Equations and Inequalities

1
Chapter 1
Equations and Inequalities
2
1.1
Linear Equations
3
Equations

• An equation is a statement that two expressions
  are equal.
• x 3 11 9x 3x 6x x2 4x
  8 0
• To solve an equation means to find all numbers
  that make the equation true.
• The numbers are called 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.

4
Addition and Multiplication Properties of Equality

• For real numbers a, b, and c
• a b and a c b c
• are equivalent
• That is, the same number may be added to both
  sides of an equation without changing the
  solution set.

• For real numbers a, b, and c
• If c ? 0,
• then a b and ac bc
• are equivalent.
• That is, both sides of an equation may be
  multiplied by the same nonzero number without
  changing the solution set.

5
Solving Linear Equations

• Linear Equations 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 a ?
  0
• A linear equation is also called a first-degree
  equation since the greatest degree of the
  variable is one.

6
Solving a Linear Equation

• Example Solve 6p 2 ? 10(p 2) 12p 14.
• Solution

6p 2 ? 10(p 2) 12p 14
6p 2 ? 10p 20 12p 14
-4p 18 12p 14
-4p 4p 18 12p 4p 14
18 14 16p 14 14
32 16p
2 p
7
Solving a Linear Equation continued

• Check
• 6p 2 ? 10(p 2) 12p 14
• 6(?2) 2 ? 10(?2 2) 12(?2) 14
• ?12 2 ?24 14
• ?10 ? 10
• Since replacing p with ?2 results in a true
  statement, ?2 is the solution of the given
  equation. The solution set is ?2.

8
Clearing Fractions

• If an equation has fractions as coefficients,
  begin by multiplying both sides by the least
  common denominator to clear the fractions.

Example Solve
9
Clearing Fractions continued
Check
This solution checks!
10
Identities, Conditional Equations and
Contradictions

• 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.
  
• If solving a linear equation leads to a single
  solution such as x 3, the equation is
  conditional. The solution set consists of a
  single element.
• If solving a linear equation leads to a false
  statement such as 10 ?8, the equation is a
  contradiction. The equation then has no solution.

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Examples of Identifying Types of Equations

• ?4(x ? 2) 6x 2(x 4)
• ?4x 8 6x 2x 8
• 2x 8 2x 8
• 0 0
  Type? ___________
• 6x 5 23
• 6x 18
• x 3
  Type? ___________
• 5(3x 2) 15x ?8
• 15x 10 15x ?8
• 10 ? 8 Type?
  ___________

identity
conditional
contradiction
12
Solving for a Specified Variable

• A formula is an example of a literal equation.
  The methods used to solve a linear equation can
  be used to solve some literal equations for a
  specified variable.
• Example Solve p 6x 2z for z.
• Solution

p 6x 6x 6x 2z
p 6x 2z
13
Simple Interest Formula

• The formula for simple interest is I Prt.
• Example Stanley has 8000 to invest in a mutual
  fund. The funds with higher rates of returns also
  have greater risk. Stanley wants a return of at
  least 300 every six months. What is the lowest
  rate of return that will meet his goal?
• Solution

300 (8000) r (0.5)
I Interest earned, P Principal, r rate, t
time in years.
300 4000 r
0.075 r
7.5 r
14
Homework

• Pg 90
• 3 59 eoo

15
1.2
Applications and Modeling with Linear Equations
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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.
• Step 3 Write an equation using the variable
  expression(2).

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Solving an Applied Problem continued

• 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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Example--Geometry

• The perimeter of a rectangle is 108 cm. The
  length is 8 cm longer than the width. Find the
  dimensions of the rectangle.
• Step 1 Read the problem.
• We must find the dimensions of the
  rectangle.
• Step 2 Assign a variable.

Let x width of the rectangle x 8 length of
the rectangle
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Example--Geometry continued

• Step 3 Write an equation.
• Step 4 Solve the equation.
• 108 x x x 8 x 8
• 108 4x 16
• 92 4x
• 23 x

P w w l l
108 (x) (x) (x 8) (x 8)
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Example--Geometry continued

• Step 5 State the answer.
• The width of the rectangle is 23 cm, the length
  of the rectangle is 23 8 31.
• Step 6 Check.
• The perimeter is 108 cm.
• 31 31 23 23 108

21
Motion Problems

• 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.) These variables are related by the
  equation
• and its related forms

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Example

• Nicholas and Catherine are traveling to a math
  conference. The trip takes 1 ½ hours for
  Catherine and 2 ½ hours for Nicholas, since he
  lives 75 miles farther away. Nicholas travels 6
  mph faster than Catherine. Find their average
  rates.
• Step 1 Read the problem.
• We are looking for Catherines and Nicholas
  average rate.

