Solution

Given information: System of equations $y=x^2-1;xy+6=0$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}y=x^2-1----(1)\\ xy+6=0----(2)\end{array}$

From equation (2)

  $\begin{array}{l}xy=-6\\ y=-\frac{6}{x}\end{array}$

Substituting the value in equation (1)

  $\begin{array}{l}-\frac{6}{x}=x^2-1\\ -6=x^3-x\\ x^3-x+6=0\\ \left(x+2\right)\left(x^2-2x+3\right)=0\\ x=-2,1+\sqrt{2}i,1-\sqrt{2}i\end{array}$

Thus, real roots is $y=-3$

When $y=-3$

  $\begin{array}{l}y=\frac{6}{-2}\\ y=-3\end{array}$

Conclusion:

Hence, solution is $\left(-2,-3\right)$