1
Equations and
Inequalities
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
1
1.5
Applications and Modeling with
Quadratic Equations
• Geometry Problems
• The Pythagorean Theorem
• Height of a Projected Object
• Modeling with Quadratic Equations
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Example 1
SOLVING A PROBLEM INVOLVING
VOLUME
A piece of machinery produces rectangular
sheets of metal such that the length is three
times the width. Equal-sized squares
measuring 5 in. on a side can be cut from the
corners so that the resulting piece of metal
can be shaped into an open box by folding up
the flaps. If specifications call for the volume
of the box to be 1435 in.3, find the dimensions
of the original piece of metal.
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3
Example 1
SOLVING A PROBLEM INVOLVING
VOLUME
Solution
Step 1 Read the problem. We must find the
dimensions of the original piece of metal.
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Example 1
SOLVING A PROBLEM INVOLVING
VOLUME
Solution
Step 2 Assign a variable. We know that the length
is three times the width. Let x = the width (in inches)
and thus, 3x = the length.
The box is formed by cutting 5 + 5 = 10 in. from both
the length and the width.
3x
5
3x – 10
x – 10
x
5 5
3x – 10
55
x – 10
5 5
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5
SOLVING A PROBLEM INVOLVING
VOLUME
Example 1
Solution
Step 3 Write an equation. The formula for the
volume of a box is V = lwh.
Volume
=
length
width height
(Note that the dimensions of the box must be positive
numbers, so 3x – 10 and x – 10 must be
greater than 0, which implies
and
These are both satisfied when x > 10.)
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Example 1
SOLVING A PROBLEM INVOLVING
VOLUME
Solution
Step 4 Solve the equation.
Multiply.
Subtract 1435 from
each side.
Divide each side by
5.
Factor.
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Example 1
SOLVING A PROBLEM INVOLVING
VOLUME
Solution
Step 4 Solve the equation.
or
Zero-factor property
or
Solve each equation.
The width
cannot be
negative.
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Example 1
SOLVING A PROBLEM INVOLVING
VOLUME
Solution
Step 5 State the answer. Only 17
satisfies the restriction x > 10. Thus, the
dimensions of the original piece should be
17 in. by 3(17) = 51 in.
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Example 1
SOLVING A PROBLEM INVOLVING
VOLUME
Solution
Step 6 Check. The length of the bottom of
the box is 51 – 2(5) = 41 in. The width is
17 – 2(5) = 7 in. The height is 5 in. (the
amount cut on each corner), so the volume
of the box is
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Using the Pythagorean Theorem
Example 2 requires the use of the
Pythagorean theorem for right triangles.
Recall that the legs of a right triangle form
the right angle, and the hypotenuse is the
side opposite the right angle.
Leg a
Hypotenuse
c
Leg b
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Pythagorean Theorem
In a right triangle, the sum of the
squares of the lengths of the legs is
equal to the square of the length of the
hypotenuse.
Hypotenuse
c
Leg a
Leg b
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Example 2
APPLYING THE PYTHAGOREAN
THEOREM
A piece of property has the shape of a right
triangle. The longer leg is 20 m longer than
twice the length of the shorter leg. The
hypotenuse is 10 m longer than the length of
the longer leg. Find the lengths of the sides
of the triangular lot.
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Example 2
APPLYING THE PYTHAGOREAN
THEOREM
Solution
Step 1 Read the problem. We must find the
lengths of the three sides.
Step 2 Assign a variable.
Let x = the length of the shorter leg (in meters).
Then 2x + 20 = the length of the longer leg, and
(2x + 20) + 10 or 2x + 30 = the length of the
hypotenuse.
x
2x + 30
2x + 20
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Example 2
APPLYING THE PYTHAGOREAN
THEOREM
Solution
Step 3 Write an equation.
The
hypotenuse is
c.
Substitute into the Pythagorean theorem.
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APPLYING THE PYTHAGOREAN
THEOREM
Example 2
Solution
Step 4 Solve the equation.
Remember the
middle term
here.
Remember the
middle term
here.
Square the binomials.
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APPLYING THE PYTHAGOREAN
THEOREM
Example 2
Solution
Step 4 Solve the equation.
Standard form
Factor.
or
or
Zero-factor property
Solve each equation.
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Example 2
APPLYING THE PYTHAGOREAN
THEOREM
Solution
Step 5 State the answer. Since x represents
a length, – 10 is not reasonable. The lengths
of the sides of the triangular lot are 50 m,
2(50) + 20 = 120 m, and 2(50) + 30 = 130 m.
Step 6 Check. The lengths 50, 120, and 130
satisfy the words of the problem, and also
satisfy the Pythagorean theorem.
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Height of a Projected Object
If air resistance is neglected, the height s (in
feet) of an object projected directly upward from
an initial height of s0 feet, with initial velocity v0
feet per second is given by the equation
Here t represents the number of seconds after
the object is projected. The coefficient of t 2,
– 16, is a constant based on the gravitational
force of Earth. This constant varies on other
surfaces, such as the moon and other planets.
