Chapter 16.3, Problem 5MRE

To determine

To evaluate: the real solution of the system.

Answer to Problem 5MRE

The solution of the given equations is $\left(5,3\right),\left(-5,-3\right)$ .

Explanation of Solution

Given:

A system of equation is given as

  $\begin{array}{l}xy-15=0\\ y^2-x^2=-16\end{array}$

Consider the system.

  $\begin{array}{l}xy-15=0\mathrm{......}(1)\\ y^2-x^2=-16\mathrm{......}(2)\end{array}$

Substituting the value of ‘ y’ from equation (1) in equation (2).

  $\begin{array}{l}xy=15\\ y=\frac{15}{x}\\ {\left(\frac{15}{x}\right)}^2-x^2=-16\\ 225-x^4=-16x^2\\ x^4-16x^2-225=0\\ x^4-25x^2+9x^2-225=0\\ x^2\left(x^2-25\right)+9\left(x^2-25\right)=0\\ \left(x^2+9\right)\left(x^2-25\right)=0\\ x^2=25\\ x=\pm 5,x^2\ne -9\end{array}$

When $x=5$ ,

  $\begin{array}{c}\left(5\right)y=15\\ y=\frac{15}{5}\\ y=3\end{array}$

When $x=-5$ ,

  $\begin{array}{c}\left(-5\right)y=15\\ y=\frac{15}{-5}\\ y=-3\end{array}$

So, the solution of the given equations is $\left(5,3\right),\left(-5,-3\right)$ .