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COLLEGE ALGEBRA

COLLEGE ALGEBRA. LIAL HORNSBY SCHNEIDER. 5.5. Nonlinear Systems of Equations. Solving Nonlinear Systems with Real Solutions Solving Nonlinear Equations with Nonreal Complex Solutions Applying Nonlinear Systems. Solving Nonlinear Systems with Real Solutions.

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COLLEGE ALGEBRA

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  1. COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

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

  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. SOLVING A NONLINEAR SYSTEMS BY SUBSTITUTION Example 1 Solve the system. (1) (2) Solution 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. Solve equation (2) for y. Substitute this result for y in equation (1).

  5. SOLVING A NONLINEAR SYSTEMS BY SUBSTITUTION Example 1 (1) Let y = –2 – x. Distributive property Standard form Factor. Zero factor property

  6. SOLVING A NONLINEAR SYSTEMS BY SUBSTITUTION Example 1 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. 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. SOLVING A NONLINEAR SYSTEM BY ELIMINATION Example 2 Solve the system. (1) (2) Solution 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 y2.

  9. SOLVING A NONLINEAR SYSTEM BY ELIMINATION Example 2 (1) (2) Add. Divide by 3. Remember to find both square roots. Square root property

  10. SOLVING A NONLINEAR SYSTEM BY ELIMINATION Example 2 Find y by substituting back into equation (1). If x = 2, then If x = −2, then The solutions are (2, 0) and (–2, 0) so the solution set is {(2, 0), (–2, 0)}.

  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 x2 and y2. In other words, the system is linear in x2 and y2.To see this, substitute u for x2and v for y2. The resulting system is linear in u and v.

  12. SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Example 3 Solve the system. (1) (2) Solution (1) Multiply (2) by –1 Add. (3) Solve for y(x ≠ 0).

  13. SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Example 3 Now substitute for y in either equation (1) or (2). We use equation (2). Let y = in (2). Multiply and square. Multiply x2 to clear fractions.

  14. SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Example 3 This equation is in quadratic form. Subtract 6x2. Factor. Zero factor property. Square root property; For each equation, include both square roots.

  15. SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Example 3 Substitute these x-values into equation (4) to find corresponding values of y. If , then If , then If , then If , then

  16. SOLVING A NONLINEAR SYSTEM BY A COMBINATION OF METHODS Example 3 The solution set of the system is If , then If , then If , then If , then

  17. SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION Example 4 Solve the system. (1) (2) Solution Use the substitution method. Solving equation (2) for x gives (3) Since x  0 for all x, 4 – y 0 and thus y 4. In equation (1), the first term is x2, which is the same as x2.

  18. SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION Example 4 Therefore, Substitute 4 – y in (1). Square the binomial. Combine terms. Remember the middle term. Factor. Zero factor property.

  19. SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION Example 4 To solve for the corresponding values of x, use either equation (1) or (2). We use equation (1). If y = 0, then If y = 4, then

  20. SOLVING A NONLINEAR SYSTEM WITH AN ABSOLUTE VALUE EQUATION Example 4 The solution set, {(4,0), (–4, 0), (0, 4)}, includes the points of intersection shown in the graph.

  21. SOLVING A NONLINEAR SYSTEM WITH NONREAL COMPLEX NUMBERS IN ITS SOLUTIONS Example 5 Solve the system. (1) (2) Solution Multiply (1) by –3 . (2) Add. Square root property

  22. SOLVING A NONLINEAR SYSTEM WITH NONREAL COMPLEX NUMBERS IN ITS SOLUTIONS Example 5 To find the corresponding values of y, substitute into equation (1). If x = 2i, then If x = –2i, then i2= –4

  23. SOLVING A NONLINEAR SYSTEM WITH NONREAL COMPLEX NUMBERS IN ITS SOLUTIONS Example 5 Checking the solutions in the given system shows that the solution set is Note that these solutions with nonreal complex number components do not appear as intersection points on the graph of the system.

  24. USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Example 6 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. What are the dimensions of the box? Solution Step 1 Readthe problem. We must find the dimensions (width, length, and height) of the box. Step 2 Assign variables. Let x represent the length and width of the square base, and let y represent the height.

  25. USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Example 6 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. USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Example 6 The surface consists of the base, whose area is x2, and four sides, each having area xy. The total surface area is 85 in.2, so a second equation is Sum of areas of base and sides The system to solve is (1) (2)

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

  28. USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Example 6 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. Coefficients of a quadratic polynomial factor

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

  30. USING A NONLINEAR SYSTEM TO FIND THE DIMENSIONS OF A BOX Example 6 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 61.

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