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Example 3
SOLVING A PROBLEM INVOLVING
PROJECTILE HEIGHT
If a projectile is launched vertically upward
from the ground with an initial velocity of 100
ft per sec, neglecting air resistance, its
height s (in feet) above the ground t
seconds after projection is given by
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Example 3
SOLVING A PROBLEM INVOLVING
PROJECTILE HEIGHT
(a) After how many seconds will it be 50 ft
above the ground?
Solution
We must find value(s) of t so that height s is
50 ft.
Let s = 50 in the given
equation.
Standard form
Divide by −2.
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Example 3
SOLVING A PROBLEM INVOLVING
PROJECTILE HEIGHT
(a) After how many seconds will it be 50 ft
above the ground?
Solution
Quadratic formula
Simplify.
or
Use a calculator.
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Example 3
SOLVING A PROBLEM INVOLVING
PROJECTILE HEIGHT
(a) After how many seconds will it be 50 ft
above the ground?
Solution
or
Both solutions are acceptable, since the
projectile reaches 50 ft twice—once on its
way up (after 0.55 sec) and once on its way
down (after 5.70 sec).
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Example 3
SOLVING A PROBLEM INVOLVING
PROJECTILE HEIGHT
(b) How long will it take for the projectile to
return to the ground?
Solution When the projectile returns to the
ground, the height s will be 0 ft.
Let s = 0.
Factor.
or
Zero-factor property
or
Solve each
equation.
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Example 3
SOLVING A PROBLEM INVOLVING
PROJECTILE HEIGHT
(b) How long will it take for the projectile to
return to the ground?
Solution
The first solution, 0, represents the time at
which the projectile was on the ground prior to
being launched, so it does not answer the
question. The projectile will return to the
ground 6.25 sec after it is launched.
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Example 4
ANALYZING TROLLEY
RIDERSHIP
The I-Ride Trolley service
carries passengers along the
International Drive Resort Area
of Orlando, Florida. The bar
graph shows I-Ride Trolley
ridership in millions. The
quadratic equation
models ridership from 2000 to 2013, where y
represents ridership in millions and x = 0
represents 2000, x = 1 represents 2001, and so on.
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Example 4
ANALYZING TROLLEY
RIDERSHIP
(a) Use the model to determine ridership in
2011. Compare the result to the actual
ridership figure of 2.1 million.
Solution
Since x = 0 represents the year 2000, use x = 11
to represent 2011.
Given model
Let x = 11.
Use a calculator.
The prediction is about 0.1 million (that is, 100,000)
less than the actual figure of 2.1 million.
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ANALYZING TROLLEY
RIDERSHIP
Example 4
(b) According to the model, in what year did
ridership reach 1.8 million?
Solution
Let y = 1.9 in
the model.
Standard form
Quadratic formula
or
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Use a calculator.
28
Example 4
ANALYZING TROLLEY
RIDERSHIP
(b) According to the model, in what year did
ridership reach 1.8 million?
Solution
The year 2002 corresponds to x = 2.0. Thus,
according to the model, ridership reached 1.8
million in the year 2002. This outcome closely
matches the bar graph and seems reasonable.
The year 2015 corresponds to x = 15.4. The
model predicts that ridership will be 1.8 million
again in the year 2015.
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1
Equations and
Inequalities
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
1
1.4
Quadratic Equations
• The Zero-Factor Property
• The Square Root Property
• Completing the Square
• The Quadratic Formula
• Solving for a Specified Variable
• The Discriminant
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Quadratic Equation in One
Variable
An equation that can be written in the
form
where a, b, and c are real numbers
with a ≠ 0, is a quadratic equation.
The given form is called standard
form.
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Second-degree Equation
A quadratic equation is a second-degree
equation—that is, an equation with a
squared variable term and no terms of
greater degree.
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Zero-Factor Property
If a and b are complex numbers with
ab = 0, then a = 0 or b = 0 or both
equal zero.
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Example 1
USING THE ZERO-FACTOR
PROPERTY
Solve
Solution
Standard form
Factor.
Zero-factor property
Solve each equation.
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Square Root Property
A quadratic equation of the form x2 = k can
also be solved by factoring.
Subtract k.
Factor.
Zero-factor property.
Solve each equation.
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Square Root Property
If x2 = k, then
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Square-Root Property
That is, the solution set of
which may be abbreviated
Both solutions are real if k > 0, and both are
pure imaginary if k < 0. If k < 0, we write the
solution set as
If k = 0, then there is only one distinct solution,
0, sometimes called a double solution.
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USING THE SQUARE ROOT
PROPERTY
Example 2
Solve each quadratic equation.
(a)
Solution
By the square root property, the solution set
of
is
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Example 2
USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
(b)
Solution
Since
the solution set of x2 = −25
is
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Example 2
USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
(c)
Solution
Generalized square
root property
Add 4.
