Learning ObjectivesIn this section, you will: Show
Halley’s Comet (Figure 1) 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. Solving a System of Nonlinear Equations Using SubstitutionA 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 Ax+By+C=0.Ax+By+C=0. 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. Intersection of a Parabola and a LineThere are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.
Possible Types of Solutions for Points of Intersection of a Parabola and a LineFigure 2 illustrates possible solution sets for a system of equations involving a parabola and a line.
Figure 2
How ToGiven a system of equations containing a line and a parabola, find the solution.
Example 1Solving a System of Nonlinear Equations Representing a Parabola and a LineSolve the system of equations. x−y=−1y=x2+1x−y=−1y=x2+1
Q&ACould we have substituted values for yy into the second equation to solve for xx in Example 1? Yes, but because xx is squared in the second equation this could give us extraneous solutions for x.x. For y=1y=1 y=x2+11=x2+1x2=0 x=±0=0y=x2+11=x 2+1x2=0x=±0=0 This gives us the same value as in the solution. For y=2y=2 y=x2+1 2=x2+1x2=1x=±1=±1y=x2+12=x2+1x2=1x=±1=±1 Notice that −1−1 is an extraneous solution.
Try It #1Solve the given system of equations by substitution. 3x-y=-22x2-y=0 3x-y=-22x2-y=0 Intersection of a Circle and a LineJust as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.
Possible Types of Solutions for the Points of Intersection of a Circle and a LineFigure 4 illustrates possible solution sets for a system of equations involving a circle and a line.
Figure 4
How ToGiven a system of equations containing a line and a circle, find the solution.
Example 2Finding the Intersection of a Circle and a Line by SubstitutionFind the intersection of the given circle and the given line by substitution. x2+y 2=5y=3x−5x2+y2=5 y=3x−5
Try It #2Solve the system of nonlinear equations. x2+y2=10x-3y=-10 x2+y2=10x-3y=-10 Solving a System of Nonlinear Equations Using EliminationWe 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.
Possible Types of Solutions for the Points of Intersection of a Circle and an EllipseFigure 6 illustrates possible solution sets for a system of equations involving a circle and an ellipse.
Figure 6
Example 3Solving a System of Nonlinear Equations Representing a Circle and an EllipseSolve the system of nonlinear equations. x2+y2 =26(1)3x2+25y2=100(2)x2+y2=26(1)3x2+25y2= 100(2)
Try It #3Find the solution set for the given system of nonlinear equations. 4x2+y2=13x2+y2 =104x2+y2=13x2+y2=10 Graphing a Nonlinear InequalityAll 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, y>a,y>a, or less than, y<a,y<a, the graph is drawn with a dashed line. When the inequality is greater than or equal to, y≥a,y≥a, or less than or equal to, y≤a, y≤a, 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. See Figure 8.
Figure 8 (a) an example of y>a;y>a; (b) an example of y≥a;y≥a; (c) an example of y<a;y<a; (d) an example of y≤ay≤a
How ToGiven an inequality bounded by a parabola, sketch a graph.
Example 4Graphing an Inequality for a ParabolaGraph the inequality y>x2+1.y>x2+1. Graphing a System of Nonlinear InequalitiesNow 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.
How ToGiven a system of nonlinear inequalities, sketch a graph.
