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BLITZER LEARNING GUIDE SAMPLE - Pearson

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BLITZER LEARNING GUIDE
SAMPLE
Section 5.1
Systems of Linear Equations in Two Variables
Procrastination makes you sick!
Researchers compared college students who were procrastinators and
nonprocrastinators. Early in the semester, procrastinators reported fewer
symptoms of illness, but late in the semester, they reported more symptoms
than their nonprocrastinating peers.
In this section of the textbook, you will identify when both groups have the
same number of symptoms as the point of intersection of two lines.
Objective #1: Decide whether an ordered pair is a solution of a linear system.
Solved Problem #1
1.
Determine if the ordered pair (7, 6) is a solution of
Pencil Problem #1
1.
пѓ¬ 2 x пЂ­ 3 y пЂЅ пЂ­4
the system: пѓ­
пѓ® 2x пЂ« y пЂЅ 4
Determine if the ordered pair (2, 3) is a solution of
пѓ¬ x пЂ« 3 y пЂЅ 11
the system: пѓ­
пѓ® x пЂ­ 5 y пЂЅ пЂ­13
To determine if (7, 6) is a solution to the system, replace
x with 7 and y with 6 in both equations.
2 x пЂ­ 3 y пЂЅ пЂ­4
2(7) пЂ­ 3(6) пЂЅ пЂ­4
14 пЂ­ 18 пЂЅ пЂ­4
пЂ­4 пЂЅ пЂ­4, true
2x пЂ« y пЂЅ 4
2(7) пЂ« 6 пЂЅ 4
14 пЂ« 6 пЂЅ 4
20 пЂЅ 4, false
The ordered pair does not satisfy both equations, so it is
not a solution to the system.
Objective #2: Solve linear systems by substitution.
Solved Problem #2
2.
пѓ¬3 x пЂ« 2 y пЂЅ 4
Solve by the substitution method: пѓ­
пѓ® 2x пЂ« y пЂЅ 1
Pencil Problem #2
2.
пѓ¬x пЂ« y пЂЅ 4
Solve by the substitution method: пѓ­
y пЂЅ 3x
пѓ®
Solve 2 x пЂ« y пЂЅ 1 for y.
2x пЂ« y пЂЅ 1
y пЂЅ пЂ­2 x пЂ« 1
Copyright В© 2014 Pearson Education Inc.
1
College Algebra 6e
3x пЂ« 2 y
Substitute:
пЂЅ4
y
пЂ¶пЂ·пЂё
3x пЂ« 2(пЂ­2 x пЂ« 1) пЂЅ 4
3x пЂ­ 4 x пЂ« 2 пЂЅ 4
пЂ­x пЂ« 2 пЂЅ 4
пЂ­x пЂЅ 2
x пЂЅ пЂ­2
Find y.
y пЂЅ пЂ­2 x пЂ« 1
y пЂЅ пЂ­2(пЂ­2) пЂ« 1
yпЂЅ5
The solution is пЂЁ пЂ­2,5пЂ© .
The solution set is
пЃ»пЂЁ пЂ­2,5пЂ©пЃЅ .
Objective #3: Solve linear systems by addition.
Solved Problem #3
3.
пѓ¬4 x пЂ« 5 y пЂЅ 3
Solve the system: пѓ­
пѓ®2x пЂ­ 3y пЂЅ 7
Pencil Problem #3
3.
пѓ¬3x пЂ­ 4 y пЂЅ 11
Solve the system: пѓ­
пѓ® 2 x пЂ« 3 y пЂЅ пЂ­4
Multiply each term of the second equation by –2 and add
the equations to eliminate x.
4x пЂ« 5y пЂЅ 3
пЂ­4 x пЂ« 6 y пЂЅ пЂ­14
11 y пЂЅ пЂ­11
y пЂЅ пЂ­1
Back-substitute into either of the original equations to
solve for x.
2x пЂ­ 3y пЂЅ 7
2 x пЂ­ 3( пЂ­1) пЂЅ 7
2x пЂ« 3 пЂЅ 7
2x пЂЅ 4
xпЂЅ2
The solution set is
2
пЃ»пЂЁ 2, пЂ­1пЂ©пЃЅ .
Copyright В© 2014 Pearson Education Inc.
Section 5.1
Objective #4: Identify systems that do not have exactly one ordered-pair solution.
Solved Problem #4
пѓ¬ 5x пЂ­ 2 y пЂЅ 4
4a. Solve the system: пѓ­
пѓ®пЂ­10 x пЂ« 4 y пЂЅ 7
Pencil Problem #4
x пЂЅ 9пЂ­ 2y
пѓ¬
4a. Solve the system: пѓ­
пѓ® x пЂ« 2 y пЂЅ 13
Multiply the first equation by 2, and then add the
equations. 10 x пЂ­ 4 y пЂЅ 8
пЂ­10 x пЂ« 4 y пЂЅ 7
0 пЂЅ 15
Since there are no pairs пЂЁ x, y пЂ© for which 0 will equal 15,
the system is inconsistent and has no solution.
The solution set is пѓ† or пЃ» пЃЅ .
x пЂЅ 4y пЂ­8
пѓ¬
4b. Solve the system: пѓ­
5
20
x
y
пЂ­
пЂЅ пЂ­40
пѓ®
y пЂЅ 3x пЂ­ 5
пѓ¬
4b. Solve the system: пѓ­
21
пЂ­
35
пЂЅ 7y
x
пѓ®
Substitute 4 y пЂ­ 8 for x in the second equation.
5 x пЂ­ 20 y пЂЅ пЂ­40
x
пЃЅ
5(4 y пЂ­ 8) пЂ­ 20 y пЂЅ пЂ­40
20 y пЂ­ 40 пЂ­ 20 y пЂЅ пЂ­40
пЂ­40 пЂЅ пЂ­40
Since пЂ­40 пЂЅ пЂ­40 for all values of x and y, the system is
dependent.