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Solving a Quadratic Equation
by Completing the Square
To solve ax2 + bx + c = 0, where a ≠ 0, by completing the
square, use these steps.
Step 1 If a ≠ 1, divide both sides of the equation by a.
Step 2 Rewrite the equation so that the constant term is
alone on one side of the equality symbol.
Step 3 Square half the coefficient of x, and add this square
to each side of the equation.
Step 4 Factor the resulting trinomial as a perfect square
and combine like terms on the other side.
Step 5 Use the square root property to complete the
solution.
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Example 3
USING COMPLETING THE
SQUARE (a = 1)
Solve x2 – 4x – 14 = 0.
Solution x2 – 4x – 14 = 0
Step 1 This step is not necessary since a = 1.
Step 2
Add 14 to each side.
Step 3
add 4 to each side.
Step 4
Factor. Combine
like terms.
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Example 3
USING COMPLETING THE
SQUARE (a = 1)
Solve x2 – 4x – 14 = 0.
Solution
Step 5
Square root property.
Take both
roots.
Add 2 to each side.
Simplify the radical.
The solution set is
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Example 4
USING COMPLETING THE
SQUARE (a ≠ 1)
Solve 9x2 – 12x + 9 = 0.
Solution
Divide by 9. (Step 1)
Subtract 1 from each
side. (Step 2)
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Example 4
USING COMPLETING THE
SQUARE (a ≠ 1)
Solve 9x2 – 12x + 9 = 0.
Solution
Factor. Combine like
terms. (Step 4)
Square root property
(Step 5)
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Example 4
USING COMPLETING THE
SQUARE (a ≠ 1)
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
Add
to each side.
The solution set is
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The Quadratic Formula
If we start with the equation ax2 + bx + c = 0,
for a > 0, and complete the square to solve
for x in terms of the constants a, b, and c,
the result is a general formula for solving
any quadratic equation.
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Quadratic Formula
The solutions of the quadratic equation
ax2 + bx + c = 0, where a ≠ 0, are given
by the quadratic formula.
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Caution Remember to extend the
fraction bar in the quadratic formula
extends under the – b term in the
numerator.
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Example 5
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2.
Solution
Write in standard
form. Here a = 1,
b = – 4, c = 2.
Quadratic formula
The fraction
bar extends
under – b.
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Example 5
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2
Solution
Simplify.
Factor out 2 in the
numerator.
Factor first,
then divide.
Lowest terms
The solution set is
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Example 6
USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution
Write in standard form.
Quadratic formula
with a = 2, b = – 1,
c=4
Use parentheses and
substitute carefully to
avoid errors.
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Example 6
USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution
Simplify.
The solution set is
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Cubic Equation
The equation x3 + 8 = 0 is called a cubic
equation because the greatest degree of
the terms is 3.
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Example 7
SOLVING A CUBIC EQUATION
Solve
using factoring and the
quadratic formula.
Solution
Factor as a sum of
cubes.
or
Zero-factor property
or
Quadratic formula with a = 1, b = – 2, c = 4
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Example 7
SOLVING A CUBIC EQUATION
Solve
Solution
Simplify.
Simplify the radical.
Factor out 2 in the
numerator.
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Example 7
SOLVING A CUBIC EQUATION
Solve
Solution
Lowest terms
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Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve each equation for the specified variable.
Use when taking square roots.
(a)
Solution
Goal: Isolate d,
the specified
variable.
Multiply each side by 4.
Divide each side by .
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Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve each equation for the specified variable.
Use when taking square roots.
(a)
Solution
Square root property
See the Note
following this
example.
Multiply by
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Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve each equation for the specified variable.
Use when taking square roots.
(a)
Solution
Multiply numerators.
Multiply denominators.
Simplify the radical.
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Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve each equation for the specified variable.
Use when taking square roots.
(b)
Solution
Because rt2 – st = k has terms with t2 and
t, use the quadratic formula.
Write in standard form.
Quadratic formula
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Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve each equation for the specified variable.
Use when taking square roots.
(b)
Solution
Here, a = r,
b = – s, and
c = – k.
Simplify.
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Solving for a Specified Variable
Note In Example 8, we took both
positive and negative square roots.
However, if the variable represents time or
length in an application, we consider only
the positive square root.
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The Discriminant
The Discriminant The quantity under the
radical in the quadratic formula, b2 – 4ac, is
called the discriminant.
Discriminant
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The Discriminant
Number of
Solutions
Type of
Solutions
Positive, perfect
square
Two
Rational
Positive, but not
a perfect square
Two
Irrational
One
(a double solution)
Rational
Two
Nonreal
complex
Discriminant
Zero
Negative
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Caution The restriction on a, b, and
c is important. For example,
has discriminant b2 – 4ac = 5 + 4 = 9,
which would indicate two rational solutions
if the coefficients were integers. By the
quadratic formula, the two solutions
are irrational numbers.
