1. Solving Nonlinear Systems
12-7
Solving Nonlinear Systems
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt Algebra 2

2. Warm Up Solve by substitution. Solve by elimination. 3x + 4y = 15 1.2.
5x – 4y = – 1

3. Objective
Solve systems of equations in two variables that contain at least one second-degree equation.

4. Vocabulary
nonlinear system of equations

5. A nonlinear system of equations is a system in which at least one of the equations is not linear. You have been studying one class of nonlinear equations, the conic sections.
The solution set of a system of equations is the set of points that make all of the equations in the system true, or where the graphs intersect. For systems of nonlinear equations, you must be aware of the number of possible solutions.

6. You can use your graphing calculator to find solutions to systems of nonlinear equations and to check algebraic solutions.

7. Example 1: Solving a Nonlinear System by Graphing
x2 + y2 = 25
Solve by graphing.
4x2 + 9y2 = 145
The graph of the first equation is a circle, and the graph of the second equation is an ellipse, so there may be as many as four points of intersection.

8. Example 1 Continued
Step 1 Solve each equation for y.
Solve the first equation for y.
Solve the second equation for y.
Step 2 Graph the system on your calculator, and use the intersect feature to find the solution set.
The points of intersection are (–4, –3), (–4, 3), (4, –3), (4, 3).

9. Check It Out! Example 1 3x + y = 4.5 Solve by graphing. y = (x – 3)2
The graph of the first equation is a straight line, and the graph of the second equation is a parabola, so there may be as many as two points of intersection.

10. Check It Out! Example 1 Continued
Step 1 Solve each equation for y.
y = –3x + 4.5
Solve the first equation for y.
y = (x – 3)2 1 2
Step 2 Graph the system on your calculator, and use the intersect feature to find the solution set.
The point of intersection is (0, 4.5).

11. The substitution method for solving linear systems can also be used to solve nonlinear systems algebraically.

12. Example 2: Solving a Nonlinear System by Substitution
x2 + y2 = 100
Solve by substitution.
y = x2 – 26 1 2
The graph of the first equation is a circle, and the graph of the second equation is a parabola, so there may be as many as four points of intersection.

13. Example 2 Continued
Step 1 It is simplest to solve for x2 because both equations have x2 terms.
Solve for x2 in the second equation.
x2 = 2y + 52
Step 2 Use substitution.
Substitute this value into the first equation.
(2y + 52) + y2 = 100
y2 + 2y – 48 = 0
Simplify, and set equal to 0.
(y + 8) (y – 6) = 0
Factor.
y = –8 or y = 6

14. Example 2 Continued
Step 3 Substitute –8 and 6 into x2 = 2y + 52 to find values of x.
x2 = 2(–8) + 52 = 36
x = ±6
(6, –8) and (–6, –8) are solutions.
x2 = 2(6) + 52 = 64
x = ±8
(8, 6) and (–8, 6) are solutions.
The solution set of the system is {(6, –8) (–6, –8), (8, 6), (–8, 6)}.

15. Example 2 Continued
Check Use a graphing calculator. The graph supports that there are four points of intersection.

16. Check It Out! Example 2a
Solve the system of equations by using the substitution method.
x + y = –1
x2 + y2 = 25
The graph of the first equation is a line, and the graph of the second equation is a circle, so there may be as many as two points of intersection.

17. Check It Out! Example 2a Continued
Step 1 Solve for x.
x = –y – 1
Solve for x in the first equation.
Step 2 Use substitution.
Substitute this value into the second equation.
(–y – 1)2 + y2 = 25
y2 + 2y y2 – 25 = 0
Simplify and set equal to 0.
2y2 +2y – 24 = 0
2[(y2 + y – 12)] = 0
Factor.
2(y + 4)(y – 3) = 0
y = –4 or y = 3

18. Check It Out! Example 2a Continued
Step 3 Substitute –4 and 3 into x + y = –1 to find values of x.
x + (–4) = –1
x = 3
(3, –4) is a solution.
x + (3) = –1
x = –4
(–4, 3) is a solution.
The solution set of the system is {(3, –4), (–4, 3)}.

19. Check It Out! Example 2a Continued
Check Use a graphing calculator. The graph supports that there are two points of intersection.

20. Check It Out! Example 2b
Solve the system of equations by using the substitution method.
x2 + y2 = 25
y – 5 = –x2
The graph of the first equation is a circle, and the graph of the second equation is a parabola, so there may be as many as four points of intersection.

21. Check It Out! Example 2b Continued
Step 1 It is simplest to solve for x2 because both equations have x2 terms.
Solve for x2 in the second equation.
x2 = –y + 5
Step 2 Use substitution.
Substitute this value into the first equation.
(–y + 5) + y2 = 25
y2 – y – 20 = 0
Simplify, and set equal to 0.
(y + 4) (y – 5) = 0
Factor.
y = –4 or y = 5

22. Check It Out! Example 2b Continued
Step 3 Substitute –4 and 5 into x2 + y2 = 25 to find values of x.
x2 + (–4)2 = 25
x = ±3
(3, –4) and (–3, –4) are solutions.
x2 +(5)2 = 25
x = 0
(0, 5) is a solution.
The solution set of the system is {(3, –4), (–3, –4), (0, 5)}.

