By the end of this section, you will be able to:
Halley’s Comet orbits the sun about once every 75 years. Its path can be considered to be a very elongated ellipse. Other comets follow similar paths in space. These orbital paths can be studied using systems of equations. These systems, however, are different from the ones we considered in the previous section because the equations are not linear.
Figure 1. Halley’s Comet (credit: “NASA Blueshift”/Flickr)
In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations.
A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form [latex]Ax+By+C=0[/latex]. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.
There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.
Figure 2 illustrates possible solution sets for a system of equations involving a parabola and a line.
Solve the system of equations.
Show SolutionSolve the first equation for [latex]x[/latex] and then substitute the resulting expression into the second equation.
Expand the equation and set it equal to zero.
Solving for [latex]y[/latex] gives [latex]y=2[/latex] and [latex]y=1[/latex]. Next, substitute each value for [latex]y[/latex] into the first equation to solve for [latex]x[/latex]. Always substitute the value into the linear equation to check for extraneous solutions.
[latex]\beginx-y=-1 \\ x-\left(2\right)=-1 \\ x=1 \\ \hfill \\ x-\left(1\right)=-1 \\ x=0 \end[/latex]
The solutions are [latex]\left(1,2\right)[/latex] and [latex]\left(0,1\right),\text<>[/latex] which can be verified by substituting these [latex]\left(x,y\right)[/latex] values into both of the original equations.
Yes, but because [latex]x[/latex] is squared in the second equation this could give us extraneous solutions for [latex]x[/latex].
This gives us the same value as in the solution.
[latex]\beginNotice that [latex]-1[/latex] is an extraneous solution.
Solve the given system of equations by substitution.
Show Solution[latex]\left(-\frac,\frac\right)[/latex] and [latex]\left(2,8\right)[/latex]
Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.
Figure 4 illustrates possible solution sets for a system of equations involving a circle and a line.
Find the intersection of the given circle and the given line by substitution.
Show SolutionOne of the equations has already been solved for [latex]y[/latex]. We will substitute [latex]y=3x - 5[/latex] into the equation for the circle.
Now, we factor and solve for [latex]x[/latex].
[latex]\begin10\left(^-3x+2\right)=0 \\ 10\left(x - 2\right)\left(x - 1\right)=0 \\ x=2 \\ x=1 \end[/latex]
Substitute the two x-values into the original linear equation to solve for [latex]y[/latex].
[latex]\beginy&=3\left(2\right)-5 \\ &=1\hfill \\ y&=3\left(1\right)-5 \\ &=-2 \end[/latex]
The line intersects the circle at [latex]\left(2,1\right)[/latex] and [latex]\left(1,-2\right)[/latex], which can be verified by substituting these [latex]\left(x,y\right)[/latex] values into both of the original equations.
Solve the system of nonlinear equations.
Show SolutionWe have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse.
Figure 6 illustrates possible solution sets for a system of equations involving a circle and an ellipse.
Solve the system of nonlinear equations.
[latex]\begin\hfill ^+^=26& \hfill \left(1\right)\\ \hfill 3^+25^=100& \hfill \left(2\right)\end[/latex]
Show SolutionLet’s begin by multiplying equation (1) by [latex]-3,\text<>[/latex] and adding it to equation (2).
After we add the two equations together, we solve for [latex]y[/latex].
Substitute [latex]y=\pm 1[/latex] into one of the equations and solve for [latex]x[/latex].
There are four solutions: [latex]\left(5,1\right),\left(-5,1\right),\left(5,-1\right),\text\left(-5,-1\right)[/latex].
Find the solution set for the given system of nonlinear equations.
Show SolutionAll of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality.
Recall that when the inequality is greater than, [latex]y>a[/latex], or less than, [latex]y[/latex] the graph is drawn with a dashed line. When the inequality is greater than or equal to, [latex]y\ge a,\text<>[/latex] or less than or equal to, [latex]y\le a,\text<>[/latex] the graph is drawn with a solid line. The graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade.
or equal to x squared minus 4 the parabola is solid, and the region above it is shaded. For y less than x squared minus 4 the parabola is dotted, and the region below it is shaded. For y less than or equal to x squared minus 4 the parabola is solid, and the region below it is shaded." width="975" height="469" />
Figure 8. (a) an example of [latex]y>a[/latex]; (b) an example of [latex]y\ge a[/latex]; (c) an example of [latex]y
Graph the inequality [latex]y>^+1[/latex]. First, graph the corresponding equation [latex]y=^+1[/latex]. Since [latex]y>^+1[/latex] has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points [latex]\left(0,2\right)[/latex] and [latex]\left(2,0\right)[/latex]. One point is clearly inside the parabola and the other point is clearly outside. The graph is shown in Figure 9. We can see that the solution set consists of all points inside the parabola, but not on the graph itself. Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region. Graph the given system of inequalities. These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable [latex]y[/latex] will be eliminated. Then we solve for [latex]x[/latex]. [latex]\begin x^−y&=0 \\ 2x^+y&=12 \\ \hline 3x^&=12 \\ \hfill \\ x^&=4 \\ x&=\pm 2\end[/latex] Substitute the x-values into one of the equations and solve for [latex]y[/latex]. The two points of intersection are [latex]\left(2,4\right)[/latex] and [latex]\left(-2,4\right)[/latex]. Notice that the equations can be rewritten as follows. Graph each inequality. The feasible region is the region between the two equations bounded by [latex]2^+y\le 12[/latex] on the top and [latex]^-y\le 0[/latex] on the bottom. Graph the given system of inequalities. Shade the area bounded by the two curves, above the quadratic and below the line. feasible region the solution to a system of nonlinear inequalities that is the region of the graph where the shaded regions of each inequality intersect nonlinear inequality an inequality containing a nonlinear expression system of nonlinear equations a system of equations containing at least one equation that is of degree larger than one system of nonlinear inequalities a system of two or more inequalities in two or more variables containing at least one inequality that is not linear 1. Explain whether a system of two nonlinear equations can have exactly two solutions. What about exactly three? If not, explain why not. If so, give an example of such a system, in graph form, and explain why your choice gives two or three answers. 2. When graphing an inequality, explain why we only need to test one point to determine whether an entire region is the solution? 3. When you graph a system of inequalities, will there always be a feasible region? If so, explain why. If not, give an example of a graph of inequalities that does not have a feasible region. Why does it not have a feasible region? 4. If you graph a revenue and cost function, explain how to determine in what regions there is profit. 5. If you perform your break-even analysis and there is more than one solution, explain how you would determine which x-values are profit and which are not. For the following exercises, solve the system of nonlinear equations using substitution. For the following exercises, solve the system of nonlinear equations using elimination. For the following exercises, use any method to solve the system of nonlinear equations. For the following exercises, use any method to solve the nonlinear system.How To: Given an inequality bounded by a parabola, sketch a graph.
Example 4: Graphing an Inequality for a Parabola
Graphing a System of Nonlinear Inequalities
How To: Given a system of nonlinear inequalities, sketch a graph.
Example 5: Graphing a System of Inequalities
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Key Concepts
(1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points.Glossary
Section 8.2 Homework Exercises