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Example 9
USING THE DISCRIMINANT
Evaluate the discriminant for each equation. Then use it to
determine the number of distinct solutions, and tell whether
they are rational, irrational, or nonreal complex numbers.
(a)
Solution
For 5x2 + 2x – 4 = 0, use a = 5, b = 2, and
c = – 4.
b2 – 4ac = 22 – 4(5)(– 4) = 84
The discriminant 84 is positive and not a perfect
square, so there are two distinct irrational solutions.
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Example 9
USING THE DISCRIMINANT
Evaluate the discriminant for each equation. Then use it to
determine the number of distinct solutions, and tell whether
they are rational, irrational, or nonreal complex numbers.
(b)
Solution
First, write the equation in standard form as
x2 – 10x + 25 = 0. Thus, a = 1, b = – 10, and
c = 25.
b2 – 4ac =(–10 )2 – 4(1)(25) = 0
There is one distinct rational solution, a double solution.
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Example 9
USING THE DISCRIMINANT
Evaluate the discriminant for each equation. Then use it to
determine the number of distinct solutions, and tell whether
they are rational, irrational, or nonreal complex numbers.
(c)
Solution
For 2x2 – x + 1 = 0, use a = 2, b = –1, and c = 1.
b2 – 4ac = (–1)2 – 4(2)(1) = –7.
There are two distinct nonreal complex solutions.
(They are complex conjugates.)
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1
Equations and
Inequalities
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1
1.3
Complex Numbers
• Basic Concepts of Complex Numbers
• Operations on Complex Numbers
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2
Basic Concepts of Complex
Numbers
There is no real number solution of the
equation
since no real number, when squared, gives
To extend the real number system to include
solutions of equations of this type, the
number i is defined to have the following
property.
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Basic Concepts of Complex
Numbers
If a and b are real numbers, then any
number of the form a + bi is a complex
number.
In the complex number a + bi , a is the real
part and b is the imaginary part.
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Basic Concepts of Complex
Numbers
Two complex numbers a + bi and c + di are
equal provided that their real parts are equal
and their imaginary parts are equal; that is
if and only if
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and
5
Basic Concepts of Complex
Numbers
For complex number a + bi, if b = 0, then
a + bi = a.
Thus, the set of real numbers is a subset of
the set of complex numbers.
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Basic Concepts of Complex
Numbers
If a = 0 and b ≠ 0, the complex number is said
to be a pure imaginary number.
A pure imaginary number, or a number like
7 + 2i with a ≠ 0 and b ≠ 0, is a nonreal
complex number.
A complex number written in the form a + bi (or
a + ib) is in standard form.
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2017, 2009,
2013, 2005
2009 Pearson Education, Inc.
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THE EXPRESSION
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Example 1
WRITING
AS
Write as the product of a real number and i,
using the definition of
(a)
Solution
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Example 1
WRITING
AS
Write as the product of a real number and i,
using the definition of
(b)
Solution
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Example 1
WRITING
AS
Write as the product of a real number and i,
using the definition of
(c)
Solution
Product rule
for radicals
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Operations on Complex Numbers
Products or quotients with negative
radicands are simplified by first rewriting
for a positive number a.
Then the properties of real numbers and the
fact that
are applied
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Operations on Complex Numbers
Caution When working with negative
radicands, use the definition
before using any of the other rules for
radicals.
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Operations on Complex Numbers
Caution In particular, the rule
is valid only when c and d are not both
negative.
is correct,
while
is incorrect.
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FINDING PRODUCTS AND
QUOTIENTS INVOLVING
Example 2
Multiply or divide, as indicated. Simplify each
answer.
(a)
Solution
First write all square
roots in terms of i.
i 2 = −1
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Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING
Multiply or divide, as indicated. Simplify each
answer.
(b)
Solution
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Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING
Multiply or divide, as indicated. Simplify each
answer.
(c)
Solution
Quotient
rule for
radicals
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Example 2
FINDING PRODUCTS AND
QUOTIENTS INVOLVING
Multiply or divide, as indicated. Simplify each
answer.
(d)
Solution
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Example 3
Write
SIMPLIFYING A QUOTIENT
INVOLVING
in standard form a + bi.
Solution
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Example 3
Write
SIMPLIFYING A QUOTIENT
INVOLVING
in standard form a + bi.
Solution
Be sure to
factor before
simplifying
Factor.
Lowest terms
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Addition and Subtraction of
Complex Numbers
For complex numbers a + bi and c + di,
and
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Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference. Write answers
in standard form.
(a)
Solution
Add real
parts.
Add
imaginary
parts.
Commutative, associative,
distributive properties
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Example 4
ADDING AND SUBTRACTING
COMPLEX NUMBERS
Find each sum or difference.
(b)
Solution
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Multiplication of Complex Numbers
The product of two complex numbers is
found by multiplying as though the numbers
were binomials and using the fact that
i2 = – 1, as follows.
