Question

How many different ordered pairs satisfy both x^{2} + y^{2} = 100 and x^{2} + 2y^{2} = 108?

Answer 1

$\begin{cases}x^2+y^2=100\\x^2+2y^2=108\end{cases}\\\\\\ \begin{cases}x^2+y^2=100\\x^2+y^2+y^2=108\end{cases}\\\\\\100+y^2=108\\\\y^2=108-100\\\\y^2=8\qquad|\sqrt{(\ldots)}\\\\y=-\sqrt{8}\qquad\vee\qquad y=\sqrt{8}\\\\\boxed{y=-2\sqrt{2}\qquad\vee\qquad y=2\sqrt{2}}$

We know that $y^2=8$ so:

$x^2+y^2=100\\\\x^2+8=100\\\\x^2=100-8\\\\x^2=92\qquad|\sqrt{(\ldots)}\\\\ x=-\sqrt{92}\qquad\vee\qquad x=\sqrt{92}\\\\x=-\sqrt{4\cdot23}\qquad\vee\qquad x=\sqrt{4\cdot23}\\\\\boxed{x=-2\sqrt{23}\qquad\vee\qquad x=2\sqrt{23}}$

As we see there are 4 such pairs:

$x=-2\sqrt{23}\qquad y=-2\sqrt{2}\\\\x=2\sqrt{23}\qquad y=-2\sqrt{2}\\\\ x=-2\sqrt{23}\qquad y=2\sqrt{2}\\\\x=2\sqrt{23}\qquad y=2\sqrt{2}$