Transcript Chapter_02 - USF Math Center

Section 2.1
The Addition &
Subtraction Properties of
Equality
OBJECTIVES
A
Determine whether a
number satisfies an
equation.
OBJECTIVES
B
Use the addition and
subtraction properties of
equality to solve
equations.
OBJECTIVES
C
Use both properties
together to solve an
equation.
SOLUTIONS
The replacements of the
variable that make the
equation a true statement.
“Solved the equation.”
SOLUTIONS
A number that is a solution of
an equation satisfies the
equation.
SOLUTIONS
For example
3 is a solution of 2x = 6
because:
2 3=6
3 satisfies the equation.
PROCEDURE:
Solving Equations by Adding
or Subtracting
1. Simplify both sides if
necessary.
PROCEDURE:
Solving Equations by Adding
or Subtracting
2. Add or subtract the same
numbers on both sides so
that one side contains only
variables.
PROCEDURE:
Solving Equations by Adding
or Subtracting
3. Add or subtract the same
expressions on both sides
so that the other side
contains only numbers.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.1
Exercise #1
Does the number 3 satisfy the equation
6=9–x?
6 =9–x
Does the number 3 satisfy the equation
6=9–x?
6 =9–3
Yes.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.1
Exercise #2
2 3
Solve x – = .
7 7
2
Add :
7
2 2 3 2
x– + = +
7 7 7 7
5
Simplify: x =
7
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.1
Exercise #3
7
5 5
Solve – 2x + + 3x – = .
8
8 8
Simplify:
2 5
x+ =
8 8
2
Subtract :
8
2 2 5 2
x+ – = –
8 8 8 8
3
x=
8
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.1
Exercise #4
Solve 2 = 3 x – 1 + 5 – 2x.
Use Distributive Property:
2 = 3x – 3 + 5 – 2x
Simplify:
2 =x +2
Subtract 2: 2 – 2 = x + 2 – 2
0 =x
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.1
Exercise #5
Solve 2 + 5  x + 1 = 8 + 5x.
Use Distributive Property:
2 + 5x + 5 = 8 + 5x
Simplify:
7 + 5x = 8 + 5x
Subtract 5x: 7 + 5x – 5x = 8 + 5x – 5x
7=8
Impossible. No solution.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.1
Exercise #6
Solve – 3 – 2  x – 1 = – 1 – 2x.
Use Distributive Property:
– 3 – 2x + 2 = – 1 – 2x
Simplify:
Identity:
– 1 – 2x = – 1 – 2x
All real numbers.
Section 2.2
The Multiplication and
Division Properties of
Equality
OBJECTIVES
A
Use the multiplication
and division properties
of equality to solve
equations.
OBJECTIVES
B
Multiply by reciprocals to
solve equations.
OBJECTIVES
C
Multiply by LCMs to
solve equations.
OBJECTIVES
D
Solve applications
involving percents.
PROCEDURE:
Clearing Fractions
Multiply both sides of the
equation by the LCM of the
denominators.
Or, multiply each term by the
LCM.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.2
Exercise #7
2
Solve x = – 4
3
2
3 • x = –4•3
3
2x = – 12
2
2
x = –6
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.2
Exercise #8
2
Solve – x = – 6
3
–3 –2
–3
•
x = –6•
2
3
2
–3
x = –6•
12
3
x =9
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.2
Exercise #9
x 2x
Solve +
= 11
4
3
LCD = 12
 x  4  2x 
12   + 12   = 11 • 12
 4
 31 
3
1
3x + 8x = 132
11x = 132
x = 12
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.2
Exercise #10
x x
Solve – = 2
3 5
LCD = 15
 
3
x
x
15
– 15
= 15 • 2
3
5
5
1
1
5x – 3x = 30
2x = 30
x = 15
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.2
Exercise #11
Solve.
8
x –2
5
–
x +1
8
LCD = 40
=0
   
x –2 5 x +1
40
– 40
= 40 • 0
51
8 1
8  x – 2  – 5  x + 1 = 0
8x – 16 – 5x – 5 = 0
3x – 21 = 0
3x = 21
x =7
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.2
Exercise #12
What percent of 55 is 11?
x • 55 = 11
55x = 11
x = 0.20
x = 20%
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.2
Exercise #13
Nine is 36% percent of what number?
9 = 0.36 • x
9 = 0.36x
25 = x
The number is 25.
Section 2.3
Linear Equations
OBJECTIVES
A
Solve linear equations in
one variable.
OBJECTIVES
B
Solve a literal equation for
one of the unknowns.
PROCEDURE:
Solving Linear Equations
x – 1 = 7  x – 2
4 6 12
1. Clear fractions (Multiply by
the LCM of the denominator).
PROCEDURE:
Solving Linear Equations
x – 1 = 7  x – 2
4 6 12
12
x – 12 1 = 12
4
6
 7

 12
 x – 2



PROCEDURE:
Solving Linear Equations
2. Remove parentheses and
simplify.
3x – 2 = 7(x – 2)
3x – 2 = 7x – 14
PROCEDURE:
Solving Linear Equations
3. Isolate the variable.
3x – 2 = 7x – 14
3x – 2 + 2 = 7x – 14 + 2
3x = 7x – 12
PROCEDURE:
Solving Linear Equations
3x = 7x – 12
3x – 7x = 7x – 7x – 12
– 4x = – 12
PROCEDURE:
Solving Linear Equations
4. If coefficient is NOT 1, divide
both sides by coefficient
– 4x = – 12
– 4x = – 12
–4
–4
PROCEDURE:
Solving Linear Equations
– 4x = – 12
–4
–4
x =3
PROCEDURE:
Solving Linear Equations
5. Check your answer in the
original equation.
3 – 1 = 7  3 – 2
4 6 12
PROCEDURE:
Solving Linear Equations
3 – 1 = 7  3 – 2
4 6 12
9 – 2 = 7
12 12 12
7 = 7
12 12


Literal Equation
An equation that contains
several variables.
D = RT and I = Prt are examples.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.3
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.3
Exercise #14
1 x 23  x + 3 
Solve. – =
LCD
=
15
5 3
15
3 – 5 x = 23  x + 5 
3 – 5x = 23x + 115
– 5x = 23x + 112
– 28x = 112
x = –4
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.3
Exercise #15
1 2
Solve for h in S = r h
3
3S = 1r 2 h
3S
1r 2
=h
h=
3S
r
2
Section 2.4
Problem Solving: Integer,
General, and Geometry
Problems
OBJECTIVES
Use RSTUV method to solve:
A
Integer problems
B
General word problems
C
Geometry word problems
PROCEDURE:
RSTUV Method for Solving
Word Problems
1. Read the problem and
decide what is asked for
(the unknown).
PROCEDURE:
RSTUV Method for Solving
Word Problems
2. Select a variable to
represent this unknown.
PROCEDURE:
RSTUV Method for Solving
Word Problems
3. Think of a plan to help you
write an equation.
PROCEDURE:
RSTUV Method for Solving
Word Problems
4. Use algebra to solve the
resulting equation.
PROCEDURE:
RSTUV Method for Solving
Word Problems
5. Verify the answer.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.4
Exercise #16
The sum of two numbers is 75. If one of the numbers
is 15 more than the other, what are the numbers?
Let x = smaller number
x + 15 = larger number
x +  x + 15  = 75
x + x + 15 = 75
2x + 15 = 75
2x = 60
The sum of two numbers is 75. If one of the numbers
is 15 more than the other, what are the numbers?
Let x = smaller number
x + 15 = larger number
2x = 60
x = 30
x + 15 = 45
The numbers are 30 and 45.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.4
Exercise #17
A man invested a certain amount of money in stocks
and bonds. His annual return from these investments
is $840. If the stocks produce $230 more in
returns than the bonds, how much money
does he receive annually from
each investment?
Let x = return from stocks
840 – x = return from bonds
x =  840 – x  + 230
x =  840 – x  + 230
x = 840 – x + 230
x = 1070 – x
2x = 1070
x = 535
840 – x = 305
$535 from stocks and $305 from bonds.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.4
Exercise #18
Find the measure of an angle whose supplement is 50°
less than 3 times its complement.
Let x = measure of the angle
90 – x = complement of the angle
180 – x = supplement of the angle
180 – x = 3  90 – x  – 50
Find the measure of an angle whose supplement is 50°
less than 3 times its complement.
180 – x = 3  90 – x  – 50
180 – x = 270 – 3x – 50
180 – x = 220 – 3x
180 + 2x = 220
2x = 40
x = 20
The angle measures 20°.
Section 2.5
Problem Solving: Motion,
Mixture, and Investment
Problems
OBJECTIVES
Use the RSTUV method to solve:
A
Motion problems
B
Mixture problems
C
Investment problems
PROCEDURE:
RSTUV Method for Solving
Word Problems
1. Read the problem and
decide what is asked for
(the unknown).
PROCEDURE:
RSTUV Method for Solving
Word Problems
2. Select a variable to
represent this unknown.
PROCEDURE:
RSTUV Method for Solving
Word Problems
3. Think of a plan to help you
write an equation.
PROCEDURE:
RSTUV Method for Solving
Word Problems
4. Use algebra to solve the
resulting equation.
PROCEDURE:
RSTUV Method for Solving
Word Problems
5. Verify the answer.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.5
Exercise #19
A freight train leaves a station traveling at 30 miles
per hour. Two hours later, a passenger train leaves
the same station traveling in the same direction
at 42 miles per hour. How long does
it take for the passenger train to
catch the freight train?
Let x = time of the passenger train r = 42
r = 30
x + 2 = time of the freight train
D =r t
D = 42x
D = 30  x + 2 
D = 42x
D = 30  x + 2 
30  x + 2  = 42x
30x + 60 = 42x
30x = 42x – 60
– 12x = – 60
x =5
The passenger train catches
the freight train in 5 hours.
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.5
Exercise #21
An investor bought some municipal bonds yielding 5%
annually and some certificates of deposit yielding 7%
annually. If her total investment amounts to
$20,000 and her annual return is $1160, how
much money is invested in bonds and how
much in certificates of deposit?
Let x = amount invested in bonds at 5%
20,000 – x = amount invested in CD's at 7%
I = RT
interest = 0.05x
interest = 0.07  20,000 – x 
Total interest = 1160
Let x = amount invested in bonds at 5%
20,000 – x = amount invested in CD's at 7%
Total interest = 1160
0.05 x + 0.07  20,000 – x  = 1160
5 x + 7  20,000 – x  = 116,000
5x + 140,000 – 7x = 116,000
140,000 – 2x = 116,000
Let x = amount invested in bonds at 5%
20,000 – x = amount invested in CD's at 7%
140,000 – 2x = 116,000
I = RT
– 2x = – 24,000
x = 12,000
20,000 – x = 8,000
$12,000 is invested in bonds at 5%,
$8000 is invested in CD's at 7%
Section 2.6
Formulas and Geometry
Applications
OBJECTIVES
A
Solve a formula for one
variable and use the
result to solve a
problem.
OBJECTIVES
B
Solve problems
involving geometric
formulas.
OBJECTIVES
C
Solve geometric
problems involving
angle measurement.
OBJECTIVES
D
Solve an application.
DEFINITION
Angle of measure 1
One complete revolution around
a circle is 360°, and
1 of a complete revolution is 1°
360
DEFINITION
Complementary Angles
Two angles whose sum is 90°
are called complementary
angles.
DEFINITION
Complementary Angles
90 – x
x
30° + 60° = 90°
DEFINITION
Supplementary Angles
Two angles whose sum is 180°
are called supplementary
angles.
DEFINITION
Supplementary Angles
180 – x
x
50° + 130° = 180°
DEFINITION
Vertical Angles
Are formed in opposite sides
of two intersecting lines and
have equal measures.
DEFINITION
Vertical Angles
1
2
4
3
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.6
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.6
Exercise #22
The cost C of riding a taxi is C = 1.95 + 0.85m, where m
is the number of miles (or fraction) you travel.
a. Solve for m.
b. How many miles did you travel if the cost
of the ride was $20.65?
The cost C of riding a taxi is C = 1.95 + 0.85m, where m
is the number of miles (or fraction) you travel.
a. Solve for m.
C = 1.95 + 0.85m
C – 1.95 = 0.85m
C – 1.95
=m
0.85
C – 1.95
m=
0.85
The cost C of riding a taxi is C = 1.95 + 0.85m, where m
is the number of miles (or fraction) you travel.
b. How many miles did you travel if the cost
of the ride was $20.65?
C – 1.95
m=
; C = 20.65
0.85
20.65 – 1.95 18.70
= 22
=
m=
0.85
0.85
22 miles were traveled.
Section 2.7
Properties of Inequalities
OBJECTIVES
A
Determine which of two
numbers is greater.
OBJECTIVES
B
Solve and graph linear
inequalities.
OBJECTIVES
C
Write, solve, and graph a
compound inequality.
OBJECTIVES
D
Solve an application.
DEFINITIONS
On the number line, greater
numbers are always to the
right of smaller ones.
–3
–2
–1
0
1
2
3 > 1 or 1 < 3
3
DEFINITIONS
Addition and Subtraction
Properties of Inequalities.
You can add or subtract the
same number on both sides of
an inequality and obtain an
equivalent inequality.
DEFINITIONS
Addition and Subtraction
Properties of Inequalities.
If
x <y
then x + a< y + a
or
x – b< y – b
DEFINITIONS
Addition and Subtraction
Properties of Inequalities.
If
x >y
then x + a> y + a
or
x – b> y – b
DEFINITIONS
Multiplication and Division
Properties of Inequalities for
Positive Numbers
You can multiply or divide both
sides of an inequality by any
positive number and obtain an
equivalent inequality.
DEFINITIONS
Multiplication and Division
Properties of Inequalities for
Positive Numbers
If
x <y
then ax <ay
y
x
or
<
a a
DEFINITIONS
Multiplication and Division
Properties of Inequalities for
Positive Numbers
If
x >y
then ax >ay
y
x
or
>
a a
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.7
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.7
Exercise #24
Fill in the blank with < or > to make the resulting
statement true.
a. – 3
–5
–5 –3
0
1
b. –
3
3
–
1
3
0
3
Fill in the blank with < or > to make the resulting
statement true.
a. – 3
>
–5
–5 –3
1
b. –
3
0
<
3
–
1
3
0
3
Chapter 2
Equations, Problem Solving,
and Inequalities
Section 2.7
Exercise #25
Solve and graph the inequality.
x
x
x +2
LCD = 4
a. –
+

2
4
4
2
x
x
x +2
4 –
+4
 4
2
4
4
    
–2 x + x  x + 2
–x  x + 2
–2 x  2
x  –1

Solve and graph the inequality.
x
x
x +2
a. –
+

2
4
4
x  –1
–1 0
Solve and graph the inequality.
b. x + 1 Š 3 and – 2x < 6
x Š 2 and – 2x < 6
x Š 2 and
x > –3
–3 < x Š 2
–3
0
2