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### Section 7-5 : Nonlinear Systems

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

\[\begin{align*}y & = 1 - 3x\\ \frac{{{x^2}}}{4} + {y^2} & = 1\end{align*}\]Show All Steps Hide All Steps

Start SolutionBefore 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 the first equation is in the form “\(y\) =” and so we can just plug this directly into the second equation. Doing this gives,

\[\frac{{{x^2}}}{4} + {\left( {1 - 3x} \right)^2} = 1\]Be careful with the parenthesis when plugging the first equation in. We had \({y^2}\) and so we need to make sure and square the whole \(y\) form the first equation when we plugged that in. In other words, we need the parenthesis in there to make sure we deal with the exponent properly.

Show Step 2Now, this is just a quadratic equation (which admittedly needs some simplification) and by this point we should be able to solve that so here is the solution work for the quadratic.

\[\begin{align*}\frac{{{x^2}}}{4} + {\left( {1 - 3x} \right)^2} & = 1\\ \frac{{{x^2}}}{4} + 1 - 6x + 9{x^2} - 1 & = 0\\ \frac{{37}}{4}{x^2} - 6x & = 0\\ x\left( {\frac{{37}}{4}x - 6} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = 0,\,\,\,\,\,\,\frac{{37}}{4}x - 6 = 0\end{align*}\]From this we see we have two values of \(x\) for our solution : \(x = 0\) and \(x = \frac{{24}}{{37}}\) .

Show Step 3We now have two values of \(x\) and so all we need to do is plug into either of the original equations (the line would be easier) to determine the corresponding values of \(y\) for each \(x\).

\[\begin{align*} & x = 0\,\,\,\,:\,\,\,\,y = 1 - 3\left( 0 \right) = 1 & \Rightarrow \hspace{0.25in} & \left( {0,1} \right)\\ & x = \frac{{24}}{{37}}\,\,:\,\,\,\,y = 1 - 3\left( {\frac{{24}}{{37}}} \right) = - \frac{{35}}{{37}} & \Rightarrow \hspace{0.25in} & \left( {\frac{{24}}{{37}}, - \frac{{35}}{{37}}} \right)\end{align*}\]

Note that with this system we have also run into a potential problem. We found corresponding \(y\)’s by plugging our \(x\)’s into the line. What if we had plugged them into the ellipse?

Let’s use \(x = 0\) as an example since it will be a little easier to do the work for. Plugging this into the equation for the ellipse gives,

\[\frac{{{{\left( 0 \right)}^2}}}{4} + {y^2} = 1\,\,\,\,\, \to \,\,\,\,\,\,{y^2} = 1\,\,\,\,\,\,\, \to \,\,\,\,\,\,y = \pm 1\]This implies that there should be two solutions corresponding to \(x = 0\) and not the single solution we found above! So, which is correct? Well recall that whatever solution we get must satisfy ** both** equations and only one of those values of \(y\) will satisfy the line when using \(x = 0\). Therefore, the only solution is the one we got from the line. We would have a similar issue with the second value of \(x\).

The problem arose when we plugged the line into the ellipse and squared it. The process of squaring it introduced potentially “bad” solutions. We saw similar issues when we solved equations with radicals several chapters back and the problem arose there for the same reason, we squared something.

The nice thing about these problems however is that if we use the equation we plugged in (the line in this case) to find the second values we don’t need to worry about the “bad” solutions since they only arise from the equation that we plugged into (the ellipse in this case).

This in fact is the real reason we used the line to find the corresponding \(y\), although it was also the easier of the two equations to use.

So, for this system of equations we have two solutions : \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {0,1} \right)}}\) and \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\frac{{24}}{{37}}, - \frac{{35}}{{37}}} \right)}}\) .