Chapter 9.8, Problem 8WE

To determine

To calculate: To find the real solutions

Answer to Problem 8WE

Solutions are $\left(3,-2\right),\left(2,-3\right)$

Explanation of Solution

Given information: System of equations $xy+6=0;x-y=5$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}xy+6=0-----(1)\\ x-y=5------(2)\end{array}$

From equation (1),

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

Substituting the value in equation (2)

  $\begin{array}{l}x-\left(-\frac{6}{x}\right)=5\\ x+\frac{6}{x}=5\\ x^2+6=5x\\ x^2-5x+6=0\\ x^2-3x-2x+6=0\\ \left(x-3\right)\left(x-2\right)=0\\ x=3,2\end{array}$

When $x=3$

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

When $x=2$

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

Conclusion:

Hence, solutions are $\left(3,-2\right),\left(2,-3\right)$