FOIL
Distributive property;
i2 = – 1
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Multiplication of Complex
Numbers
For complex numbers a + bi and c + di,
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Example 5
MULTIPLYING COMPLEX
NUMBERS
Find each product.
(a)
Solution
FOIL
Multiply.
Combine like terms;
i2 = −1
Standard form
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27
Example 5
MULTIPLYING COMPLEX
NUMBERS
Find each product.
(b)
Solution
Remember to add twice the
product of the two terms.
Square of a binomial
Multiply.
i 2 = −1
Standard form
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28
Example 5
MULTIPLYING COMPLEX
NUMBERS
Find each product.
(c)
Solution
Product of the sum
and difference of
two terms
i 2 = −1
Multiply.
Standard form
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29
Example 5(c) showed that
The numbers 6 + 5i and 6 – 5i differ only in the
sign of their imaginary parts and are called
complex conjugates. The product of a
complex number and its conjugate is
always a real number. This product is the
sum of the squares of real and imaginary parts.
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30
Property of Complex
Conjugates
For real numbers a and b,
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31
Example 6
DIVIDING COMPLEX NUMBERS
Find each quotient. Write answers in
standard form.
(a)
Solution
Multiply by the complex
conjugate of the
denominator in both
the numerator and the
denominator.
Multiply.
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32
Example 6
DIVIDING COMPLEX NUMBERS
Find each quotient.
(a)
Solution
Combine like terms; i 2 = −1
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33
Example 6
DIVIDING COMPLEX NUMBERS
Find each quotient.
(a)
Solution
Write in lowest terms and standard
form.
Check
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34
Example 6
DIVIDING COMPLEX NUMBERS
Find each quotient. Write answers in
standard form.
(b)
Solution
– i is the conjugate of i.
Multiply.
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35
Example 6
DIVIDING COMPLEX NUMBERS
Find each quotient.
(b)
Solution
i 2 = −1(−1) = 1
Standard form
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
36
Simplifying Powers of i
Powers of i can be simplified using the facts
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
37
Powers of i
and so on.
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38
Powers of i cycle through the same four
outcomes (i, –1, –i, and 1) since i4 has the
same multiplicative property as 1. Also,
any power of i with an exponent that is a
multiple of 4 has value 1. As with real
numbers, i0 = 1.
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39
Example 7
SIMPLIFYING POWERS OF i
Simplify each power of i.
(a)
Solution
Since i4 = 1, write the given power as a
product involving i4.
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40
Example 7
SIMPLIFYING POWERS OF i
Simplify each power of i.
(b)
Solution
Multiply
by 1 in the form of i4 to create the
least positive exponent for i.
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41
1
Equations and
Inequalities
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
1
1.2
Applications and Modeling with
Linear Equations
• Solving Applied Problems
• Geometry Problems
• Motion Problems
• Mixture Problems
• Modeling with Linear Equations
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2
SOLVING AN APPLIED PROBLEM
Step 1 Read the problem carefully until you
understand what is given and what is to be
found.
Step 2 Assign a variable to represent the unknown
value, using diagrams or tables as needed.
Write down what the variable represents. If
necessary, express any other unknown
values in terms of the variable.
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3
SOLVING AN APPLIED PROBLEM
Step 3 Write an equation using the variable
expression(s).
Step 4 Solve the equation.
Step 5 State the answer to the problem. Does it
seem reasonable?
Step 6 Check the answer in the words of the
original problem.
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4
EXAMPLE 1
FINDING THE DIMENSIONS
OF A SQUARE
If the length of each side of a square is
increased by 3 cm, the perimeter of the new
square is 40 cm more than twice the length
of the original square. Find the dimensions
of the original square.
Solution
Step 1 Read the problem. We must find
the length of the original square.
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5
FINDING THE DIMENSIONS
OF A SQUARE
Step 2 Assign a variable. Since the length of a
side of the original square is to be found, let the
variable represent this length.
Original
x
square
EXAMPLE 1
The length of a side of the
new square is 3 cm more
than the length of a side of
the old square.
x
Side is
increased
by 3.
x+3
x+3
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6
EXAMPLE 1
FINDING THE DIMENSIONS
OF A SQUARE
Step 2 Assign a variable.
Write a variable expression for the perimeter
of the new square. The perimeter of a square
is 4 times the length of a side.
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7
EXAMPLE 1
FINDING THE DIMENSIONS
OF A SQUARE
Step 3 Write an equation. Translate
the English sentence that follows into
its equivalent algebraic expression.
The new
perimeter
is
40
more
than
twice the length of a side
of the original square.
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8
EXAMPLE 1
FINDING THE DIMENSIONS
OF A SQUARE
Step 4 Solve the equation.
Distributive property
Subtract 2x and 12.
Divide by 2.