Example 5Graphing a System of InequalitiesGraph the given system of inequalities. x2−y≤0 2x2+y≤12x2−y≤02x2+y ≤12
Try It #4Graph the given system of inequalities. y≥x2−1x−y≥− 1y≥x2−1x−y≥−1
7.3 Section ExercisesVerbal1. 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. AlgebraicFor the following exercises, solve the system of nonlinear equations using substitution. 6. x+y=4 x2+y2=9 x+y=4x 2+y2=9 7. y=x−3x2+y2=9 y=x−3x2+y2=9 8. y=xx2+y2=9 y=xx2+y2=9 9. y =−xx2+y2=9 y=−xx2+y2=9 10. x=2x2−y2=9 x=2x2−y2=9 For the following exercises, solve the system of nonlinear equations using elimination. 11. 4x2−9y2=36 4x2+9y2=364x2−9y2=36 4x2+9y2=36 12. x2+y2 =25x2−y2=1x2+y2=25x2−y2=1 13. 2 x2+4y2=42x2−4y2=25x−102x 2+4y2=42x2−4y2=25x−10 14. y2−x2=93x2+2y2=8y2−x2 =93x2+2y2=8 15. x2+y2+116=2500 y=2x2x2+y2+116=2500y=2x2 For the following exercises, use any method to solve the system of nonlinear equations. 16. −2x2+y=−5 6x−y=9−2x2+y=−5 6x−y=9 17. −x2+ y=2−x+y=2−x2+y=2−x+y =2 18. x2+y2=1 y=20x2−1x2+y2=1 y=20x2−1 19. x2+y2=1 y=−x2x2+y2=1 y=−x2 20. 2x3−x2=y y=12−x2x3−x2=y y=12−x 21. 9x2+25y2 =225(x−6)2+y2=19x2+25 y2=225(x−6)2+y2=1 22. x4−x2=y x2+y=0x4−x2=y x2+y=0 23. 2x3−x 2=y x2+y=02x3−x2=y x2+y=0 For the following exercises, use any method to solve the nonlinear system. 24. x2+y2=9 y=3−x2x2+y2=9 y=3−x2 25. x2− y2=9 x=3x2−y2=9 x=3 26. x2−y2=9 y=3x2 −y2=9 y=3 27. x2−y2=9 x−y=0x2−y2=9 x−y=0 28. −x2+y=2 −4x+y=−1−x2+y=2−4x+y= −1 29. −x2+y=2 2y=−x− x2+y=2 2y=−x 30. x2+y2=25x2−y2=36x2+y2=25 x2−y2=36 31. x2+y2=1 y2=x2x2+y2=1 y2=x2 32. 16x2−9y2+144=0 y2+x2=1616x2−9y2+144=0 y2+x2=16 33. 3x2−y2=12(x−1)2+ y2=1 3x2−y2=12(x −1)2+y2=1 34. 3x2−y2=12(x−1)2+y2=4 3x2−y2=12(x−1)2+y2=4 35. 3x2 −y2=12 x2+y2=163x2− y2=12 x2+y2=16 36. x2−y2−6x−4y−11=0 − x2+y2=5x2−y2−6x−4y−11=0 −x2+y2=5 37. x2+y2−6y =7 x2+y=1x2+y2−6y =7 x2+y=1 38. x2+y2=6 xy=1x2+y2=6 xy=1 GraphicalFor the following exercises, graph the inequality. 40. x2+y2<4x2+y2<4 For the following exercises, graph the system of inequalities. Label all points of intersection. 41. x2+y<1y>2xx2+y<1y>2x 42. x2+y<−5y>5x+10 x2+y<−5y>5x+10 43. x2+y2<253x 2−y2>12x2+y2<253x2−y2>12 44. x2−y2>−4x2+y2<12x2−y2>−4x2+y2<12 45. x2+3y2>163 x2−y2<1x2+3y2>163x2−y2<1 ExtensionsFor the following exercises, graph the inequality. 46. y≥ex y≤ln(x)+5y≥exy≤ln (x)+5 47. y≤−log(x)y≤exy≤−log(x) y≤ex For the following exercises, find the solutions to the nonlinear equations with two variables. 48. 4x2 +1y2=245x2−2y2+4=04x2 +1y2=245x2−2y2+4=0 49. 6x2−1y2=8 1x2−6y2=186x2−1y2=8 1x2−6y2=18 50. x2−xy+y2−2=0x+3y=4x2 −xy+y2−2=0x+3y=4 51. x2−xy−2y2 −6=0x2+y2=1x2−xy−2y2−6= 0x2+y2=1 52. x2+4xy−2y2−6=0x=y+2x2 +4xy−2y2−6=0x=y+2 TechnologyFor the following exercises, solve the system of inequalities. Use a calculator to graph the system to confirm the answer. 54. x2+y<3y>2xx2+y<3y>2x Real-World ApplicationsFor the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions. 55. Two numbers add up to 300. One number is twice the square of the other number. What are the numbers? 56. The squares of two numbers add to 360. The second number is half the value of the first number squared. What are the numbers? 57. A laptop company has discovered their cost and revenue functions for each day: C(x)=3 x2−10x+200C(x)=3x2−10x+200 and R(x)=−2 x2+100x+50.R(x)=−2x2+100x+50. If they want to make a profit, what is the range of laptops per day that they should produce? Round to the nearest number which would generate profit. 58. A cell phone company has the following cost and revenue functions: C(x)=8x2−600x+21,500C(x)=8x2−600x+21,500 and R(x)=−3x2+480x. R(x)=−3x2+480x. What is the range of cell phones they should produce each day so there is profit? Round to the nearest number that generates profit. |