Section 2.1 Linear Equations in One Variable. OBJECTIVES A Determine whether a number is a solution...

Section 2.1

Linear Equations in One Variable

OBJECTIVES

A Determine whether a number is a solution of a given equation.

OBJECTIVES

B Solve linear equations using the properties of equality.

OBJECTIVES

C Solve linear equations in one variable using the six-step procedure(CRAM).

OBJECTIVES

D Solve linear equations involving decimals.

DEFINITION

For real numbers a, b, and c.

PROPERTIES OF EQUALITIES

1. a = a Reflexive

2. If a = b, then b = a Symmetric

3. If a = b and b = c, then a = c Transitive

DEFINITION

An equation that can be written in the form:

LINEAR EQUATIONS

ax + b = cwhere , , and are realnumbers and 0.

a b ca

DEFINITION

Replacements of the variable that make the equation a true statement.

SOLUTIONS OF AN EQUATION

DEFINITION

Two equations that have the same solution set.

EQUIVALENT EQUATIONS

Clear fractions/decimals

Remove parentheses/simplify

Add/Subtract to get variable isolated

Multiply/Divide to make coefficient 1

PROCEDURE

DEFINITION

No solutions(contradictions):

EQUATIONS WITH NO SOLUTIONS AND INFINITELY MANY SOLUTIONS

x + 4 = x – 2Infinitely many solutions(identities):

2x + 8 = 2(x +4)

Section 2.1Exercise #5

Chapter 2Linear Equationsand Inequalities

Solve.

65

+ 3x15

= x + 4

10

6 3 + 4 • + • =

5 15 13

00 30 30

x x

LCD = 30

6 2 3

1 1 1

36 + 6x = 3x + 12

36 – 36 + 6x = 3x + 12 – 36

3

3

x = – 8

6x – 3x = 3x – 3x – 24

3x = – 24

Solve.

0.06P + 0.07 1500 – P = 96

6P + 7 1500 – P = 9600

6P + 10,500 – 7P = 9600

10,500 – 1P = 9600

– 1P = – 900

P = 900

Section 2.2

Formulas, Geometry and Problem Solving

OBJECTIVES

A Solve a formula for a specified variable and then evaluate the answer for given values of the variables.

OBJECTIVES

B Write a formula for a given situation that has been described in words.

OBJECTIVES

C Solve problems about angle measures.

SOLVE FOR A SPECIFIED VALUEPROCEDURE

1.Add or Subtract the same quantity on both sides.

2.Use the distributive property.3.Use CRAM.

Section 2.2


H = 2.75h + 71.48

a. Solve for h.

b. Find h if H = 140.23.

H = 2.75h + 71.48

H = 2.75h + 71.48

H – 71.48 = 2.75h

a. Solve for h.

H – 71.482.75

= h

H = 2.75h + 71.48

b. Find h if H = 140.23.

=

140.23 – 71.482.75

=

68.752.75

= 25

h =

H – 71.482.75

The perimeter of a rectangle is P = 2L + 2W , where L is thelength and W is the width.

a. Solve for L.b. If the perimeter is 100 ft and the length is 20 ft more than the width, what are the dimensions of the rectangle?


a. Solve for L.

P = 2L + 2W

P – 2W = 2L

P – 2W2

= L


b. If the perimeter is 100 ft and the length is 20 ft more than the width, what are the dimensions of the rectangle?

100 ft = Perimeter, P

(w + 20) = length, L

Let w = width, W


100 = 2 w + 20 + 2w

100 = 2w + 40 + 2w

100 = 4w + 40

60 = 4w

15 = w

w + 20 = 35



Let w = width, W


w = 15

w + 20 = 35



Let w = width, W

The dimensions are 15 ft by 35 ft .

b. If the perimeter is 100 ft and the length is 20 ft more than the width, what are the dimensions of the rectangle?

If L1 and L2 are parallel lines, find x and the measure ofthe unknown angles.

L1

L2

10x – 24 °

8x + 6 °

These are the alternate exterior anglesand they are equal.


10 – 24 = 8 + 6x x

10x – 24 = 8x + 6

2x – 24 = 6

2x = 30

x = 15


10 – 24 = 10 15 – 24x

8 + 6 = 8 15 + 6x

= 120 + 6 = 126

= 150 – 24 = 126

Each of the angles is 126°.

Section 2.3

Problem Solving: Integers and Geometry

OBJECTIVES

A Translate a word expression into a mathematical expression.

OBJECTIVES

B Solve word problems of a general nature.

OBJECTIVES

C Solve word problems about integers.

OBJECTIVES

D Solve word problems about geometric formulas and angles.

PROCEDURE:

Read Select Think Use Verify

RSTUV Method for Solving Word Problems

Section 2.3


The bill for repairing an appliance totaled $72.50. If the repair shop charges $35 for the service call, plus$25 for each hour of labor, how many hourslabor did the repair take?

Let B = bill for repairs h = number of hours of labor

B = 35 + 25h If B = 72.50, find h.

72.50 = 35 + 25h

37.50 = 25h

1.5 = h 1.5 hours

Section 2.4

Problem Solving: Percent, Investment, Motion, and Mixture Problems

OBJECTIVES

A Solve percent problems.

OBJECTIVES

B Solve investment problems.

OBJECTIVES

C Solve uniform motion problems.

OBJECTIVES

D Solve mixture problems.

PROCEDURE:

Read Select Think Use Verify

RSTUV Method for Solving Word Problems

Section 2.4


An investor bought some municipal bonds yielding 5 percent annually and some certificates of deposit yielding 7 percent. If his total investmentamounts to $20,000 and his annual interestis $1100, how much money is invested inbonds and how much in certificates ofdeposit?

