Exercise 18 Page 643

We want to solve the given system of equations. We can see that, in Equation (I), the $y\text{-}$variable can be isolated by dividing by $2$ both sides of the equation. Let's do this and then use the Substitution Method. The $y\text{-}$variable is isolated in Equation (I). This allows us to substitute $\frac{1}{2}x+5$ for $y$ in Equation (II). Notice that Equation (II) is a quadratic equation in terms of only the $x\text{-}$variable. We can substitute $a=1,$ $b=\text{-}8,$ and $c=\text{-}20$ into the Quadratic Formula. This result tells us that we have two solutions for $x.$ One solution uses the positive sign and the other the negative sign.

$x=\frac{8\pm 12}{2}$
$x_1=\frac{8+12}{2}$	$x_2=\frac{8-12}{2}$
$x_1=\frac{20}{2}$	$x_2=\frac{\text{-}4}{2}$
$x_1=10$	$x_2=\text{-}2$

Now, consider Equation (I). We can substitute $x=10$ and $x=\text{-}2$ into the above equation to find the values for $y.$ Let's start with $x=10.$ We found that $y=10$ when $x=10.$ Therefore, one solution to the system is the point $(10,10).$ To find the other solution, we will substitute $\text{-}2$ for $x$ in Equation (I) again. We found that $y=4$ when $x=\text{-}2.$ Therefore, our second solution is the point $(\text{-}2,4).$