пЃ»пЂЁ x, y пЂ© x пЂЅ 4 y пЂ­ 8пЃЅ or
пЃ»пЂЁ x, y пЂ© 5x пЂ­ 20 y пЂЅ пЂ­40пЃЅ.
The solution set is
Objective #5: Solve problems using systems of linear equations.
Solved Problem #5
5.
A company that manufactures running shoes has a
fixed cost of $300,000. Additionally, it costs $30 to
produce each pair of shoes. The shoes are sold at
$80 per pair.
5a. Write the cost function, C, of producing x pairs of
running shoes.
Pencil Problem #5
5.
A company that manufactures small canoes has a
fixed cost of $18,000. Additionally, it costs $20 to
produce each canoe. The selling price is $80 per
canoe.
5a. Write the cost function, C, of producing x canoes.
per pair
fixed
costs
пЂ¶
пЂґпЂ·пЂґ
пЂё $30пЃЅ
C ( x ) пЂЅ 300,000 пЂ« 30 x
Copyright В© 2014 Pearson Education Inc.
3
College Algebra 6e
5b. Write the revenue function, R, from the sale of x
pairs of running shoes.
5b. Write the revenue function, R, from the sale of x
canoes.
$80 per pair
R( x ) пЂЅ
пЃЅ
80 x
5c. Determine the break-even point. Describe what this
means.
5c. Determine the break-even point. Describe what this
means.
пѓ¬ y пЂЅ 300,000 пЂ« 30 x
The system is пѓ­
пѓ® y пЂЅ 80 x
The break-even point is where R( x ) пЂЅ C ( x ).
R( x ) пЂЅ C ( x )
80 x пЂЅ 300,000 пЂ« 30 x
50 x пЂЅ 300,000
x пЂЅ 6000
Back-substitute to find y: y пЂЅ 80 x
y пЂЅ 80(6000)
y пЂЅ 480, 000
The break-even point is (6000, 480,000).
This means the company will break even when it
produces and sells 6000 pairs of shoes. At this level, both
revenue and costs are $480,000.
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1. The ordered pair is a solution to the system. (5.1 #1)
2.
пЃ»пЂЁ1,3пЂ©пЃЅ
3.
пЃ»пЂЁ1, пЂ­2пЂ©пЃЅ
4a. пѓ† or
4b.
(5.1 #5)
(5.1 #27)
пЃ» пЃЅ
пЃ»пЂЁ x, y пЂ©
(5.1 #31)
пЃЅ
y пЂЅ 3x пЂ­ 5 or
5a. C пЂЁ x пЂ© пЂЅ 18,000 пЂ« 20 x
пЃ»пЂЁ x, y пЂ©
21x пЂ­ 35 пЂЅ 7 y
(5.1 #61a)
пЃЅ
(5.1 #33)
5b. R пЂЁ x пЂ© пЂЅ 80 x (5.1 #61b)
5c. Break-even point: (300, 24,000). Which means when 300 canoes are produced the company
will break-even with cost and revenue at $24,000. (5.1 #61c)
4
Copyright В© 2014 Pearson Education Inc.
Section 5.2
Systems of Linear Equations in Three Variables
Hit the BRAKES!
Did you know that a mathematical model can be used to describe the relationship between
the number of feet a car travels once the brakes are applied and the number of seconds the car
is in motion after the brakes are applied?
In the Exercise Set of this section, using data collected
by a research firm, you will be asked to write the mathematical model that describes this
situation.
Objective #1: Verify the solution of a system of linear equations in three variables.
Solved Problem #1
1.
Show that the ordered triple (пЂ­1, пЂ­4,5) is
a solution of the system:
пѓ¬ x пЂ­ 2 y пЂ« 3 z пЂЅ 22
пѓЇ
пѓ­ 2x пЂ­ 3y пЂ­ z пЂЅ 5
пѓЇ 3x пЂ« y пЂ­ 5z пЂЅ пЂ­32
пѓ®
Pencil Problem #1
1.
Determine if the ordered triple (2, пЂ­1, 3) is
a solution of the system:
пѓ¬ x пЂ« yпЂ«zпЂЅ 4
пѓЇ
пѓ­x пЂ­ 2 y пЂ­ z пЂЅ 1
пѓЇ 2 x пЂ­ y пЂ­ z пЂЅ пЂ­1
пѓ®
Test the ordered triple in each equation.
x пЂ­ 2 y пЂ« 3 z пЂЅ 22
(пЂ­1) пЂ­ 2 ( пЂ­4) пЂ« 3 (5) пЂЅ 22
22 пЂЅ 22, true
2x пЂ­ 3y пЂ­ z пЂЅ 5
2(пЂ­1) пЂ­ 3( пЂ­4) пЂ­ (5) пЂЅ 5
5 пЂЅ 5, true
3x пЂ« y пЂ­ 5z пЂЅ пЂ­32
3( пЂ­1) пЂ« ( пЂ­4) пЂ­ 5(5) пЂЅ пЂ­32
пЂ­32 пЂЅ пЂ­32, true
The ordered triple (–1, –4, 5) makes all three equations
true, so it is a solution of the system.
Copyright В© 2014 Pearson Education Inc.
1
College Algebra 6e
Objective #2: Solve systems of linear equations in three variables.
Solved Problem #2
2.
пѓ¬ x пЂ« 4 y пЂ­ z пЂЅ 20
пѓЇ
Solve the system: пѓ­ 3x пЂ« 2 y пЂ« z пЂЅ 8
пѓЇ2 x пЂ­ 3 y пЂ« 2 z пЂЅ пЂ­16
пѓ®
Pencil Problem #2
2.
пѓ¬ 4 x пЂ­ y пЂ« 2 z пЂЅ 11
пѓЇ
Solve the system: пѓ­ x пЂ« 2 y пЂ­ z пЂЅ пЂ­1
пѓЇ 2 x пЂ« 2 y пЂ­ 3 z пЂЅ пЂ­1
пѓ®
Add the first two equations to eliminate z.
x пЂ« 4 y пЂ­ z пЂЅ 20
3x пЂ« 2 y пЂ« z пЂЅ 8
4x пЂ« 6 y
пЂЅ 28
Multiply the first equation by 2 and add it to the third
equation to eliminate z again.
