Question

How do I solve this algebra two btw

Answer 1

$\bf \begin{cases} x^2+y^2=400\\ \boxed{y}=x-28 \end{cases}\qquad \stackrel{\textit{substituting on the 1st equation}}{x^2+\left( \boxed{x-28} \right)^2=400} \\\\\\ x^2+(x^2-56x+28^2)=400\implies x^2+x^2-56x+28^2=400 \\\\\\ 2x^2-56x+784=400\implies 2x^2-56x+384=0 \\\\\\ 2(x^2-28x+192)=0\implies x^2-28x+192=0 \\\\\\ (x-16)(x-12)=0\implies x = \begin{cases} 16\\ 12 \end{cases}$

now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then

$\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill$