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Lesson 7.4, page 746 Nonlinear Systems of Equations

Lesson 7.4, page 746 Nonlinear Systems of Equations. Objective : To solve a nonlinear system of equations. Review – What?. System – 2 or more equations together Solution of system – any ordered pair that makes all equations true Possible solutions: One point More than one point

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Lesson 7.4, page 746 Nonlinear Systems of Equations

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  1. Lesson 7.4, page 746Nonlinear Systems of Equations Objective: To solve a nonlinear system of equations.

  2. Review – What? • System – 2 or more equations together • Solution of system – any ordered pair that makes all equations true • Possible solutions: • One point • More than one point • No solution • Infinite solutions

  3. Review - How? • What methods have we used to solve linear systems of equations? • Graphing • Substitution • Elimination

  4. ReviewSteps for using SUBSTITUTION • Solve one equation for one variable. (Hint: Look for an equation already solved for a variable or for a variable with a coefficient of 1 or -1.) • Substitute into the other equation. • Solve this equation to find a value for the variable. • Substitute again to find the value of the other variable. • Check.

  5. 2x – y = 6 y = 5x ReviewSolve using Substitution.

  6. ReviewSTEPS for ELIMINATION

  7. ReviewSolve using elimination. • 3x + 5y = 11 • 2x + 3y = 7

  8. What’s New? • A non-linear system is one in which one or more of the equations has a graph that is not a line. • With non-linear systems, the solution could be one or more points of intersection or no point of intersection. • We’ll solve non-linear systems using substitution or elimination. • A graph of the system will show the points of intersection.

  9. An Example… • Solve the following system of equations:

  10. An Example… • We use the substitution method. First, we solve equation (2) for y.

  11. Next, we substitute y = 2x 3 in equation (1) and solve for x: An Example…

  12. An Example… … • Now, we substitute these numbers for x in equation (2) and solve for y. • x = 0 x = 12 / 5 y = 2x 3 y = 2(0)  3 y = 3 SOLUTIONS (0, 3) and

  13. Check: (0, 3) Check: Visualizing the Solution An Example…

  14. See Example 1, page 747 Check Point 1: Solve by substitution. • x2 = y – 1 • 4x – y = -1

  15. See Example 2, page 748 • Check Point 2: Solve by substitution. x + 2y = 0 (x – 1)2+ (y – 1)2 = 5

  16. Solve the following system of equations: xy = 4 3x + 2y = 10 Another example to watch…

  17. Solve xy = 4 for y. Substitute into 3x + 2y = 10.

  18. Use the quadratic formula (or factor) to solve:

  19. 3x + 2y = 10 x = 4/3 x = 2 The solutions are (4/3, 3) and (2, 2). Visualizing the Solution Substitute values of x to find y.

  20. Need to watch another one? • Solve the system of equations:

  21. Solve by elimination. Multiply equation (1) by 2 and add to eliminate the y2 term.

  22. Substituting x = 1 in equation (2) gives us: x = 1 x = -1 • The possible solutions are (1, 3), (1, 3), (1, 3) and (1, 3).

  23. All four pairs check, so they are the solutions. Visualizing the Solution

  24. See Example 3, page 749 Check Point 3: Solve by elimination. • 3x2 + 2y2 = 35 • 4x2 + 3y2 = 48

  25. See Example 4, page 750 Check Point 4: Solve by elimination. • y = x2 + 5 • x2 + y2 = 25

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