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9
EXAMPLE 1
FINDING THE DIMENSIONS
OF A SQUARE
Step 5 State the answer. Each side of the
original square measures 14 cm.
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10
EXAMPLE 1
FINDING THE DIMENSIONS
OF A SQUARE
Step 6 Check. Go back to the words of the
original problem to see that all necessary
conditions are satisfied.
The length of a side of the new square would
be 14 + 3 = 17 cm.
The perimeter of the new square would be
4(17)= 68 cm.
Twice the length of a side of the original
square would be 2(14) = 28 cm.
Since 40 + 28 = 68, the answer checks.
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11
Motion Problems
Problem-Solving Hint In a motion
problem, the components distance, rate,
and time are denoted by the letters d, r, and
t, respectively. (The rate is also called the
speed or velocity. Here, rate is understood
to be constant.)
and its related forms
and
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12
Example 2
SOLVING A MOTION PROBLEM
Maria and Eduardo are traveling to a
business conference. The trip takes 2 hr
for Maria and 2.5 hr for Eduardo, since he
lives 40 mi farther away. Eduardo travels
5 mph faster than Maria. Find their
average rates.
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13
Example 2
SOLVING A MOTION PROBLEM
Solution
Step 1 Read the
problem. We must
find Maria’s and
Eduardo’s average
rates.
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14
SOLVING A MOTION PROBLEM
Example 2
Step 2 Assign a variable. Since average
rates are to be found, we let the variables
represent one of these rates.
Let x = Maria’s rate. Then x + 5 = Eduardo’s
rate. Summarize the given information in a
table.
Maria
Eduardo
r
x
x+5
t
2
2.5
d
2x
2.5(x + 5)
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
Use
d = rt.
15
Example 2
SOLVING A MOTION PROBLEM
Step 3 Write an equation. Eduardo’s
distance traveled exceeds Maria’s distance by
40 mi. Translate this relationship into its
algebraic form.
Eduardo’s
distance
is
40 more
than Maria’s.
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16
SOLVING A MOTION PROBLEM
Example 2
Step 4 Solve.
Distributive property
Subtract 2x and 12.5.
Divide by 0.5
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17
Example 2
SOLVING A MOTION PROBLEM
Step 5 State the answer. Maria’s rate of
travel is 55 mph, and Eduardo’s rate is
55 + 5 = 60 mph.
Step 6 Check. The diagram shows that the
conditions of the problem are satisfied.
Distance traveled by Maria:
2(55) = 110 mi
150 – 110 = 40
Distance traveled by Eduardo:
2.5(60) = 150 mi
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18
Mixture Problems
Problem-Solving Hint In mixture
problems involving solutions, the rate
(percent) of concentration is multiplied by
the quantity to get the amount of pure
substance present. The concentration
of the final mixture must be between
the concentrations of the two solutions
making up the mixture.
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Example 3
SOLVING A MIXTURE PROBLEM
A chemist needs a 20% solution of alcohol.
She has a 15% solution on hand, as well as
a 30% solution. How many liters of the 15%
solution should she add to 3L of the 30%
solution to obtain her 20% solution?
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Example 3
SOLVING A MIXTURE PROBLEM
Solution
Step 1 Read the problem. We must find
the required number of liters of 15% alcohol
solution.
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21
Example 3
SOLVING A MIXTURE PROBLEM
Step 2 Assign a variable.
Let x = the number of liters of 15% solution
to be added.
Strength
15%
30%
20%
Liters of
Solution
x
3
x+3
Liters of
Pure
Alcohol
0.15x
0.30(3)
0.20(x + 3)
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
Sum
must
equal
22
SOLVING A MIXTURE PROBLEM
Example 3
Step 3 Write an equation. The number of
liters of pure alcohol in the 15% solution plus
the number of liters in the 30% solution must
equal the number of liters in the final 20%
solution.
Liters of pure
alcohol in 15%
+
Liters of pure
alcohol in 30%
=
Liters of pure
alcohol in 20%
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SOLVING A MIXTURE PROBLEM
Example 3
Step 4 Solve.
Distributive property
Subtract 0.60 and 0.15x.
Divide by 0.05.
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Example 3
SOLVING A MIXTURE PROBLEM
Step 5 State the answer. Thus, 6 L of 15%
solution should be mixed with 3 L of 30%
solution, giving 6 + 3 = 9 L of 20% solution.
Step 6 Check. The answer checks since the
amount of alcohol in the two solutions is equal
to the amount of alcohol in the mixture.
Solutions
Mixture
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Mixture Problems
Problem-Solving Hint In mixed
investment problems, multiply each
principal by the interest rate and the time
in years to find the amount of interest
earned.
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Example 4
SOLVING AN INVESTMENT
PROBLEM
An artist has sold a painting for $410,000.
He invests a portion of the money for 6
months at 2.65% and the rest for a year at
2.91%. His broker tells him the two
investments will earn a total of $8761. How
much should be invested at each rate to
obtain that amount of interest?
