[tex] \bf \begin{cases}
\cfrac{x^2}{4}+\cfrac{y^2}{16}=1\\\\
x^2+y^2=4
\end{cases}\\\\
-------------------------------\\\\
\cfrac{x^2}{4}+\cfrac{y^2}{16}=1\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{16}}{4x^2+y^2=16}\implies \boxed{y^2=16-4x^2}
\\\\\\
\stackrel{\textit{pluging in the above in the second equation}}{x^2+\boxed{16-4x^2}=4}\implies -3x^2=-12
\\\\\\
x^2=\cfrac{-12}{-3}\implies x^2=4\implies x=\pm\sqrt{4}\implies x\pm 2 [/tex]
and since now we know that "x" is, let's do some substitution on the second equation, either will do, but we'll use the second one, and plug in that "x" value.
[tex] \bf (\pm 2)^2+y^2=4\implies 4+y^2=4\implies y^2=0\implies y=\sqrt{0}\implies y=0\\\\
-------------------------------\\\\
~~~~~~~~~~\stackrel{solutions}{(\stackrel{x}{2},\stackrel{y}{0})\qquad \qquad (\stackrel{x}{-2},\stackrel{y}{0})} [/tex]
recall that a solution for a system of equations is where the graphs meet, or intersect, namely where their x,y pairs equal each other.
[tex] \bf \begin{cases}
\cfrac{x^2}{4}+\cfrac{y^2}{16}=1\\\\
x^2+y^2=4
\end{cases}\\\\
-------------------------------\\\\
\cfrac{x^2}{4}+\cfrac{y^2}{16}=1\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{16}}{4x^2+y^2=16}\implies \boxed{y^2=16-4x^2}
\\\\\\
\stackrel{\textit{pluging in the above in the second equation}}{x^2+\boxed{16-4x^2}=4}\implies -3x^2=-12
\\\\\\
x^2=\cfrac{-12}{-3}\implies x^2=4\implies x=\pm\sqrt{4}\implies x\pm 2 [/tex]
and since now we know that "x" is, let's do some substitution on the second equation, either will do, but we'll use the second one, and plug in that "x" value.
[tex] \bf (\pm 2)^2+y^2=4\implies 4+y^2=4\implies y^2=0\\\\\\ y=\sqrt{0}\implies y=0\\\\
-------------------------------\\\\
~~~~~~~~~~\stackrel{solutions}{(\stackrel{x}{2},\stackrel{y}{0})\qquad \qquad (\stackrel{x}{-2},\stackrel{y}{0})} [/tex]
recall that a solution for a system of equations is where the graphs meet, or intersect, namely where their x,y pairs equal each other.