2 x пЂ« 8 y пЂ­ 2 z пЂЅ 40
2 x пЂ­ 3 y пЂ« 2 z пЂЅ пЂ­16
4x пЂ« 5y
пЂЅ 24
Solve the system of two equations in two variables.
4 x пЂ« 6 y пЂЅ 28
4 x пЂ« 5 y пЂЅ 24
Multiply the second equation by –1 and add the
equations.
4 x пЂ« 6 y пЂЅ 28
пЂ­4 x пЂ­ 5 y пЂЅ пЂ­24
yпЂЅ4
Back-substitute 4 for y to find x.
4 x пЂ« 6 y пЂЅ 28
4 x пЂ« 6(4) пЂЅ 28
4 x пЂ« 24 пЂЅ 28
4x пЂЅ 4
x пЂЅ1
Back-substitute into an original equation.
3x пЂ« 2 y пЂ« z пЂЅ 8
3(1) пЂ« 2(4) пЂ« z пЂЅ 8
11 пЂ« z пЂЅ 8
z пЂЅ пЂ­3
The solution is пЂЁ1, 4, пЂ­3пЂ©
and the solution set is
2
пЃ»пЂЁ1, 4, пЂ­3пЂ©пЃЅ.
Copyright В© 2014 Pearson Education Inc.
Section 5.2
Objective #3: Solve problems using systems in three variables.
Solved Problem #3
3.
Find the quadratic function y пЂЅ ax 2 пЂ« bx пЂ« c whose
graph passes through the points (1, 4), (2,1), and
(3, 4).
Pencil Problem #3
3.
Find the quadratic function y пЂЅ ax 2 пЂ« bx пЂ« c whose
graph passes through the points (пЂ­1, 6), (1, 4), and
(2, 9).
Use each ordered pair to write an equation.
y пЂЅ ax 2 пЂ« bx пЂ« c
(1, 4) :
4 пЂЅ a пЂЁ1пЂ© пЂ« b пЂЁ1пЂ© пЂ« c
2
4 пЂЅ aпЂ«bпЂ«c
(2,1) :
y пЂЅ ax 2 пЂ« bx пЂ« c
1 пЂЅ a пЂЁ2пЂ© пЂ« b пЂЁ2пЂ© пЂ« c
2
1 пЂЅ 4 a пЂ« 2b пЂ« c
(3, 4) :
y пЂЅ ax 2 пЂ« bx пЂ« c
4 пЂЅ a пЂЁ 3пЂ© пЂ« b пЂЁ 3пЂ© пЂ« c
2
4 пЂЅ 9a пЂ« 3b пЂ« c
The system of three equations in three variables is:
пѓ¬ a пЂ«bпЂ«c пЂЅ 4
пѓЇ
пѓ­ 4 a пЂ« 2b пЂ« c пЂЅ 1
пѓЇ 9a пЂ« 3b пЂ« c пЂЅ 4
пѓ®
пѓ¬ a пЂ«bпЂ«c пЂЅ 4
пѓЇ
Solve the system: пѓ­4a пЂ« 2b пЂ« c пЂЅ 1
пѓЇ 9a пЂ« 3b пЂ« c пЂЅ 4
пѓ®
Multiply the first equation by –1 and add it to the second
equation:
пЂ­ a пЂ­ b пЂ­ c пЂЅ пЂ­4
4 a пЂ« 2b пЂ« c пЂЅ 1
3a пЂ« b
пЂЅ пЂ­3
(continued on next page)
Copyright В© 2014 Pearson Education Inc.
3
College Algebra 6e
Multiply the first equation by –1 and add it to the third
equation:
пЂ­ a пЂ­ b пЂ­ c пЂЅ пЂ­4
9a пЂ« 3b пЂ« c пЂЅ 4
8a пЂ« 2b
пЂЅ 0
Solve this system of two equations in two variables.
3a пЂ« b пЂЅ пЂ­3
8a пЂ« 2b пЂЅ 0
Multiply the first equation by –2 and add to the second
equation: пЂ­6a пЂ­ 2b пЂЅ 6
8a пЂ« 2b пЂЅ 0
2a пЂЅ 6
aпЂЅ3
Back-substitute to find b:
3a пЂ« b пЂЅ пЂ­3
3(3) пЂ« b пЂЅ пЂ­3
9 пЂ« b пЂЅ пЂ­3
b пЂЅ пЂ­12
Back-substitute into an original equation to find c:
aпЂ«bпЂ«c пЂЅ 4
(3) пЂ« (пЂ­12) пЂ« c пЂЅ 4
пЂ­9 пЂ« c пЂЅ 4
c пЂЅ 13
The quadratic function is y пЂЅ 3x 2 пЂ­ 12 x пЂ« 13 .
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1. not a solution (5.2 #1)
2. {(2, пЂ­1,1)} (5.2 #7)
3. y пЂЅ 2 x 2 пЂ­ x пЂ« 3 (5.2 #19)
4
Copyright В© 2014 Pearson Education Inc.
Section 5.3
Partial Fractions
Where’s the “UNDO” Button?
We have learned how to write a sum or difference of rational expressions as a
single rational expression and saw how this skill is necessary when solving
rational inequalities. However, in calculus, it is sometimes necessary to write a
single rational expression as a sum or difference of simpler rational expressions,
undoing the process of adding or subtracting.
In this section, you will learn how to break up a rational expression into sums or
differences of rational expressions with simpler denominators.
Objective #1: Decompose
P
, where Q has only distinct linear factors.
Q
Solved Problem #1
1.
Find the partial fraction decomposition of
5x пЂ­ 1
.
( x пЂ­ 3)( x пЂ« 4)
Write a constant over each distinct linear factor in the denominator.
