Okay, in this case we have a hyperbola (the first equation, although it isn’t in standard form) and a rational function (the second equation if we solved for \(y\)). As with the first example we can’t use elimination on this system so we will have to use substitution.

The best way is to solve the second equation for either \(x\) or \(y\). Either one will give us pretty much the same work so we’ll solve for \(y\) since that is probably the one that will make the equation look more like those that we’ve looked at in the past. In other words, the new equation will be in terms of \(x\) and that is the variable that we are used to seeing in equations.

\[y = \frac{2}{x}\]
\[\begin{align*}{x^2} - 2{\left( {\frac{2}{x}} \right)^2} & = 2\\ {x^2} - 2\frac{4}{{{x^2}}} & = 2\\ {x^2} - \frac{8}{{{x^2}}} & = 2\end{align*}\]

The first step towards solving this equation will be to multiply the whole thing by *x*^{2} to clear out the denominators.

\[\begin{align*}{x^4} - 8 & = 2{x^2}\\ {x^4} - 2{x^2} - 8 & = 0\end{align*}\]

Now, this is quadratic in form and we know how to solve those kinds of equations. If we define,

\[u = {x^2}\hspace{0.25in} \Rightarrow \hspace{0.25in}{u^2} = {\left( {{x^2}} \right)^2} = {x^4}\]

and the equation can be written as,

\[\begin{align*}{u^2} - 2u - 8 & = 0\\ \left( {u - 4} \right)\left( {u + 2} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}u = - 2,\,\,\,u = 4\end{align*}\]

In terms of \(x\) this means that we have the following,

\[\begin{align*}{x^2} & = 4\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \pm \,2\\ {x^2} & = - 2\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \pm \,\sqrt 2 \,\,i\end{align*}\]

So, we have four possible values of \(x\) and two of them are complex. To determine the values of \(y\) we can plug these into our substitution.

\[\begin{align*}x & = 2\hspace{0.25in} \Rightarrow \hspace{0.25in}y = \frac{2}{2} = 1\\ x & = - 2\hspace{0.25in} \Rightarrow \hspace{0.25in}y = \frac{2}{{ - 2}} = - 1\end{align*}\]
\[\begin{align*}x & = \sqrt 2 \,i\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}y = \frac{2}{{\sqrt 2 \,i}} = \frac{2}{{\sqrt 2 \,i}}\frac{i}{i} = \frac{{2i}}{{\sqrt 2 \,{i^2}}} = - \frac{{2i}}{{\sqrt 2 }}\\ x & = - \sqrt 2 \,i\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}y = - \frac{2}{{\sqrt 2 \,i}} = - \frac{2}{{\sqrt 2 \,i}}\frac{i}{i} = - \frac{{2i}}{{\sqrt 2 \,{i^2}}} = \frac{{2i}}{{\sqrt 2 }}\end{align*}\]

For the complex solutions, notice that we made sure the \(i\) was in the numerator. The for solutions are then,

\[\begin{align*}& x = 2,\,\,\,y = 1\hspace{0.25in}{\mbox{and}}\hspace{0.25in}x = - 2,y = - 1\hspace{0.25in}{\mbox{and}}\\ & x = \sqrt 2 \,i,\,\,\,y = - \frac{{2i}}{{\sqrt 2 }}\hspace{0.25in}{\mbox{and}}\hspace{0.25in}x = - \sqrt 2 \,i,\,\,\,y = \frac{{2i}}{{\sqrt 2 }}\end{align*}\]

Two of the solutions are real and so represent intersection points of the graphs of these two equations. The other two are complex solutions and while solutions will not represent intersection points of the curves.

For reference purposes, here is a sketch of the two curves.

Note that there are only two intersection points of these two graphs as suggested by the two real solutions. Complex solutions never represent intersections of two curves.