section 2.1 linear equations in one variable. objectives a determine whether a number is a solution...
TRANSCRIPT
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Section 2.1
Linear Equations in One Variable
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OBJECTIVES
A Determine whether a number is a solution of a given equation.
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OBJECTIVES
B Solve linear equations using the properties of equality.
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OBJECTIVES
C Solve linear equations in one variable using the six-step procedure(CRAM).
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OBJECTIVES
D Solve linear equations involving decimals.
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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
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DEFINITION
An equation that can be written in the form:
LINEAR EQUATIONS
ax + b = cwhere , , and are realnumbers and 0.
a b ca
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DEFINITION
Replacements of the variable that make the equation a true statement.
SOLUTIONS OF AN EQUATION
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DEFINITION
Two equations that have the same solution set.
EQUIVALENT EQUATIONS
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Clear fractions/decimals
Remove parentheses/simplify
Add/Subtract to get variable isolated
Multiply/Divide to make coefficient 1
PROCEDURE
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DEFINITION
No solutions(contradictions):
EQUATIONS WITH NO SOLUTIONS AND INFINITELY MANY SOLUTIONS
x + 4 = x – 2Infinitely many solutions(identities):
2x + 8 = 2(x +4)
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Section 2.1Exercise #5
Chapter 2Linear Equationsand Inequalities
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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
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Section 2.1Exercise #6
Chapter 2Linear Equationsand Inequalities
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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
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Section 2.2
Formulas, Geometry and Problem Solving
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OBJECTIVES
A Solve a formula for a specified variable and then evaluate the answer for given values of the variables.
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OBJECTIVES
B Write a formula for a given situation that has been described in words.
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OBJECTIVES
C Solve problems about angle measures.
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SOLVE FOR A SPECIFIED VALUEPROCEDURE
1.Add or Subtract the same quantity on both sides.
2.Use the distributive property.3.Use CRAM.
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Section 2.2
Chapter 2Linear Equationsand Inequalities
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Section 2.2Exercise #7
Chapter 2Linear Equationsand Inequalities
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H = 2.75h + 71.48
a. Solve for h.
b. Find h if H = 140.23.
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H = 2.75h + 71.48
H = 2.75h + 71.48
H – 71.48 = 2.75h
a. Solve for h.
H – 71.482.75
= h
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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
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Section 2.2Exercise #9
Chapter 2Linear Equationsand Inequalities
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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?
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The perimeter of a rectangle is P = 2L + 2W , where L is thelength and W is the width.
a. Solve for L.
P = 2L + 2W
P – 2W = 2L
P – 2W2
= L
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The perimeter of a rectangle is P = 2L + 2W , where L is thelength and W is the width.
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
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The perimeter of a rectangle is P = 2L + 2W , where L is thelength and W is the width.
100 = 2 w + 20 + 2w
100 = 2w + 40 + 2w
100 = 4w + 40
60 = 4w
15 = w
w + 20 = 35
100 ft = Perimeter, P
(w + 20) = length, L
Let w = width, W
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The perimeter of a rectangle is P = 2L + 2W , where L is thelength and W is the width.
w = 15
w + 20 = 35
100 ft = Perimeter, P
(w + 20) = length, L
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?
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Section 2.2Exercise #10
Chapter 2Linear Equationsand Inequalities
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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.
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If L1 and L2 are parallel lines, find x and the measure ofthe unknown angles.
10 – 24 = 8 + 6x x
10x – 24 = 8x + 6
2x – 24 = 6
2x = 30
x = 15
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If L1 and L2 are parallel lines, find x and the measure ofthe unknown angles.
10 – 24 = 10 15 – 24x
8 + 6 = 8 15 + 6x
= 120 + 6 = 126
= 150 – 24 = 126
Each of the angles is 126°.
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Section 2.3
Problem Solving: Integers and Geometry
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OBJECTIVES
A Translate a word expression into a mathematical expression.
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OBJECTIVES
B Solve word problems of a general nature.
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OBJECTIVES
C Solve word problems about integers.
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OBJECTIVES
D Solve word problems about geometric formulas and angles.