5x пЂ­ 1
A
B
пЂЅ
пЂ«
( x пЂ­ 3)( x пЂ« 4) x пЂ­ 3 x пЂ« 4
Multiply by the LCD, ( x пЂ­ 3)( x пЂ« 4), to eliminate fractions. Then simplify and rearrange terms.
5x пЂ­ 1
A
B
( x пЂ­ 3)( x пЂ« 4)
пЂЅ ( x пЂ­ 3)( x пЂ« 4)
пЂ« ( x пЂ­ 3)( x пЂ« 4)
( x пЂ­ 3)( x пЂ« 4)
xпЂ­3
xпЂ«4
5 x пЂ­ 1 пЂЅ A( x пЂ« 4) пЂ« B( x пЂ­ 3)
5 x пЂ­ 1 пЂЅ Ax пЂ« 4 A пЂ« Bx пЂ­ 3B
5 x пЂ­ 1 пЂЅ ( A пЂ« B ) x пЂ« (4 A пЂ­ 3B )
Equating the coefficients of x and equating the constant terms, we obtain a system of equations.
пѓ¬ AпЂ« B пЂЅ 5
пѓ­
пѓ®4 A пЂ­ 3B пЂЅ пЂ­1
Multiplying the first equation by 3 and adding it to the second equation, we obtain 7A = 14, so A = 2. Substituting 2
for A in either equation, we obtain B = 3. The partial fraction decomposition is
5x пЂ­ 1
2
3
пЂЅ
пЂ«
.
( x пЂ­ 3)( x пЂ« 4) x пЂ­ 3 x пЂ« 4
Copyright В© 2014 Pearson Education Inc.
1
College Algebra 6e
Pencil Problem #1
1.
Find the partial fraction decomposition of
3x пЂ« 50
.
( x пЂ­ 9)( x пЂ« 2)
Objective #2: Decompose
P
, where Q has repeated linear factors.
Q
Solved Problem #2
2.
Find the partial fraction decomposition of
xпЂ«2
.
x ( x пЂ­ 1)2
Include one fraction for each power of x в€’ 1.
xпЂ«2
A
B
C
пЂЅ пЂ«
пЂ«
x ( x пЂ­ 1)2 x x пЂ­ 1 ( x пЂ­ 1)2
Multiply by the LCD, x ( x пЂ­ 1)2 , to eliminate fractions. Then simplify and rearrange terms.
x ( x пЂ­ 1) 2
xпЂ«2
A
B
C
пЂЅ x( x пЂ­ 1)2 пЂ« x( x пЂ­ 1)2
пЂ« x ( x пЂ­ 1)2
2
x
x пЂ­1
x ( x пЂ­ 1)
( x пЂ­ 1)2
x пЂ« 2 пЂЅ A( x пЂ­ 1)2 пЂ« Bx( x пЂ­ 1) пЂ« Cx
x пЂ« 2 пЂЅ A( x 2 пЂ­ 2 x пЂ« 1) пЂ« Bx ( x пЂ­ 1) пЂ« Cx
x пЂ« 2 пЂЅ Ax 2 пЂ­ 2 Ax пЂ« A пЂ« Bx 2 пЂ­ Bx пЂ« Cx
2
0 x 2 пЂ« x пЂ« 2 пЂЅ ( A пЂ« B ) x 2 пЂ« ( пЂ­2 A пЂ­ B пЂ« C ) x пЂ« A
Copyright В© 2014 Pearson Education Inc.
Section 5.3
Equating the coefficients of like terms, we obtain a system of equations.
AпЂ« B пЂЅ 0
пѓ¬
пѓЇ
пѓ­ пЂ­2 A пЂ­ B пЂ« C пЂЅ 1
пѓЇ
AпЂЅ 2
пѓ®
We see immediately that A = 2. Substituting 2 for A in the first equation, we obtain B = в€’2. Substituting these values
into the second equation, we obtain C = 3. The partial fraction decomposition is
xпЂ«2
2 пЂ­2
3
2
2
3
пЂЅ пЂ«
пЂ«
or пЂ­
пЂ«
.
2
2
x
x
пЂ­
1
x
x
пЂ­
1
x ( x пЂ­ 1)
( x пЂ­ 1)
( x пЂ­ 1)2
Pencil Problem #2
2.
Find the partial fraction decomposition of
x2
.
( x пЂ­ 1)2 ( x пЂ« 1)
Copyright В© 2014 Pearson Education Inc.
3
College Algebra 6e
Objective #3: Decompose
P
, where Q has a nonrepeated prime quadratic factor.
Q
Solved Problem #3
8 x 2 пЂ« 12 x пЂ­ 20
.
( x пЂ« 3)( x 2 пЂ« x пЂ« 2)
Use a constant over the linear factor and a linear expression over the prime quadratic factor.
A
Bx пЂ« C
8 x 2 пЂ« 12 x пЂ­ 20
пЂЅ
пЂ« 2
2
( x пЂ« 3)( x пЂ« x пЂ« 2) x пЂ« 3 x пЂ« x пЂ« 2
3.
Find the partial fraction decomposition of
Multiply by the LCD, ( x пЂ« 3)( x 2 пЂ« x пЂ« 2), to eliminate fractions. Then simplify and rearrange terms.
( x пЂ« 3)( x 2 пЂ« x пЂ« 2)
8 x 2 пЂ« 12 x пЂ­ 20
A
Bx пЂ« C
пЂЅ ( x пЂ« 3)( x 2 пЂ« x пЂ« 2)
пЂ« ( x пЂ« 3)( x 2 пЂ« x пЂ« 2) 2
2
xпЂ«3
( x пЂ« 3)( x пЂ« x пЂ« 2)
x пЂ« xпЂ«2
8 x 2 пЂ« 12 x пЂ­ 20 пЂЅ A( x 2 пЂ« x пЂ« 2) пЂ« ( Bx пЂ« C )( x пЂ« 3)
8 x 2 пЂ« 12 x пЂ­ 20 пЂЅ Ax 2 пЂ« Ax пЂ« 2 A пЂ« Bx 2 пЂ« 3Bx пЂ« Cx пЂ« 3C
8 x 2 пЂ« 12 x пЂ­ 20 пЂЅ ( A пЂ« B ) x 2 пЂ« ( A пЂ« 3B пЂ« C ) x пЂ« (2 A пЂ« 3C )
Equating the coefficients of like terms, we obtain a system of equations.
