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Section 7-5 : Nonlinear Systems

3. Find the solution to the following system of equation.

\[\begin{align*}xy & = 4\\ \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} & = 1\end{align*}\]

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Start Solution

Before we get too far into the solution we first should mention that there is no one correct solution path to these. Many of these types of problems will have multiple paths that we can take to find the solution. However, regardless of the path we take the solution to the system will be the same.

Okay on to the problem. In this case we can solve the first equation for either \(x\) or \(y\) and plug this into the second equation. For no reason other than we had equations in \(x\) for the first two practice problems for this section we’ll solve the first equation for \(x\) and plug this into the second equation. The result will be an equation involving only \(y\)’s.

Here is that work.

\[\begin{align*}x = \frac{4}{y}\hspace{0.25in} \to \hspace{0.25in}\frac{{{{\left( {\frac{4}{y}} \right)}^2}}}{4} + \frac{{{y^2}}}{{25}} & = 1\\ \hspace{0.25in}\frac{{\frac{{16}}{{{y^2}}}}}{4} + \frac{{{y^2}}}{{25}} & = 1\\ \hspace{0.25in}\frac{4}{{{y^2}}} + \frac{{{y^2}}}{{25}} & = 1\end{align*}\] Show Step 2

Now, let’s multiply both sides of this by \(25{y^2}\) to clear denominators.

\[\begin{align*}\,100 + {y^4} & = 25{y^2}\\ {y^4} - 25{y^2} + 100 & = 0\end{align*}\] Show Step 3

This is quadratic in form so we can define \(u = {y^2}\) (and so \({u^2} = {\left( {{y^2}} \right)^2} = {y^4}\)). Using this substitution the equation becomes,

\[\begin{align*}{u^2} - 25u + 100 & = 0\\ \left( {u - 5} \right)\left( {u - 20} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}u = 5,\,\,\,\,\,\,\,u = 20\end{align*}\] Show Step 4

So, we got two values of \(u\) and each of these correspond to the following equation in terms of \(y\) (i.e. using the substitution above).

\[\begin{align*} & u = 5\,\,\,\,:\,\,\,\,\,\,{y^2} = 5 & \to \hspace{0.25in} & y = \pm \sqrt 5 \\ & u = 20:\,\,\,\,\,\,{y^2} = 20 & \to \hspace{0.25in} & y = \pm \sqrt {20} = \pm 2\sqrt 5 \end{align*}\]

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We have four values of \(y\) that we need to find corresponding values of \(x\) for. We’ll plug these into the first equation (much easier to plug these into that equation).

\[\begin{align*} & y = \sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{\sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( {\frac{4}{{\sqrt 5 }},\sqrt 5 } \right)\\ & y = - \sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{ - \sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( { - \frac{4}{{\sqrt 5 }}, - \sqrt 5 } \right)\\ & y = 2\sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{2\sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( {\frac{2}{{\sqrt 5 }},2\sqrt 5 } \right)\\ & y = - 2\sqrt 5 \,\,\,\,:\,\,\,\,x = \frac{4}{{ - 2\sqrt 5 }} & \Rightarrow \hspace{0.25in} & \left( { - \frac{2}{{\sqrt 5 }}, - 2\sqrt 5 } \right)\end{align*}\]

So, for this system of equations we have the four solutions listed above.