Exercise 7 Page 642

We want to solve the given system of equations. Let's use the Substitution Method. To do this, we will start by isolating $x^2$ in Equation (II). Now, $x^2$ is isolated in Equation (II). This allows us to substitute $y^2-8$ for $x^2$ in Equation (I). Notice that in Equation (I), we have a quadratic equation in terms of only the $y\text{-}$variable. We can substitute $a=1,$ $b=2,$ and $c=\text{-}15$ into the Quadratic Formula. This result tells us that we have two solutions for $y.$ One solution uses the positive sign and the other the negative sign.

$y=\frac{\text{-}2\pm 8}{2}$
$y_1=\frac{\text{-}2+8}{2}$	$y_2=\frac{\text{-}2-8}{2}$
$y_1=\frac{6}{2}$	$y_2=\frac{\text{-}10}{2}$
$y_1=3$	$y_2=\text{-}5$

Now, consider Equation (II). We can substitute $y=3$ and $y=\text{-}5$ into the above equation to find the values for $x.$ Let's start with $y=3.$ We found that $x=\pm 1$ when $y=3.$ Therefore, we found two solutions to the system, which are $(\text{-}1,3)$ and $(1,3).$ To find more solutions, we will substitute $\text{-}5$ for $y$ in Equation (II) again. We found that $x=\pm \sqrt{17}$ when $y=\text{-}5.$ We found two more solutions to the given system of equations. These are $\left(\text{-}\sqrt{17},\text{-}5\right)$ and $\left(\sqrt{17},5\right).$