AпЂ« B пЂЅ 8
пѓ¬
пѓЇ
пѓ­ A пЂ« 3B пЂ« C пЂЅ 12
пѓЇ 2 A пЂ« 3C пЂЅ пЂ­20
пѓ®
Multiply the second equation by в€’3 and add to the third equation to obtain в€’A в€’ 9B = в€’56. Add this result to the first
equation in the system above to obtain в€’8B = в€’48, so B = 6. Substituting this value into the first equation, we obtain
A = 2. Substituting the value of A into the third equation, we obtain C = в€’8.
The partial fraction decomposition is
8 x 2 пЂ« 12 x пЂ­ 20
2
6x пЂ­ 8
.
пЂЅ
пЂ« 2
2
( x пЂ« 3)( x пЂ« x пЂ« 2) x пЂ« 3 x пЂ« x пЂ« 2
Pencil Problem #3
3. Find the partial fraction decomposition of
4
5x 2 пЂ« 6 x пЂ« 3
.
( x пЂ« 1)( x 2 пЂ« 2 x пЂ« 2)
Copyright В© 2014 Pearson Education Inc.
Section 5.3
Objective #4:
P
, where Q has a prime, repeated quadratic factor
Q
Solved Problem #4
4. Find the partial fraction decomposition of
2 x3 пЂ« x пЂ« 3
.
( x 2 пЂ« 1)2
Include one fraction with a linear numerator for each power of x 2 пЂ« 1.
2 x 3 пЂ« x пЂ« 3 Ax пЂ« B Cx пЂ« D
пЂЅ 2
пЂ«
( x 2 пЂ« 1)2
x пЂ« 1 ( x 2 пЂ« 1)2
Multiply by the LCD, ( x 2 пЂ« 1)2 , to eliminate fractions. Then simplify and rearrange terms.
( x 2 пЂ« 1)2
Ax пЂ« B
Cx пЂ« D
2 x3 пЂ« x пЂ« 3
пЂЅ ( x 2 пЂ« 1)2 2
пЂ« ( x 2 пЂ« 1)2 2
2
2
x пЂ«1
( x пЂ« 1)
( x пЂ« 1)2
2 x 3 пЂ« x пЂ« 3 пЂЅ ( Ax пЂ« B )( x 2 пЂ« 1) пЂ« Cx пЂ« D
2 x 3 пЂ« x пЂ« 3 пЂЅ Ax 3 пЂ« Ax пЂ« Bx 2 пЂ« B пЂ« Cx пЂ« D
2 x 3 пЂ« x пЂ« 3 пЂЅ Ax 3 пЂ« Bx 2 пЂ« ( A пЂ« C ) x пЂ« ( B пЂ« D )
Equating the coefficients of like terms, we obtain a system of equations.
AпЂЅ2
пѓ¬
пѓЇ
B
пЂЅ0
пѓЇ
пѓ­
пѓЇ AпЂ«C пЂЅ1
пѓЇпѓ® B пЂ« D пЂЅ 3
We immediately see that A = 2 and B = 0. By performing appropriate substitutions, we obtain C = в€’1 and D = 3. The
partial fraction decomposition is
2 x3 пЂ« x пЂ« 3
2x
пЂ­x пЂ« 3
.
пЂЅ 2
пЂ« 2
2
2
( x пЂ« 1)
x пЂ« 1 ( x пЂ« 1) 2
Copyright В© 2014 Pearson Education Inc.
5
College Algebra 6e
Pencil Problem #4
4.
Find the partial fraction decomposition of
x3 пЂ« x2 пЂ« 2
.
( x 2 пЂ« 2)2
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1.
2.
3.
4.
6
7
4
(5.3 #11)
пЂ­
xпЂ­9 xпЂ«2
1
3
1
пЂ«
пЂ«
4( x пЂ« 1) 4( x пЂ­ 1) 2( x пЂ­ 1) 2
(5.3 #27)
2
3x пЂ­ 1
пЂ«
(5.3 #31)
x пЂ« 1 x2 пЂ« 2 x пЂ« 2
x пЂ«1
2x
(5.3 #37)
пЂ­ 2
2
x пЂ« 2 ( x пЂ« 2) 2
Copyright В© 2014 Pearson Education Inc.
Section 5.4
Systems of Nonlinear Equations in Two Variables
Do you feel SAFE????
Scientists debate the probability that a “doomsday rock”
will collide with Earth. It has been estimated that
an asteroid crashes into Earth about once every
250,000 years, and that such a collision would
have disastrous results.
Understanding the path of Earth and the path of a comet
is essential to detecting threatening space debris.
Orbits about the sun are not described by linear equations. The ability to
solve systems that do not contain linear equations provides NASA scientists
watching for troublesome asteroids with a way to locate
possible collision points with Earth’s orbit.
Objective #1: Recognize systems of nonlinear equations in two variables.
Solved Problem #1
1a. True or false: A solution of a nonlinear system in
two variables is an ordered pair of real numbers that
satisfies at least one equation in the system.
Pencil Problem #1
1a. True or false: A system of nonlinear equations
cannot contain a linear equation.
False; a solution must satisfy all equations in the system.
1b. True or false: The solution of a system of nonlinear
equations corresponds to the intersection points of
the graphs in the system.
1b. True or false: The graphs of the equations in a
nonlinear system could be a parabola and a circle.
True; each solution will correspond to an intersection
point of the graphs.
Copyright В© 2014 Pearson Education Inc.
1
College Algebra 6e
Objective #2: Solve nonlinear systems by substitution.
Solved Problem #2
2.
Pencil Problem #2
Solve by the substitution method:
x пЂ« 2y пЂЅ 0
пѓ¬пѓЇ
пѓ­
2
2
пѓЇпѓ®( x пЂ­ 1) пЂ« ( y пЂ­ 1) пЂЅ 5
2.
