Question

What's the answer? Plz explain.

Answer 1

To find the solutions to the system, we need to find exactly when one expression is equal to the other. We can do this by setting both of the right sides equal to each other, so that

$(x-5)(x+6)=x+6$

Subtracting x+6 from either side, we find

$(x-5)(x+6)-(x+6)=0$

factoring out an x+6:

$(x+6)(x-5-1)=(x+6)(x-6)=0$

$x=\pm6$

The x-coordinates of our solutions will therefore be 6 and -6. Since the question only asks for the x-coordinate of the midpoint, we simply need to find the number exactly halfway between 6 and -6, which is 0.

Answer 2

$\bf \begin{cases}y=(x-5)(x+6)\\y=x+6\end{cases}\qquad \stackrel{y}{(x-5)(x+6)}=\stackrel{y}{x+6}\\\\\\\stackrel{FOIL}{x^2+x-30}=x+6\implies x^2-36=0\implies x^2=36\\\\\\x=\pm\sqrt{36}\implies \boxed{x=\pm 6}\\\\\\\stackrel{\textit{substituting that \underline{x} in the second equation}}{y=\pm 6 + 6}\implies y=\begin{cases}12\\0\end{cases}\\\\\\\stackrel{solutions}{(6,12)\qquad \qquad (-6,0)}$

now, let's find the midpoint of those two solutions

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{6}~,~\stackrel{y_1}{12})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{0}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{-6+6}{2}~~,~~\cfrac{0+12}{2} \right)\implies (\stackrel{x}{0}~,~6)$