Download Section 7.4

Chapter 7
Systems of
Equations
and Inequalities
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All rights reserved
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1
SECTION 7.4 Nonlinear Systems of Equations and Inequalities
OBJECTIVES
1
2
Solve nonlinear systems of equations.
Solve a nonlinear system of inequalities.
You will not need to know the details of
inequalities.
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2
EXAMPLE 1
Using Substitution to Solve a Nonlinear
System
Solve the system of equations by the
substitution method.
 4 x  y  3 (1)
 2
x  y  1 (2)
Solution
Step 1 Solve for one variable. Express y in
terms of x in equation (2).
2
y  x 1
Step 2 Substitute. Substitute x2 +1 for y in
equation (1).
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3
EXAMPLE 1
Using Substitution to Solve a Nonlinear
System
Solution continued
4 x  y  3
4 x   x  1  3
2
4 x  x  1  3
2
x  4x  4  0
2
Step 3 Solve the equation resulting from step (2).
x  2 x  2   0
x20
x  2
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4
EXAMPLE 1
Using Substitution to Solve a Nonlinear
System
Solution continued
Step 4 Back substitution. Substitute x = –2 in
equation (3) to obtain the corresponding
y-value.
y  x2  1
y   2   1
2
y5
Since x = –2 and y = 5, the apparent
solution set of the system is {(–2, 5)}.
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EXAMPLE 1
Using Substitution to Solve a Nonlinear
System
Solution continued
Step 5 Check. Replace x with –2 and y with 5 in
both equations (1) and (2).
4 x  y  3
 x2  y  1
4  2   5  3
  2   5 ? 1
?
?
8  5  3
3  3
2
4  5  1
11
?
Confirm the solution with a graph.
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6
EXAMPLE 1
Using Substitution to Solve a Nonlinear
System
Solution continued
The graphs of the
line 4x + y = –3
and the parabola
y = x 2 + 1 confirm
that the solution
set is {(–2, 5)}.
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8
EXAMPLE 2
Using Elimination to Solve a Nonlinear
System
Solve the system of equations by the
elimination method.
 x 2  y 2  25 (1)
 2
 x  y  5 (2)
Solution
Step 1 Adjust the coefficients. Multiply
equation (2) by –1 to eliminate x.
2
2
x  y  25 (1)
–x  y  –5 (3)
2
y  y  20
2
Step 2
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EXAMPLE 2
Using Elimination to Solve a Nonlinear
System
Solution continued
Step 3 Solve the equation obtained in Step 2.
2
y  y  20
y 2  y  20  0
y  5 y  4   0
y  5  0 or y  4  0
y  5 or y  4
Step 4 Back-substitute the values in one of the
original equations to solve for the other
variable.
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EXAMPLE 2
Using Elimination to Solve a Nonlinear
System
Solution continued
(i) Substitute y = –5 in equation (2) & solve for x.
x2  y  5
x 2   5   5
x2  5  5
x2  0
x0
Thus (0, –5) is a solution of the system.
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11
EXAMPLE 2
Using Elimination to Solve a Nonlinear
System
Solution continued
(ii) Substitute y = 4 in equation (2) & solve for x.
2
x  y5
x   4  5
2
x 45
2
x 9
x  3
2
Thus (3, 4) and (–3, 4) are the solutions of
the system.
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Using Elimination to Solve a Nonlinear
System
EXAMPLE 2
Solution continued
Step 5 Check (0, –5), (3, 4), and (–3, 4) in the
equations x2 + y2 = 25 and x2 – y = 5.
0   5   25 3  4  25
?
25  25 9  16  25
25  25
2
2
?
 0    5  5
2
?
55
2
2
?
3 45
?
945
55
2
?
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 3
 4  25
2
2
?
?
9  16  25
25  25
 3
2
?
45
?
945
55
13
EXAMPLE 2
Using Elimination to Solve a Nonlinear
System
Solution continued
The graphs of
the circle
x2 + y2 = 25
and the
parabola
y = x2 – 5
confirm that the
solution set is
{(0, –5), (3, 4), (–3, 4)}.
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15
Since the nature of
such a problem
requires a rather
lengthy list of
formulas, it is unlikely
that this would be a FE
problem.
Business majors
might wish to look at
Example 3 and this
related problem. It
seems
specialized.
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