Solve by the substitution method:
пѓ¬пѓЇ x пЂ« y пЂЅ 2
пѓ­
2
пѓЇпѓ® y пЂЅ x пЂ­ 4 x пЂ« 4
Solve the first equation for x: x пЂ« 2 y пЂЅ 0
x пЂЅ пЂ­2 y
Substitute the expression –2y for x in the second equation
and solve for y.
( x пЂ­ 1)2 пЂ« ( y пЂ­ 1)2 пЂЅ 5
x
пЃЅ
(пЂ­2 y пЂ­ 1)2 пЂ« ( y пЂ­ 1)2 пЂЅ 5
4 y2 пЂ« 4 y пЂ«1пЂ« y2 пЂ­ 2 y пЂ«1 пЂЅ 5
5y2 пЂ« 2 y пЂ­ 3 пЂЅ 0
(5 y пЂ­ 3)( y пЂ« 1) пЂЅ 0
5 y пЂ­ 3 пЂЅ 0 or y пЂ« 1 пЂЅ 0
yпЂЅ
3
5
y пЂЅ пЂ­1
or
If y пЂЅ 53 , x пЂЅ пЂ­2
пЂЁ 53 пЂ© пЂЅ пЂ­ 65 .
If y пЂЅ пЂ­1, x пЂЅ пЂ­2( пЂ­1) пЂЅ 2.
пЂЁ 2, пЂ­ 1пЂ©
Check
in both original equations.
x пЂ«2y пЂЅ 0
2 пЂ« 2( пЂ­1) пЂЅ 0
0 пЂЅ 0, true
пЂЁ
Check пЂ­ 65 , 53
( x пЂ­ 1) 2 пЂ« ( y пЂ­ 1)2 пЂЅ 5
пЂЁ 53 пЂ© пЂЅ 0
пЂЁпЂ­ 65 пЂ­ 1пЂ© пЂ« пЂЁ 35 пЂ­ 1пЂ©
2
пЂ­ пЂ« пЂЅ0
6
5
(2 пЂ­ 1)2 пЂ« (пЂ­1 пЂ­ 1)2 пЂЅ 5
1пЂ« 4 пЂЅ 5
5 пЂЅ 5, true
пЂ© in both original equations.
x пЂ« 2y пЂЅ 0
пЂ­ 65 пЂ« 2
( x пЂ­ 1)2 пЂ« ( y пЂ­ 1)2 пЂЅ 5
6
5
121
25
0 пЂЅ 0, true
2
пЂЅ5
4
пЂ« 25
пЂЅ5
125
25
пЂЅ5
5 пЂЅ 5, true
The solution set is
пЃ»пЂЁпЂ­
6 3
,
5 5
пЂ© , пЂЁ2, пЂ­ 1пЂ©пЃЅ.
Objective #3: Solve nonlinear systems by addition.
2
Copyright В© 2014 Pearson Education Inc.
Section 5.4
Solved Problem #3
3.
Solve by the addition method:
пѓ¬пѓЇ
y пЂЅ x2 пЂ« 5
пѓ­ 2
2
пѓЇпѓ® x пЂ« y пЂЅ 25
Pencil Problem #3
3.
Solve by the addition method:
пѓ¬пѓЇ x 2 пЂ« y 2 пЂЅ 13
пѓ­ 2
2
пѓЇпѓ® x пЂ­ y пЂЅ 5
Arrange the first equation so that variable terms appear
on the left, and constants appear on the right.
Add the resulting equations to eliminate the x 2 -terms
and solve for y.
пЂ­ x2 пЂ« y пЂЅ 5
x 2 пЂ« y 2 пЂЅ 25
y 2 пЂ« y пЂЅ 30
Solve the resulting quadratic equation.
y 2 пЂ« y пЂ­ 30 пЂЅ 0
( y пЂ« 6)( y пЂ­ 5) пЂЅ 0
y пЂ«6 пЂЅ 0
or
yпЂ­5пЂЅ 0
y пЂЅ пЂ­6 or
yпЂЅ5
If y пЂЅ пЂ­6,
x 2 пЂ« (пЂ­6)2 пЂЅ 25
x 2 пЂ« 36 пЂЅ 25
x 2 пЂЅ пЂ­11
When y пЂЅ пЂ­6 there is no real solution.
If y пЂЅ 5,
x 2 пЂ« (5) 2 пЂЅ 25
x 2 пЂ« 25 пЂЅ 25
x2 пЂЅ 0
xпЂЅ0
Check (0,5) in both original equations.
y пЂЅ x2 пЂ« 5
x 2 пЂ« y 2 пЂЅ 25
5 пЂЅ (0) 2 пЂ« 5
5 пЂЅ 5, true
02 пЂ« 52 пЂЅ 25
25 пЂЅ 25, true
The solution set is пЃ»(0,5)пЃЅ .
Objective #4: Solve problems using systems of nonlinear equations.
Copyright В© 2014 Pearson Education Inc.
3
College Algebra 6e
Solved Problem #4
4.
Pencil Problem #4
Find the length and width of a rectangle whose
perimeter is 20 feet and whose area is 21 square feet.
4.
The sum of two numbers is 10 and their product
is 24. Find the numbers.
пѓ¬2 x пЂ« 2 y пЂЅ 20
The system is пѓ­
xy пЂЅ 21.
пѓ®
Solve the second equation for x: xy пЂЅ 21
xпЂЅ
Substitute the expression
21
y
21
for x in the first equation
y
and solve for y.
2 x пЂ« 2 y пЂЅ 20
пѓ¦ 21 пѓ¶
2 пѓ§ пѓ· пЂ« 2 y пЂЅ 20
пѓЁ yпѓё
42
пЂ« 2 y пЂЅ 20
y
42 пЂ« 2 y 2 пЂЅ 20 y
2 y 2 пЂ­ 20 y пЂ« 42 пЂЅ 0
y 2 пЂ­ 10 y пЂ« 21 пЂЅ 0
( y пЂ­ 7)( y пЂ­ 3) пЂЅ 0
y пЂ­ 7 пЂЅ 0 or
yпЂ­3пЂЅ 0
y пЂЅ 7 or
yпЂЅ3
21
пЂЅ 3.