23. Check It Out! Example 2b Continued
Check Use a graphing calculator. The graph supports that there are three points of intersection.

24. The elimination method can also be used to solve systems of nonlinear equations.
In Example 3, you can check your work on a graphing calculator.
Remember!

25. Example 3: Solving a Nonlinear System by Elimination
4x2 + 25y2 = 41
Solve by using the
elimination method.
36x2 + 25y2 = 169
The graph of the first equation is an ellipse, and the graph of the second equation is an ellipse, There may be as many as four points of intersection.

26. Example 3 Continued
Step 1 Eliminate y2.
36x2 + 25y2 = 169
Subtract the first equation from the second.
–4x2 – 25y2 = –41
32x = 128
x2 = 4, so x = ±2
Solve for x.

27. Example 3 Continued
Step 2 Find the values for y.
4(4) + 25y2 = 41
Substitute 4 for x2.
y2 = 41
Simplify.
25y2 = 25
y = ±1
The solution set of the system is {(–2, –1), (–2, 1), (2, –1), (2, 1)}.

28. Check It Out! Example 3
25x2 + 9y2 = 225
Solve by using the
elimination method.
25x2 – 16y2 = 400
The graph of the first equation is an ellipse, and the graph of the second equation is a hyperbola, There may be as many as four points of intersection.

29. Check It Out! Example 3 Continued
Step 1 Eliminate x2.
25x2 – 16y2 = 400
Subtract the first equation from the second.
–25x2 – 9y2 = –225
–25y2 = 175
y2 = –7
Solve for y.
There is no real solution of the system.

30. Understand the Problem
Example 4: Problem-Solving Application
Suppose that the paths of two boats are modeled by 36x2 + 25y2 = 900 and y = 0.25x2 – 6. How many possible collision points are there? 1
Understand the Problem
There is a potential danger of a collision if the two paths cross. The paths will cross if the graphs of the equations intersect.

31. Understand the Problem
Example 4 Continued 1
Understand the Problem
List the important information:
36x2 + 25y2 = 900 represents the path of the first boat.
y = 0.25x2 – 6 represents the path of the second boat.

32. Example 4 Continued 2
Make a Plan
To see if the graphs intersect, solve the system
36x2 + 25y2 = 900 .
y = 0.25x2 – 6

33. Example 4 Continued
Solve 3
The graph of the first equation is an ellipse, and the graph of the second equation is a parabola. There may be as many as four points of intersection.
Solve the second equation for x2.
x2 = 4y + 24
36(4y + 24) + 25y2 = 900
Substitute this value into the first equation.
25y y – 36 = 0
Simplify, and set equal to 0.
Use the quadratic formula.

34. Example 4 Continued
Substitute y = 0.24 and y = –6 into x2 = 4y + 24 to find the values for x.
x2 = 4(0.24) + 24
x2 = 4(–6) + 24
x  ±5
x = 0
There are three real solutions to the system, (0, –6), (5, 0.24), and (–5, 0.24), so there are 3 possible collision points.

35. Example 4 Continued
Look Back 4
The graph supports that there are three points of intersection. Because the paths intersect, the boats are in danger of colliding if they arrive at one of the three points of intersection at the same time.

36. What if …? Suppose the paths of the boats can be modeled by the system
Check It Out! Example 4
What if …? Suppose the paths of the boats can be modeled by the system .
Is there any danger of collision?
36x2 + 25y2 = 900
y + 2 = – x2 1 10

37. Understand the Problem
Check It Out! Example 4 Continued 1
Understand the Problem
There is a potential danger of a collision if the two paths cross. The paths will cross if the graphs of the equations intersect.
List the important information:
36x2 + 25y2 = 900 represents the path of the first boat.
represents the path of the second boat.
y + 2 = – x2 1 10

38. Check It Out! Example 4 Continued2
Make a Plan
To see if the graphs intersect, solve the system
36x2 + 25y2 = 900 .
y + 2 = – x2 1 10

39. Check It Out! Example 4 Continued
Solve 3
The graph of the first equation is an ellipse, and the graph of the second equation is a parabola. There may be as many as four points of intersection.
Solve the second equation for x2.
x2 = –10y – 20
36(–10y – 20) + 25y2 = 900
Substitute this value into the first equation.
25y2 – 360y – 1620 = 0
Use the quadratic formula.

40. Check It Out! Example 4 Continued
Substitute y = 18 and y = –3.6 into x2 = –10y – 20 to find the values for x.
x2 = –10(18) – 20
x2 = –10(–3.6) – 20
x2 = –200
There are no real values.
x = ±4
There are two real solutions to the system, (4, –3.6) and (–4, –3.6), so there are 2 possible collision points.

41. Check It Out! Example 4 Continued
Look Back 4
The graph supports that there are two points of intersection. Because the paths intersect, the boats are in danger of colliding if they arrive at the intersections (4, –3.6) or (–4, –3.6) at the same time.

42. Lesson Quiz
4x2 – 9y2 = 108
1. Solve by graphing.
(±6, ±2)
x2 + y2 = 40
2x2 + y2 = 54
2. Solve by substitution.
(±5, ±2)
x2 – 3y2 = 13
4x2 + 4y2 = 52
3. Solve by elimination.
(±3, ±2)
9x2 – 4y2 = 65