Let x = bonds at 5%

0.05x = interest on bonds 20,000 – x = C.D.'s at 7%

0.07 20,000 – x = interest on C.D.'s

The sum of these interests = 1100

0.05x + 0.07 20,000 – x = 1100

0.05x + 1400 – 0.07x = 1100

– 0.02x + 1400 = 1100

– 0.02x = – 300

x = 15,000 Bonds

20,000 – x = 5000 C.D.'s

The sum of these interests = 1100

0.05x + 0.07 20,000 – x = 1100

0.05x + 1400 – 0.07x = 1100

– 0.02x + 1400 = 1100

– 0.02x = – 300

$15,000 bonds$5000 C.D.'s

A freight train leaves a station traveling at 40 mi/hr.Two hours later, a passenger train leaves thesame station traveling in the same directionat 60 mi/hr. How far from the station does thepassenger train overtake the freight train?

Let t = time of freight train

t – 2 = time of passenger train

Rate Time Distance

Freight

Passenger

40 mi/hr t 60 mi/hr t – 2

40t

60 – 2t

Rate Time Distance

Freight

Passenger


40t

60 – 2t

Their distances are equal:

40 = 60 – 2t t

40t = 60t – 120

– 20t = – 120

t = 6

t – 2 = 4

Freight's distance = 40 6 = 240

Passenger's distance = 60 4 = 240

Rate Time Distance

Freight

Passenger


40t

60 – 2t

Their distances are equal:

40 = 60 – 2t t

40t = 60t – 120

The passenger train overtakes thefreight train 240 miles from the station.

Section 2.5

Linear and Compound Inequalities

OBJECTIVES

A Graph linear inequalities.

OBJECTIVES

B Solve and graph linear inequalities.

OBJECTIVES

C Solve and graph compound inequalities.

OBJECTIVES

D Use the inequality symbols to translate sentences into inequalities.

DEFINITION

An inequality that can be written in the form:

LINEAR INEQUALITIES

ax + b < cwhere , , and are realnumbers and 0.

a b ca

DEFINITIONUNION OF TWO SETS

If A and B are sets, the union of A and B, denoted by A B, is the set of elements in either A or B.

DEFINITIONINTERSECTION OF TWO SETS

If A and B are two sets, the intersection of A and B, denoted by A B, is the set of elements in both A and B.

DEFINITIONEQUIVALENT STATEMENTS FOR “AND”

a < x and x < b equivalent to

a < x < b

Section 2.5


Solve, graph and write the solution set in interval notation.

24 24 – 8

• – • 24 < • 8 3 8

x x x

3 – 8 < 3 – 8x x x

– 5x < 3x – 24

– 8x < – 24

x > 3

x8

– x3

< x – 8

8 LCD = 24


x > 3

x8

– x3

< x – 8

8

0 1 2 3 4 5 6(

3,


< – 1 or x 2x

– 3 – 2 – 1 0 1 2 3[)

– , – 1 2,

< – 1 or 2x x x

– 4 – 3 – 1 0 1 2 3 – 2 4


](

– 3 < x – 3, 3

3x x > – 3

and – 3 < x

+ 1 4 and – 2 < 6x x

3

– 4 – 2 – 1 – 3 0


](

– 4 – 2 – 6 < 0 x

– 3 < – 1 x

– 4 – 2 – + 6 + 6 6 0 + < 6 x

2 – 2 < 6 x

– 1 > – 3 x

– 3,– 1

Section 2.6

Absolute-Value Equations and Inequality

OBJECTIVES

A Solve absolute-value equations.

OBJECTIVES

B Solve absolute-value inequalities of the form |ax + b| < c or |ax + b| > c, where c > 0.

DEFINITION

If a ≥ 0, the solutions of |x| = a are x = a and x = –a.

THE SOLUTIONS OF |X| = A (A ≥ 0)

STATEMENT TRANSLATION

If |expression| = a, where a ≥ 0

expression = a or –a

ABSOLUTE VALUE EQUATIONS


If |expression| = |expression|,

expression = expression

expression = –

(expression)

ABSOLUTE VALUE EQUATIONS


|x| = 2: x is exactly 2 units from 0

|x| < 2: x is less than 2 units from 0

|x| > 2: x is more than 2 units from 0

0

0

0

DEFINITION

|x| < a is

equivalent to

–a < x < a

DEFINITION

|x| > a is

equivalent to

x < –a or x > a

Section 2.6


Solve.

34

x + 2 = 5

34

x + 2 = 5 34

x + 2 = – 5 or

34

x = 3 34

x = – 7

x = 3 •

43

x = – 7 •

43

34

x + 2 + 4 = 9

x = 4 x =

– 283

or

Solve.

x – 3 = x – 7 – 3 = – – 7x xor

– 3 = – 7 x – 3 = – x + 7or

x = – x + 10

2x = 10

x = 5

x – 3 = x – 7

Solve and graph.

– 5 2 – 1 5 x

– 4 2 6 x

– 2 3 x

– 4 – 3 – 1 0 1 2 3 – 2 4][

2 – 1 5x

Solve and graph.

2x + 1 > 3 2x + 1 < – 3or

2x > 2 2x < – 4or

x > 1 x < – 2or

– 4 – 3 – 1 0 1 2 3 – 2 4()

2x + 1 > 3

Section 2.1 Linear Equations in One Variable. OBJECTIVES A Determine whether a number is a solution...

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Transcript of Section 2.1 Linear Equations in One Variable. OBJECTIVES A Determine whether a number is a solution...