7
21
If y пЂЅ 3, x пЂЅ
пЂЅ 7.
3
If y пЂЅ 7, x пЂЅ
The dimensions are 7 feet by 3 feet.
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1a. false (5.4 #1)
3.
4
1b. true (5.4 #27)
пЃ»пЂЁпЂ­3, пЂ­2пЂ© , пЂЁ пЂ­3, 2пЂ© , пЂЁ3, пЂ­2пЂ© , пЂЁ3, 2пЂ©пЃЅ
2.
(5.4 #19)
пЃ»пЂЁ1,1пЂ© , пЂЁ2,0пЂ©пЃЅ
(5.4 #3)
4. 6 and 4 (5.4 #43)
Copyright В© 2014 Pearson Education Inc.
Section 5.5
Systems of Inequalities
Does Your Weight Fit You?
This chapter opened by noting that the modern emphasis on thinness as the ideal body shape has
been suggested as a major cause of eating disorders. In this section, the
textbook will demonstrate how systems of linear inequalities in two variables can enable you to
establish a healthy weight range for your height and age.
Objective #1: Graph a linear inequality in two variables.
Solved Problem #1
1a. Graph: 4 x − 2 y ≥ 8 .
Pencil Problem #1
1a. Graph: x в€’ 2 y > 10 .
First, graph the equation 4 x в€’ 2 y = 8 with a solid line.
Find the x –intercept:
4x в€’ 2 y = 8
4 x в€’ 2(0) = 8
4x = 8
x=2
Find the y –intercept:
4x в€’ 2 y = 8
4(0) в€’ 2 y = 8
в€’2 y = 8
y = в€’4
Next, use the origin as a test point.
4x − 2 y ≥ 8
4(0) − 2(0) ≥ 8
0 ≥ 8, false
Since the statement is
false, shade the halfplane that does not
contain the test point.
Copyright В© 2013 Pearson Education Inc.
1
College Algebra 6e
1b. Graph: x ≤ 1 .
1b. Graph: y > 1 .
Graph the line y = 1 with a dashed line.
Since the inequality is of the form y > a, shade the halfplane above the line.
Objective #2: Graph a nonlinear inequality in two variables..
Solved Problem #2
2.
Graph: x 2 пЂ« y 2 п‚і 16.
Pencil Problem #2
2. Graph: x 2 пЂ« y 2 пЂѕ 25.
The graph of x 2 пЂ« y 2 пЂЅ 16 is a circle of radius 4 centered
at the origin. Use a solid circle because equality is
included in ≥.
The point (0, 0) is not on the circle, so we use it as a test
point. The result is 0 ≥ 16, which is false. Since the point
(0, 0) is inside the circle, the region outside the circle
belongs to the solution set. Shade the region outside the
circle.
2
Copyright В© 2013 Pearson Education Inc.
Section 5.5
Objective #3: Use mathematical models involving systems of linear inequalities.
Solved Problem #3
3.
The healthy weight region for men and women ages
19 to 34 can be modeled by the following system of
linear inequalities:
Pencil Problem #3
3.
The healthy weight region for men and women ages
35 and older can be modeled by the following
system of linear inequalities:
 4.9 x − y ≥ 165
пѓ­
3.7 x − y ≤ 125
5.3x − y ≥ 180
пѓ­
 4.1x − y ≤ 14
Show that (66, 130) is a solution of the system of
inequalities that describes healthy weight for this
age group.
Show that (66, 160) is a solution of the system of
inequalities that describes healthy weight for this
age group.
Substitute the coordinates of (66, 130) into both
inequalities of the system.
 4.9 x − y ≥ 165
пѓ­
3.7 x − y ≤ 125
4.9 x − y ≥ 165
4.9(66) − 130 ≥ 165
193.4 ≥ 165, true
3.7 x − y ≤ 125
3.7(66) − 130 ≤ 125
114.2 ≤ 125, true
(66, 130) is a solution of the system.
Objective #4: Graph a system of linear inequalities.
Solved Problem #4
4.
Graph the solution set of the system:
пѓ¬ x в€’ 3y < 6
пѓ­
 2 x + 3 y ≥ −6
Pencil Problem #4
4.
Graph the solution set of the system:
пѓ¬ y > 2x в€’ 3
пѓ­
пѓ® y < в€’x + 6
Graph the line x в€’ 3 y = 6 with a dashed line.
Graph the line 2 x + 3 y = в€’6 with a solid line.
For x в€’ 3 y < 6 use a test point such as (0, 0).
x в€’ 3y < 6
0 в€’ 3(0) < 6
0 < 6, true
Since the statement is true, shade the half-plane that
contains the test point.
Copyright В© 2014 Pearson Education Inc.
3
College Algebra 6e
For 2 x + 3 y ≥ −6 use a test point such as (0, 0).
2 x + 3 y ≥ −6
2(0) + 3(0) ≥ −6
0 ≥ −6, true
Since the statement is true, shade the half-plane that
contains the test point.
The solution set
of the system is
the intersection
(the overlap) of
the two halfplanes.
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1a.
(5.5 #3)
2.
(5.5 #15)
5.3 x − y ≥ 180
5.3(66) − 160 ≥ 180
3.
189.8 ≥ 180, true
1b.
(5.5 #9)
4.1x − y ≤ 14
4.1(66) − 160 ≤ 140
110.6 ≤ 140, true
(5.5 #77)
4.
4
Copyright В© 2013 Pearson Education Inc.
Section 5.6
Linear Programming
MAXIMUM Output with MINIMUM Effort!
Many situations in life involve quantities that must be maximized or minimized.
Businesses are interested in maximizing profit and minimizing costs.
