Lca10 0505

5.5 - 1
10TH
EDITION
LIAL
HORNSBY
SCHNEIDER
COLLEGE
ALGEBRA

5.5 - 25.5 - 2
5.5 Nonlinear Systems of Equations
Solving Nonlinear Systems with Real
Solutions
Solving Nonlinear Equations with Nonreal
Complex Solutions
Applying Nonlinear Systems

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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.

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Example 1
SOLVING A NONLINEAR
SYSTEMS BY SUBSTITUTION
Solve the system.
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.
Solution
2
4x y− =
2x y+ = −
(1)
(2)
2y x= − − Solve equation (2) for y.
Substitute this result for y in equation (1).

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Example 1
SOLVING A NONLINEAR
2
4x y− = (1)
Let y = –2 – x.
2
2( ) 4xx − −− =
2
2 4x x+ + = Distributive property
2
2 0x x+ − = Standard form
( 2)( 1) 0x x+ − = Factor.
2 0 or 1 0x x+ = − = Zero factor property
2 or 1x x= − =

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Example 1
SOLVING A NONLINEAR
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)}.

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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.

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Example 2 SOLVING A NONLINEAR SYSTEM
BY ELIMINATION
Solve the system.
Solution
2 2
4x y+ = (1)
(2)2 2
2 8x y− =
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
.

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BY ELIMINATION
2 2
4x y+ =
2 2
2 8x y− =
(1)
(2)
2
3 12x = Add.
2
4x = Divide by 3.
2x = ± Square root property
Remember to
find both
square roots.

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BY ELIMINATION
Find y by substituting back into equation (1).
If x = 2, then If x = −2, then
2 2
2 4y+ =
2
0y =
0.y =
2 2
42( ) y+ =−
2
0y =
0.y =
The solutions are (2, 0) and (–2, 0) so the solution
set is {(2, 0), (–2, 0)}.

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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 x2
and v for y2
. The resulting system is linear
in u and v.

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Example 3 SOLVING A NONLINEAR SYSTEM BY A
COMBINATION OF METHODS
Solve the system.
Solution
2 2
3 22x xy y+ + = (1)
(2)2 2
6x xy y− + =
2 2
3 22x xy y+ + = (1)
2 2
6x xy y− + − = − Multiply (2) by –1
4 16xy = Add. (3)
4
y
x
= Solve for y(x ≠ 0).

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Let y = in (2).
Now substitute for y in either equation (1)
or (2). We use equation (2).
4
x
2
2
6
4 4
x x
x x
   
− + = ÷  ÷
   
4
x
2
2
16
4 6x
x
− + = Multiply and square.
4 2 2
4 16 6x x x− + = Multiply x2
to clear
fractions.

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4 2
10 16 0x x− + = Subtract 6x2
.
This equation is
in quadratic form.
2 2
( 2)( 8) 0x x− − = Factor.
2 2
2 or 8x x= = Zero factor
property.
2 or 2 2x x= ± = ± Square root
property;
8 4 2 2 2± = ± = ±gFor each equation,
include both square
roots.

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Substitute these x-values into equation (4) to
find corresponding values of y.
If , then2x = If , then2x = −
2 2
4
2
.y = = 2 2
4
.
2
y = = −
−
If , then2 2x =
.2
4
2 2
y = =
If , then2 2x = −
2 2
4
2.y
−
= = −

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If , then2x = If , then2x = −
2 2
4
2
.y = = 2 2
4
.
2
y = = −
−
If , then2x =
.2
4
2 2
y = =
If , then2x = −
2 2
4
2.y
−
= = −
The solution set of the system is
( ) ( ) ( ) ( ){ }2,2 2 , 2, 2 2 , 2 2, 2 , 2 2, 2 .− − − −

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Example 4 SOLVING A NONLINEAR SYSTEM WITH
AN ABSOLUTE VALUE EQUATION
Solve the system.
Solution
2 2
16x y+ = (1)
(2)4x y+ =
Use the substitution method. Solving equation (2)
for |x| gives
4 .x y= − (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
.

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Example 4
Therefore,
2 2
(4 ) 16y y− + = Substitute 4 – y in (1).
2 2
(16 8 ) 16y y y− + + = Square the binomial.
Remember
the middle
term.
2
2 8 0y y− = Combine terms.
2 ( 4) 0y y − = Factor.
or .0 4y y= = Zero factor property.
SOLVING A NONLINEAR SYSTEM WITH

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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
2 2
160x + =
.4x = ±
2
16x =
2 2
164x + =
2
0x =
0.x =

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Example 4
The solution set, {(4,0), (–4, 0), (0, 4)}, includes
the points of intersection shown in the graph.

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Example 5
NONREAL COMPLEX NUMBERS IN ITS
SOLUTIONS
Solve the system.
Solution
2 2
5x y+ = (1)
(2)2 2
4 3 11x y+ =
2 2
3 3 15x y− − = − Multiply (1) by –3 .
2 2
4 3 11x y+ =
2
4x = −
(2)
Add.
4x = ± − Square root property
2x = ± i 4 4 2− = =i i

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Example 5
SOLUTIONS
To find the corresponding values of y, substitute
into equation (1).
If x = 2i, then If x = –2i, then
2 2
2( ) 5y+ =i
i2
= –4 2
4 5y− + =
2
9y =
.3y = ±
2 2
2( ) 5y+ =i
2
4 5y− + =
2
9y =
.3y = ±

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Example 5
SOLUTIONS
Checking the solutions in the given system
shows that the solution set is
( ) ( ) ( ) ( ){ }2 ,3 , 2 , 3 , 2 ,3 , 2 , 3 .− − − −i i i i
Note that these solutions with nonreal complex
number components do not appear as
intersection points on the graph of the system.

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Example 6 USING A NONLINEAR SYSTEM TO
FIND THE DIMENSIONS OF A BOX
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 Read the 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.

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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
.
2
75x y = Volume formula

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2
4 85x xy+ = Sum of areas of
base and sides
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
The system to solve is
2
75x y =
2
4 85x xy+ =
(1)
(2)

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2
2 7
4 85
5
x
x
x
 
+ = ÷
 
Let in (2).
Step 4 Solve the system. Solve equation (1) for y
to get and substitute into equation
(2).
2
75
,y
x
=
2
75
y
x
=
2 300
85x
x
+ = Multiply.
3
300 85x x+ = Multiply by x, x ≠ 0.
3
85 300 0x x− + = Subtract 85x.

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Coefficients of a quadratic polynomial factor
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.
5 1 0 85 300−
5 25 300−
1 5 60 0−

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Therefore, one value of x is 5 and We
must now solve
2
75
3.
5
y = =
2
5 60 0x x+ − =
for any other possible positive solutions.
Using the quadratic formula, the positive solution
is
2
5 5 4(1)( 60)
5.639
2(1)
x
− − −
≈
+
=
Quadratic formula
with a = 1, b = 5,
c = –60
This value of x leads to y ≈ 2.359.

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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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