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27
Example 4
SOLVING AN INVESTMENT
PROBLEM
Solution
Step 1 Read the problem.
We must find the amount
to be invested at each rate.
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28
Example 4
SOLVING AN INVESTMENT
PROBLEM
Step 2 Assign a variable.
Let x = dollar amount invested for 6 months at 2.65%.
410,000 – x = dollar amount invested for 1 yr at 2.91%.
Invested Amount
Interest
Rate (%)
Time
(in years)
Interest Earned
x
2.65
0.5
x(0.0265)(0.5)
410,000 – x
2.91
1
(410,000 – x)(0.0291)(1)
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SOLVING AN INVESTMENT
PROBLEM
Example 4
Step 3 Write an equation. The sum of the
two interest amounts must equal the total
interest earned.
Interest from
2.65%
investment
+
Interest from 2.91%
investment
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=
Total
interest
30
SOLVING AN INVESTMENT
PROBLEM
Example 4
Step 4 Solve.
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31
Example 4
SOLVING AN INVESTMENT
PROBLEM
Step 5 State the answer. The
artist should invest $200,000 at
2.65% for 6 months and
$410,000 – $200,000 = $210,000
at 2.91% for 1 yr to earn $8761 in
interest.
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32
Example 4
SOLVING AN INVESTMENT
PROBLEM
Step 6 Check. The 6-month investment earns
while the 1-yr investment earns
The total amount of interest earned is
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33
Modeling with Linear Equations
A mathematical model is an equation (or
inequality) that describes the relationship
between two quantities.
A linear model is a linear equation.
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34
Example 5
MODELING THE PREVENTION
OF INDOOR POLLUTANTS
If a vented range hood removes contaminants such
as carbon monoxide and nitrogen dioxide from the air
at a rate of F liters of air per second, then the percent
P of contaminants that are also removed from the
surrounding air can be modeled by the linear equation
where
What flow F must a range hood have to remove
50% of the contaminants from the air? (Source:
Proceedings of the Third International Conference
on Indoor Air Quality and Climate.)
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35
Example 5
MODELING THE PREVENTION
OF INDOOR POLLUTANTS
Solution Replace P with 50 in the linear
model, and solve for F.
Given model
Let P = 50.
Subtract 7.18.
Divide by 1.06.
Therefore, to remove 50% of the contaminants,
the flow rate must be approximately 40.40 L of
air per second.
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36
Example 6
MODELING HEALTH CARE COSTS
The projected per capita health care
expenditures in the United States, where y is in
dollars, and x is years after 2000, are given by
the linear equation.
Linear model
(Source: Centers for Medicare and Medicaid Services.)
(a) What were the per capita health care
expenditures in the year 2010?
(b) If this model continues to describe health
care expenditures, when will the per capita
expenditures reach $11,000?
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Example 6
MODELING HEALTH CARE COSTS
In part (a) we are given information to determine a
value for x and asked to find the corresponding value
of y, whereas in part (b) we are given a value for y
and asked to find the corresponding value of x.
(a) The year 2010 is 10 yr after the year 2000.
Let x = 10 and find the value of y.
Given model
Let x = 10.
Multiply and then add.
In 2010, the estimated per capita health care
expenditures were $8401.
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Example 6
MODELING HEALTH CARE COSTS
(b) Let y = 11,000 in the given model, and
find the value of x.
Let y = 11,000.
Subtract 4512.
17 corresponds
to
2000 + 17 = 2017.
Divide by 343.
The x-value of 17.9 indicates that per capita
health care expenditures are projected to
reach $11,000 during the 17th year after
2000—that is, 2017.
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39
1
Equations and
Inequalities
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
1
1.1
Linear Equations
• Basic Terminology of Equations
• Linear Equations
• Identities, Conditional Equations, and
Contradictions
• Solving for a Specified Variable (Literal
Equations)
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2
Equations
An equation is a statement that two
expressions are equal.
x + 2 =9 11x = 5x + 6x
x2 – 2x – 1 = 0
To solve an equation means to find all numbers
that make the equation a true statement. These
numbers are the solutions, or roots, of the
equation. A number that is a solution of an
equation is said to satisfy the equation, and the
solutions of an equation make up its solution
set. Equations with the same solution set are
equivalent equations.
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3
Addition and Multiplication
Properties of Equality
Let a, b, and c represent real numbers.
If a = b, then a + c = b + c.
That is, the same number may be added
to each side of an equation without
changing the solution set.
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4
Addition and Multiplication
Properties of Equality
Let a, b, and c represent real numbers.
If a = b and c ≠ 0, then ac = bc.
That is, each side of an equation may be
multiplied by the same nonzero number
without changing the solution set.
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5
Linear Equation in One
Variable
A linear equation in one variable is an
equation that can be written in the form
ax + b = 0,
where a and b are real numbers with
≠ 0.