In the Exercise Set for this section of the textbook, you
will encounter a manufacturer looking to maximize
profits from selling two models of mountain bikes.
But there is a limited amount of time available to
assemble and paint these bikes. We will learn how
to balance these constraints and determine the proper
number of each bike that should be produced.
Objective #1:
Write an objective function describing a quantity that must be maximized or minimized.
Solved Problem #1
1.
A company manufactures bookshelves and desks
for computers. Let x represent the number of
bookshelves manufactured daily and y the number
of desks manufactured daily. The company’s profits
are $25 per bookshelf and $55 per desk.
Write the objective function that models the
company’s total daily profit, z, from x bookshelves
and y desks.
Pencil Problem #1
1.
A television manufacturer makes rear-projection
and plasma televisions. The profit per unit is $125
for the rear-projection televisions and $200 for the
plasma televisions. Let x = the number of rearprojection televisions manufactured in a month and
y = the number of plasma televisions manufactured
in a month.
Write the objective function that models the total
monthly profit.
The total profit is 25 times the number of bookshelves, x,
plus 55 times the number of desks, y.
profit
пЃЅ
z =
$25 for each
bookshelf
пЃЅ
25 x
$55 for each
desk
+
пЃЅ
55 y
The objective function is z = 25 x + 55 y .
Copyright В© 2014 Pearson Education Inc.
1
College Algebra 6e
Objective #2: Use inequalities to describe limitations in a situation.
Solved Problem #2
2.
Recall that the company in Solved Problem #1
manufactures bookshelves and desks for computers.
x represents the number of bookshelves
manufactured daily and y the number of desks
manufactured daily.
2a. Write an inequality that models the following
constraint: To maintain high quality, the company
should not manufacture more than a total of 80
bookshelves and desks per day.
Pencil Problem #2
2.
Recall that the manufacturer in Pencil Problem #1
makes rear-projection and plasma televisions.
x represents the number of rear-projection
televisions manufactured monthly and y the number
of plasma televisions manufactured monthly.
2a. Write an inequality that models the following
constraint: Equipment in the factory allows for
making at most 450 rear-projection televisions in
one month.
x + y ≤ 80
2b. Write an inequality that models the following
constraint: To meet customer demand, the company
must manufacture between 30 and 80 bookshelves
per day, inclusive.
2b. Write an inequality that models the following
constraint: Equipment in the factory allows for
making at most 200 plasma televisions in one
month.
30 ≤ x ≤ 80
2c. Write an inequality that models the following
constraint: The company must manufacture at least
10 and no more than 30 desks per day.
2c. Write an inequality that models the following
constraint: The cost to the manufacturer per unit is
$600 for the rear-projection televisions and $900
for the plasma televisions. Total monthly costs
cannot exceed $360,000.
10 ≤ y ≤ 30
2d. Summarize what you have described about this
company by writing the objective function for its
profits (from Solved Problem #1) and the three
constraints.
2d. Summarize what you have described about this
company by writing the objective function for its
profits (from Pencil Problem #1) and the three
constraints.
Objective function: z = 25 x + 55 y .
 x + y ≤ 80
пѓЇ
Constraints: 30 ≤ x ≤ 80
10 ≤ y ≤ 30
пѓ®
2
Copyright В© 2014 Pearson Education Inc.
Section 5.6
Objective #3: Use linear programming to solve problems.
Solved Problem #3
3a. For the company in Solved Problems #1 and 2, how
many bookshelves and how many desks should be
manufactured per day to obtain maximum profit?
What is the maximum daily profit?
Pencil Problem #3
3a. For the company in Pencil Problems #1 and 2, how
many rear-projection and plasma televisions should
be manufactured per month to obtain maximum
profit? What is the maximum monthly profit?
Graph the constraints and find the corners, or vertices, of
the region of intersection.
Find the value of the objective function at each corner of
the graphed region.
Corner
Objective Function
( x, y )
z = 25 x + 55 y
z = 25(30) + 55(10)
(30,10)
= 750 + 550
= 1300
z = 25(30) + 55(30)
(30,30)
= 750 + 1650
= 2400
z = 25(50) + 55(30)
(50,30)
= 1250 + 1650
= 2900
(Maximum)
z = 25(70) + 55(10)
(70,10)
= 1750 + 550
= 2300
The maximum value of z is 2900 and it occurs at the
point (50, 30).
In order to maximize profit, 50 bookshelves and 30 desks
must be produced each day for a profit of $2900.
Copyright В© 2014 Pearson Education Inc.
3
College Algebra 6e
3b. Find the maximum value of the objective function
z = 3 x + 5 y subject to the constraints:
 x ≥ 0, y ≥ 0
пѓЇ
x + y ≥ 1
x + y ≤ 6
пѓ®
3b. Find the maximum value of the objective function
z = 4 x + y subject to the constraints:
 x ≥ 0, y ≥ 0
пѓЇ
 2 x + 3 y ≤ 12
x + y ≥ 3
пѓ®
Graph the region that represents the intersection of the
constraints:
Find the value of the objective function at each corner of
the graphed region.
Corner Objective Function
( x, y )
z = 3x + 5 y
(0,1)
z = 3(0) + 5(1) = 5
(1, 0)
(0, 6)
(6, 0)
z = 3(1) + 5(0) = 3
z = 3(0) + 5(6) = 30
(Maximum)
z = 3(6) + 5(0) = 18
The maximum value is 30.
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1. z = 125 x + 200 y
(5.6 #15a)
2a. x ≤ 450 (5.6 #15b)
2c. 600 x + 900 y ≤ 360, 000 (5.6 #15b)
2b. y ≤ 200 (5.6 #15b)
 x ≤ 450
пѓЇ
2d. z = 125 x + 200 y ;  y ≤ 200
(5.6 #15b)
600 x + 900 y ≤ 360, 000
пѓ®
3a. 300 rear-projection and 200 plasma televisions; Maximum profit: $77,500 (5.6 #15e)
4
Copyright В© 2014 Pearson Education Inc.
3b. 24 (5.6 #7)
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