College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson.

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Presentation transcript:

College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson

Systems of Equations and Inequalities 5

Systems of Nonlinear Equations 5.4

Systems of nonlinear equations In this section we solve systems of equations in which the equations are not all linear.

Substitution Method

To solve a system of nonlinear equations, we can use the substitution method.

E.g. 1—Substitution Method Find all solutions of the system. We start by solving for y in the second equation. y = 3x – 10

E.g. 1—Substitution Method Next, we substitute for y in the first equation and solve for x: x 2 + (3x – 10) 2 = 100 x 2 + (9x 2 – 60x + 100) = x 2 – 60x = 0 10x(x – 6) = 0 x = 0 or x = 6

E.g. 1—Substitution Method Now, we back-substitute these values of x into the equation y = 3x – 10. For x = 0: y = 3(0) – 10 = –10 For x = 6: y = 3(6) – 10 = 8 So, we have two solutions: (0, –10) and (6, 8).

E.g. 1—Substitution Method The graph of the first equation is a circle. That of the second equation is a line. The graphs intersect at the two points (0, –10) and (6, 8).

Elimination Method

To solve a system of nonlinear equations, we can use the elimination method.

E.g. 2—Elimination Method Find all solutions of the system. We choose to eliminate the x-term. So, we multiply the first equation by 5 and the second equation by –3. Then, we add the two equations and solve for y.

E.g. 2—Elimination Method Now, we back-substitute y = –11 into one of the original equations (say, 3x 2 + 2y = 26) and solve for x.

E.g. 2—Elimination Method 3x 2 + 2(–11) = 26 3x 2 = 48 x 2 = 16 x = –4 or x = 4 So, we have two solutions: (–4, –11) and (4, –11)

E.g. 2—Elimination Method The graphs of both equations are parabolas. The graphs intersect at the two points (–4, –11) and (4, –11).

Graphical Method

The graphical method is particularly useful in solving systems of nonlinear equations.

E.g. 3—Graphical Method Find all solutions of the system. Solving for y in terms of x, we get the equivalent system

E.g. 3—Graphical Method The figure shows that the graphs of these equations intersect at two points. Zooming in, we see that the solutions are (–1, –1) and (3, 7).

E.g. 4—Solving a System of Equations Graphically Find all solutions of the system, correct to one decimal place. The graph of the first equation is a circle, and the graph of the second is a parabola. To graph the circle on a graphing calculator, we must first solve for y in terms of x.

E.g. 4—Solving a System of Equations Graphically To graph the circle, we must graph both functions:

E.g. 4—Solving a System of Equations Graphically In the figure, the graph of the circle is shown in red and the parabola in blue.

E.g. 4—Solving a System of Equations Graphically The graphs intersect in quadrants I and II. Zooming in or using the Intersect command, we see that the intersection points are: (–0.559, 3.419) and (2.847, 1.974)

E.g. 4—Solving a System of Equations Graphically There also appears to be an intersection point in quadrant IV.

E.g. 4—Solving a System of Equations Graphically However, when we zoom in, we see that the curves come close to each other but don’t intersect. Thus, the system has two solutions. Rounded to the nearest tenth, they are: (–0.6, 3.4) and (2.8, 2.0)