chapter 7 systems of equations and inequalities copyright © 2014, 2010, 2007 pearson education,...
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Chapter 7Systems of Equationsand Inequalities
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7.4 Systems of NonlinearEquations in Two Variables
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• Recognize systems of nonlinear equations in two variables.
• Solve nonlinear systems by substitution.• Solve nonlinear systems by addition.• Solve problems using systems of nonlinear equations.
Objectives:
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Systems of Nonlinear Equations and Their Solutions
A system of two nonlinear equations in two variables, also called a nonlinear system, contains at least one equation that cannot be expressed in the form Ax + By = C.
A solution of a nonlinear system in two variables is an ordered pair of real numbers that satisfies both equations in the system. The solution set of the system is the set of all such ordered pairs.
As with linear systems in two variables, the solution of a nonlinear system (if there is one) corresponds to the intersection point(s) of the graphs of the equations in the system.
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Example: Solving a Nonlinear System by the Substitution Method
Solve by the substitution method:
Step 1 Solve either of the equations for one variable in terms of the other.
Step 2 Substitute the expression from step 1 into the other equation.
2 1
4 1
x y
x y
2 1x y 2 1y x
4 1x y 24 ( 1) 1x x
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Example: Solving a Nonlinear System by the Substitution Method (continued)
Solve by the substitution method:
Step 3 Solve the resulting equation containing one variable.
2 1
4 1
x y
x y
24 ( 1) 1x x 24 1 1x x
20 4x x 0 ( 4)x x
0x 4 0x 4x
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Example: Solving a Nonlinear System by the Substitution Method (continued)
Solve by the substitution method:
Step 4 Back-substitute the obtained value into one of the original equations.
2 1
4 1
x y
x y
0x 2 1y x
2(0) 1y 1y
4x 2 1y x
2(4) 1y 17y The proposed solutions are
(0, 1) and (4, 17).
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Example: Solving a Nonlinear System by the Substitution Method (continued)
Solve by the substitution method:
Step 5 Check.
2 1
4 1
x y
x y
(0,1)
2(0) 1 1? 0 0
true
4 1x y 2 1x y 4(0) (1) 1?
1 1 true
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Example: Solving a Nonlinear System by the Substitution Method (continued)
Solve by the substitution method:
Step 5 (cont) Check.
2 1
4 1
x y
x y
(4,17)2 1x y 2(4) 17 1?
16 16true
4 1x y 4(4) ( 1) 17?
16 1 17? 17 17
true
The solution set is{(0, 1),(4, 17)}.
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Example: Solving a Nonlinear System by the Addition Method
Solve the system:
Step 1 Rewrite both equations in the form
Ax2 + By2 = C.
Both equations are already in this form so we can skip this step.
2 2
2 2
3 2 35
4 3 48
x y
x y
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Example: Solving a Nonlinear System by the Addition Method (continued)
Solve the system:
Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.
2 2
2 2
3 2 35
4 3 48
x y
x y
2 2
2 2
3 2 35
4 3 48
x y
x y
2 2
2 2
9 6 105
8 6 96
x y
x y
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Example: Solving a Nonlinear System by the Addition Method (continued)
Solve the system:
Steps 3 and 4 Add the equations and solve for the remaining variable.
2 2
2 2
3 2 35
4 3 48
x y
x y
2 2
2 2
9 6 105
8 6 96
x y
x y
2 9x
3x
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Example: Solving a Nonlinear System by the Addition Method (continued)
Solve the system:
Step 5 Back-substitute and find the values for the other variable.
2 2
2 2
3 2 35
4 3 48
x y
x y
3x 2 23 2 35x y 2 23(3) 2 35y
23(9) 2 35y 227 2 35y
22 8y 2 4y
2y
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Example: Solving a Nonlinear System by the Addition Method (continued)
Solve the system:
Step 5 (cont) Back-substitute and find the values for the other variable.
2 2
2 2
3 2 35
4 3 48
x y
x y
3x 2 23 2 35x y
2 23( 3) 2 35y 23(9) 2 35y 227 2 35y
22 8y 2 4y
2y
The solution set is{(3, 2), (3, –2),(–3, 2),(–3, –2)}.
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Example: Application
Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet.
Step 1 Use variables to represent unknown quantities.
Let x = the length of the rectangle and y = the width of the rectangle.
Step 2 Write a system of equations modeling the problem’s question. 2 2 20
21
x y
xy
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Example: Application (continued)
Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet.
Step 3 Solve the system and answer the problem’s question.2 2 20
21
x y
xy
21xy 21
yx
212 2 20x
x
422 20x
x
422 20x x x
x
22 42 20x x 22 20 42 0x x
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Example: Application (continued)
Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet.
Step 3 (cont) Solve the system and answer the problem’s question.2 2 20
21
x y
xy
22 20 42 0x x 22( 10 21) 0x x
2( 3)( 7) 0x x
3 0x 3x 21xy
3 21y 7y
7 0x 7x 21xy
7 21y 3y
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Example: Application (continued)
Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet.
Step 3 (cont) Solve the system and answer the problem’s question.
Solutions are (3, 7) and (7, 3).
The rectangle could have a length of 3 feet and a width of 7 feet or a length of 7 feet and a width of 3 feet.
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Example: Application (continued)
Find the length and width of a rectangle whose perimeter is 20 feet and whose area is 21 square feet.
Step 4 Check the proposed solution in the original wording of the problem.
length = 3 feet, width = 7 feet:
Perimeter = 2(3) + 2(7) = 6 + 14 = 20 feet
Area = (3 feet)(7 feet) = 21 square feet
length = 7 feet, width = 3 feet
Perimeter = 2(7) + 2(3) = 6 +14 = 20 feet
Area = (7 feet)(3 feet) = 21 square feet.