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Example continued

• Step 2 Assign a variable.
• Step 3 Write an equation.

Let r Catherines rate r 6 Nicholas rate
r r 6
1.5 2.5
1.5 ( r) 2.5( r 6)
Nicholas distance Catherines distance 75
2.5 r 15 1.5 r 75
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Example continued

• Step 4 Solve.
• Step 5 State the answer.
• Step 6 Check

Catherines rate is 60 mph and Nicholas
rate is 60 6 66 mph.
Distance traveled by Catherine 60(1.5)
90 Distance traveled by Nicholas 66(2.5)
165 The difference in the distances 165 ? 90 75
25
Work Rate Problems

• If a job can be done in t units of time then the
  rate of work is 1/t of the job per time unit.
  Therefore, rate ? time portion of job
  completed.
• If the letters r, t, and W represent the rate at
  which work is done, the time, and the amount of
  work accomplished, respectively, then W rt.
• Amounts of work are often measured in terms of
  the number of jobs accomplished. For instance, if
  one job is accomplished in t time units, then W
  1 and r 1/t.

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• Example
• If Joe can paint a house in 6 hours and Sam can
  paint the same house in 8 hours, how long does it
  take to paint the house if they work together.
• Read the problem.
• We must find the time it takes the two of them
  to paint the house together.
• Assign a variable.
• Write an equation.

Work Rate Problems
Let x time to paint the house together
x (Joes rate) x ( Sams rate) 1 job
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Work Rate Problems continued

• Solve.
• State the answer.
• Check.

It would take Sam and Joe about 3.4 hours to
paint the house if they work together.
28
Mixture Problems

• In mixture problems, the rate (percent) of
  concentration is multiplied by the quantity to
  get the amount of pure substance present. Also,
  the concentration in the final mixture must be
  between the concentrations of the two solutions
  making up the mixture.
• Example
• How many gallons of a 10 solution acid solution
  must be mixed with 5 gallons of a 20 acid
  solution to obtain a 15 acid solution?

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Mixture Problems continued

• Read the problem.
• We must find the required number of gallons of
  10 solution.
• Assign a variable.
• Write an equation.

0.10( x )
x
5
0.20 ( 5 )
x 5
0.15 ( x 5 )
0.10 ( x ) 0.20 ( 5 ) 0.15 ( x 5 )
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Mixture Problems continued

• Solve.
• State the answer.
• Check.

5 gallons of 10 solution should be mixed with 5
gallons of 20 solution, given 5 5 10 gallons
of 15 percent solution.
.10(5) .20(5) .15(5 5) 0.5 1.0 1.5
1.5 1.5
31
Investment Problem

• Antonio sold a house for 210,000. He invests
  some of the money in an account that pays 4
  interest. He invests the remaining amount in an
  investment that pays 7. His total annual
  interest earned is 12,540. How much was invested
  in each account?
• Read.
• We must find the amount invested at each rate.

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Investment Problem continued

• Assign a variable.
• Write an equation.

Let, x investment at 4 210,000 x
investment at 7
Interest at 4 Interest at 7 Total Interest
0.04 ( x ) 0.07 ( 210,000 x) 12,540
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Investment Problem continued

• Solve.
• State the answer.
• Check.

Antonio invested 72,000 at 4 and 138,000 at 7.
72,000(.04) 138,000(.07) 12,540 2880 9660
12,540
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Homework

• Pg 100
• 3 43 eoo

35
1.3
Complex Numbers
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Basic Concepts

• a bi are called complex numbers
• a is the real part
• b is the imaginary part

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Examples

• a)
• b)
• c)
• d)
• e)

38
Addition and Subtraction of Complex Numbers

• For complex numbers a bi and c di,
• Examples

39
Multiplication of Complex Numbers

• For complex numbers a bi and c di,
• The product of two complex numbers is found by
  multiplying as if the numbers were binomials and
  using the fact that i2 ?1.

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Examples

• (2 ? 4i)(3 5i)

• (7 3i)2

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Powers of i

• i1 i i5 i i9 i
• i2 ?1 i6 ?1 i10 ?1
• i3 ?i i7 ?i i11 ?i
• i4 1 i8 1 i12 1
• and so on.

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Simplifying Examples

• i 17

• i ?4

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Property of Complex Conjugates

• For real numbers a and b,
• (a bi)(a ? bi) a2 b2.
• The product of a complex number and its conjugate
  is always a real number.

• Example

44
Dividing Complex Numbers

• Example

45
Homework

• Pg 113
• 3 91 eoo