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Section 7-5 : Nonlinear Systems

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

\[\begin{align*}y & = {x^2} + 6x - 8\\ y & = 4x + 7\end{align*}\]

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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 notice that both of the equations are in the form “\(y\) =”. This means that we can “substitute” \(y\) from one of the equations into the other. In these kinds of problems this is often called “setting the equations equal”.

So, setting the equations equal gives,

\[{x^2} + 6x - 8 = 4x + 7\] Show Step 2

Now, this is just a quadratic equation and by this point we should be able to solve that so here is the solution work for the quadratic.

\[\begin{align*}{x^2} + 6x - 8 & = 4x + 7\\ {x^2} + 2x - 15 & = 0\\ \left( {x - 3} \right)\left( {x + 5} \right) & = 0\hspace{0.25in} \to \hspace{0.25in}x = - 5,\,\,\,\,\,\,x = 3\end{align*}\] Show Step 3

We 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 \(x\).

\[\begin{align*} & x = - 5:\,\,\,\,y = 4\left( { - 5} \right) + 7 = - 13 & \Rightarrow \hspace{0.25in} & \left( { - 5, - 13} \right)\\ & x = 3\,\,\,\,:\,\,\,\,y = 4\left( 3 \right) + 7 = 19 & \Rightarrow \hspace{0.25in} & \left( {3,19} \right)\end{align*}\]

So, for this system of equations we have two solutions : \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( { - 5, - 13} \right)}}\) and \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {3,19} \right)}}\) .