Chapter 9.8, Problem 19WE

To determine

To calculate: To find the real solutions

Answer to Problem 19WE

Solutions are $\left(-6,-1\right),\left(-3,-2\right),\left(2,3\right)$

Explanation of Solution

Given information: System of equations $y^2=x+7;xy=6$

Calculation:

Given the system of equations as follows:

  $\begin{array}{l}{y}^2=x+7----(1)\\ xy=6------(2)\end{array}$

From equation (2)

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

Substituting the value in equation (1)

  $\begin{array}{l}{y}^2=\frac{6}{y}+7\\ y^3=6+7y\\ y^3-7y-6=0\\ \left(y+1\right)\left(y+2\right)\left(y-3\right)=0\\ y=-1,-2,3\end{array}$

When $y=-1$

  $\begin{array}{l}x=\frac{6}{-1}\\ x=-6\end{array}$

When $y=-2$

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

When $y=3$

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

Conclusion:

Hence, solutions are $\left(-6,-1\right),\left(-3,-2\right),\left(2,3\right)$