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a
6
Linear Equations
A linear equation is also called a firstdegree equation since the greatest degree
of the variable is 1.
Linear
equations
Nonlinear
equations
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7
Example 1
SOLVING A LINEAR EQUATION
Solve
Solution
Be careful
with signs.
Distributive property
Combine like terms.
Add x to each side.
Combine like terms.
Add 12 to each side.
Combine like terms.
Divide each side
by 7.
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8
Example 1
SOLVING A LINEAR EQUATION
Check
Original equation
Let x = 2.
A check of the
solution is
recommended.
True
The solution set is {2}.
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9
Example 2
SOLVING A LINEAR EQUATION
WITH FRACTIONS
Solve
Solution
Multiply by 12, the
LCD of the fractions.
Distribute the 12 to all
terms within
parentheses.
Distributive property
Multiply.
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10
Example 2
SOLVING A LINEAR EQUATION
WITH FRACTIONS
Solve
Solution
Distributive property
Combine like terms.
Subtract 3x; subtract 16.
Divide each side by 11.
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11
Example 2
SOLVING A LINEAR EQUATION
WITH FRACTIONS
Check
Let x = − 4.
Simplify.
True
The solution set is
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12
Identities, Conditional Equations,
and Contradictions
An equation satisfied by every number that
is a meaningful replacement for the variable
is an identity.
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13
Identities, Conditional Equations,
and Contradictions
An equation that is satisfied by some
numbers but not others is a conditional
equation.
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14
Identities, Conditional Equations,
and Contradictions
An equation that has no solution is a
contradiction.
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15
Example 3
IDENTIFYING TYPES OF EQUATIONS
Determine whether each equation is an identity,
a conditional equation, or a contradiction.
(a)
Solution
Distributive
property
Combine like
terms.
Subtract x and
add 8.
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16
Example 3
IDENTIFYING TYPES OF EQUATIONS
Determine whether each equation is an identity,
a conditional equation, or a contradiction.
(a)
Solution
Subtract x and add 8.
When a true statement such as 0 = 0 results,
the equation is an identity, and the solution
set is {all real numbers}.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
17
Example 3
IDENTIFYING TYPES OF EQUATIONS
Determine whether each equation is an identity,
a conditional equation, or a contradiction.
(b)
Solution
Add 4 to each side.
Divide each side by 5.
This is a conditional equation, and its
solution set is {3}.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
18
Example 3
IDENTIFYING TYPES OF EQUATIONS
Determine whether each equation is an identity,
a conditional equation, or a contradiction.
(c)
Solution
Distributive property
Subtract 9x.
When a false statement such as − 3 = 7 results,
the equation is a contradiction, and the solution
set is the empty set or null set, symbolized by .
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19
Identifying Types of Linear Equations
1. If solving a linear equation leads to a true
statement such as 0 = 0, the equation is an
identity. Its solution set is {all real numbers}.
2. If solving a linear equation leads to a single
solution such as x = 3, the equation is
conditional. Its solution set consists of a single
element.
3. If solving a linear equation leads to a false
statement such as − 3 = 7, then the equation is
a contradiction. Its solution set is .
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20
Solving for a Specified Variable
(Literal Equations)
A formula is an example of a linear
equation (an equation involving letters).
This is the formula for simple interest.
I is the
variable for
simple
interest
t is the
variable for
years
P is the
variable for
dollars
r is the
variable for
annual
interest rate
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Example 4
SOLVING FOR A SPECIFIED VARIABLE
Solve for t.
(a)
Solution
Goal: Isolate t on one side.
Divide each side by Pr.
or
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22
Solving for a Specified Variable
(Literal Equations)
This formula gives the future value, or
maturity value, A of P dollars invested for t
years at an annual simple interest rate r.
A is the
variable
for future
or
maturity
value
t is the
variable for
years
P is the
variable for
dollars
r is the variable
for annual simple
interest rate
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
23
Example 4
SOLVING FOR A SPECIFIED VARIABLE
Solve for P.
(b)
Goal: Isolate P, the
specified variable.
Solution
Transform so that all
terms involving P are on
one side.
Factor out P.
or
Divide by 1 + rt.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
24
Example 4
SOLVING FOR A SPECIFIED VARIABLE
Solve for x.
(c)
Solution
Solve for x.
Distributive property
Combine like terms.
Isolate the xterms on one
side.
Divide each side by 2.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
25
APPLYING THE SIMPLE INTEREST
FORMULA
Example 5
A woman borrowed $5240 for new furniture.
She will pay it off in 11 months at an annual
simple interest rate of 4.5%. How much
interest will she pay?
Solution
r = 0.045
P = 5240
t=
(year)
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
26
Example 5
APPLYING THE SIMPLE INTEREST
FORMULA
A woman borrowed $5240 for new furniture.
She will pay it off in 11 months at an annual
simple interest rate of 4.5%. How much
interest will she pay?
Solution
She will pay $216.15 interest on her
purchase.
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27

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