1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 5 Systems and Matrices Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2. 2 5.5 Nonlinear Systems of Equations Solving Nonlinear Systems with Real Solutions Solving Nonlinear Systems with Nonreal Complex Solutions Applying Nonlinear Systems

3. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Solving Nonlinear Systems with Real Solutions A system of equations in which at least one equation is not linear is called a nonlinear system. The substitution method works well for solving many such systems, particularly when one of the equations is linear, as in the next example.

4. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Example 1 SOLVING A NONLINEAR SYSTEM BY SUBSTITUTION Solve the system. When one of the equations in a nonlinear system is linear, it is usually best to begin by solving the linear equation for one of the variables. Solution (1) (2) Solve equation (2) for y. Substitute this result for y in equation (1).

5. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Example 1 SOLVING A NONLINEAR SYSTEM BY SUBSTITUTION (1) Let y = – 2 – x. Distributive property Standard form Factor. Zero factor property

6. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Example 1 SOLVING A NONLINEAR SYSTEM BY SUBSTITUTION Substituting – 2 for x in equation (2) gives y = 0. If x = 1, then y = – 3. The solution set of the given system is {(– 2, 0),(1, – 3)}.

7. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Visualizing Graphs Visualizing the types of graphs involved in a nonlinear system helps predict the possible numbers of ordered pairs of real numbers that may be in the solution set of the system.

8. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Example 2 SOLVING A NONLINEAR SYSTEM BY ELIMINATION Solve the system. Solution (1) (2) The graph of equation (1) is a circle and, as we will learn in the next chapter, the graph of equation (2) is a hyperbola. These graphs may intersect in 0, 1, 2, 3, or 4 points. We add to eliminate y 2.

9. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Example 2 SOLVING A NONLINEAR SYSTEM BY ELIMINATION (1) (2) Add. Divide by 3. Square root property Remember to find both square roots.

10. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 Example 2 SOLVING A NONLINEAR SYSTEM BY ELIMINATION Find y by substituting the values of x in either equation (1) or equation (2). Let x = 2.Let x = − 2. The proposed solutions are (2, 0) and (– 2, 0). These satisfy both equations, confirming that the solution set is {(2, 0), (– 2, 0)}.

11. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 Note The elimination method works with the system in Example 2 since the system can be thought of as a system of linear equations where the variables are x 2 and y 2. To see this, substitute u for x 2 and v for y 2. The resulting system is linear in u and v.

12. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Example 3 SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Solve the system. Solution (1) (2) (1) Multiply (2) by – 1 Add. (3) Solve for y (x ≠ 0). Begin as with the elimination method.

13. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 3 SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Let y = in (2). Now substitute for y in either equation (1) or equation (2). We use equation (2). Multiply and square. Multiply by x 2 to clear fractions.

14. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 3 SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Subtract 6x 2. This equation is quadratic in form. Factor. Zero factor property. Square root property; For each equation, include both square roots. Solve each equation.

15. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Example 3 SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Substitute these x-values into equation (4) to find corresponding values of y. Let

16. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Example 3 SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Verify these solutions by substitution in the original system. The solution set of the system is

17. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Example 4 SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION Solve the system. Solution (1) (2) Use the substitution method. Begin by solving equation (2) for  x . (3) In equation (1), the first term is x 2, which is the same as  x  2. Therefore, we substitute 4 – y for x in equation (1).

18. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Example 4 Let x = 4 – y. Square the binomial. Remember the middle term. Combine like terms. Factor. Zero factor property. SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION Divide by 2. Add 4.

19. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Example 4 To solve for the corresponding values of x, use either equation (1) or equation (2). Let y = 0.Let y = 4. SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION

20. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Example 4 The solution set, {(4,0), (– 4, 0), (0, 4)}, includes the points of intersection shown in the graph. Check the solutions in the original system. SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION

21. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Example 5 Solve the system. Solution (1) (2) Multiply (1) by – 3. (2) Add. Square root property SOLVING A NONLINEAR SYSTEM WITH NONREAL COMPLEX NUMBERS AS SOLUTIONS Begin by eliminating a variable.

22. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Example 5 SOLVING A NONLINEAR SYSTEM WITH NONREAL COMPLEX NUMBERS AS SOLUTIONS To find the corresponding values of y, substitute into equation (1). Let x = 2 i. Let x = – 2 i. i 2 = – 1

23. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Example 5 Checking the proposed solutions confirms the following solution set. Note that these solutions with nonreal complex number components do not appear as intersection points on the graph of the system. SOLVING A NONLINEAR SYSTEM WITH NONREAL COMPLEX NUMBERS AS SOLUTIONS

24. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX A box with an open top has a square base and four sides of equal height. The volume of the box is 75 in. 3, and the surface area is 85 in. 2. Find the dimensions of the box. Solution Step 1 Read the problem. We must find the box width, length, and height. Step 2 Assign variables. Let x represent the length and width of the square base, and let y represent the height.

25. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Step 3 Write a system of equations. Use the formula for the volume of a box, V = LWH, to write one equation using the given volume, 75 in. 3. Volume formula

26. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Sum of areas of base and sides The surface consists of the base, whose area is x 2, and four sides, each having area xy. The total surface area of 85 in. 2 is used to write a second equation. The two equations form this system. (1) (2)

27. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Let in (2). Step 4 Solve the system. Solve equation (1) for y to get and substitute into equation (2). Multiply. Multiply by x, x ≠ 0. Subtract 85x.

28. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Coefficients of a quadratic polynomial factor We are restricted to positive values for x, and considering the nature of the problem, any solution should be relatively small. By the rational zeros theorem, factors of 300 are the only possible rational solutions. Using synthetic division, we see that 5 is a solution.

29. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Therefore, one value of x is 5 and We must now solve for any other possible positive solutions. Use the quadratic formula to find the positive solution. Quadratic formula with a = 1, b = 5, c = – 60 This value of x leads to y ≈ 2.359.

30. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Example 6 USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Step 5 State the answer. There are two possible answers. First answer: length = width = 5 in.; height = 3 in. Second answer: length = width ≈ 5.639 in.; height ≈ 2.359 in. Step 6 Check. The check is left for Exercise 63.