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|√((\ldots))\\\\y=-√(8)\qquad\vee\qquad y=√(8)\\\\\boxed{y=-2√(2)\qquad\vee\qquad y=2√(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|√((\ldots))\\\\ x=-√(92)\qquad\vee\qquad x=√(92)\\\\x=-√(4\cdot23)\qquad\vee\qquad x=√(4\cdot23)\\\\\boxed{x=-2√(23)\qquad\vee\qquad x=2√(23)}`

As we see there are 4 such pairs:


`x=-2√(23)\qquad y=-2√(2)\\\\x=2√(23)\qquad y=-2√(2)\\\\ x=-2√(23)\qquad y=2√(2)\\\\x=2√(23)\qquad y=2√(2)`