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PROCEDURE:
Read Select Think Use Verify
RSTUV Method for Solving Word Problems
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Section 2.3
Chapter 2Linear Equationsand Inequalities
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Section 2.3Exercise #11
Chapter 2Linear Equationsand Inequalities
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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
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Section 2.4
Problem Solving: Percent, Investment, Motion, and Mixture Problems
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OBJECTIVES
A Solve percent problems.
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OBJECTIVES
B Solve investment problems.
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OBJECTIVES
C Solve uniform motion problems.
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OBJECTIVES
D Solve mixture problems.
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PROCEDURE:
Read Select Think Use Verify
RSTUV Method for Solving Word Problems
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Section 2.4
Chapter 2Linear Equationsand Inequalities
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Section 2.4Exercise #14
Chapter 2Linear Equationsand Inequalities
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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
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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
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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
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Section 2.4Exercise #15
Chapter 2Linear Equationsand Inequalities
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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
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Rate Time Distance
Freight
Passenger
40 mi/hr t 60 mi/hr t – 2
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
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Rate Time Distance
Freight
Passenger
40 mi/hr t 60 mi/hr t – 2
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.
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Section 2.5
Linear and Compound Inequalities
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OBJECTIVES
A Graph linear inequalities.
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OBJECTIVES
B Solve and graph linear inequalities.
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OBJECTIVES
C Solve and graph compound inequalities.
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OBJECTIVES
D Use the inequality symbols to translate sentences into inequalities.
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DEFINITION
An inequality that can be written in the form:
LINEAR INEQUALITIES
ax + b < cwhere , , and are realnumbers and 0.
a b ca
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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.
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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.
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DEFINITIONEQUIVALENT STATEMENTS FOR “AND”
a < x and x < b equivalent to
a < x < b
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Section 2.5
Chapter 2Linear Equationsand Inequalities
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Section 2.5Exercise #18
Chapter 2Linear Equationsand Inequalities
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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
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Solve, graph and write the solution set in interval notation.
x > 3
x8
– x3
< x – 8
8
0 1 2 3 4 5 6(
3,
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Section 2.5Exercise #19
Chapter 2Linear Equationsand Inequalities
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Solve, graph and write the solution set in interval notation.
< – 1 or x 2x
– 3 – 2 – 1 0 1 2 3[)
– , – 1 2,
< – 1 or 2x x x
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Section 2.5Exercise #20
Chapter 2Linear Equationsand Inequalities
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– 4 – 3 – 1 0 1 2 3 – 2 4
Solve, graph and write the solution set in interval notation.
](
– 3 < x – 3, 3
3x x > – 3
and – 3 < x
+ 1 4 and – 2 < 6x x
3
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Section 2.5Exercise #21
Chapter 2Linear Equationsand Inequalities
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– 4 – 2 – 1 – 3 0
Solve, graph and write the solution set in interval notation.
](
– 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
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Section 2.6
Absolute-Value Equations and Inequality
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OBJECTIVES
A Solve absolute-value equations.
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OBJECTIVES
B Solve absolute-value inequalities of the form |ax + b| < c or |ax + b| > c, where c > 0.
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DEFINITION
If a ≥ 0, the solutions of |x| = a are x = a and x = –a.
THE SOLUTIONS OF |X| = A (A ≥ 0)
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STATEMENT TRANSLATION
If |expression| = a, where a ≥ 0
expression = a or –a
ABSOLUTE VALUE EQUATIONS
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STATEMENT TRANSLATION
If |expression| = |expression|,
expression = expression
expression = –
(expression)
ABSOLUTE VALUE EQUATIONS
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STATEMENT TRANSLATION
|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
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DEFINITION
|x| < a is
equivalent to
–a < x < a
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DEFINITION
|x| > a is
equivalent to
x < –a or x > a
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Section 2.6
Chapter 2Linear Equationsand Inequalities
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Section 2.6Exercise #22
Chapter 2Linear Equationsand Inequalities
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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
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Section 2.6Exercise #23
Chapter 2Linear Equationsand Inequalities
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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
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Section 2.6Exercise #24
Chapter 2Linear Equationsand Inequalities
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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
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Section 2.6Exercise #25
Chapter 2Linear Equationsand Inequalities
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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