Chapter 1
Tools of Algebra
Lesson 1-1
Properties of Real Numbers
Real Numbers
Rational Numbers
Irrational Numbers
1
3 1
, 0.50,1, 3 , , 0.3
2
5 3
 3, , 5,2.12479...
Integers
..., 2, 1,0,1,2,...
Whole Numbers
0,1, 2, 3,...
Natural Numbers
1,2,3,....
Example 1 – Page 8, #2, 4, 6
To which sets of numbers does the number belong to?
2
6
4  6
6  0.6
Irrational Numbers, Real Numbers
Integers, Rational Numbers, Real Numbers
Rational Numbers, Real Numbers
Example 1 – Page 8, #10
Which set of numbers best describes the value of each
variable?
the year y; the median selling price p for a house that year
Natural numbers, rational numbers
Example 2 – Page 8, #12, 14, 16
Graph each number on a number line.
12  0
14  2
2
16) 4
3
-4
-2
0
Example 3 – Page 8, #18, 20
Replace each
sentence true.
18  3
20 
6
with the symbol <, >, = to make the


< less
> greater
3
10
Opposite or Reciprocal


The opposite or additive inverse of any
number a is – a.
The reciprocal or multiplicative
inverse of any nonzero number a is 1/a.

The product of reciprocals is 1
Example 4 – Page 9, #34
Find the opposite and the reciprocal of the number.
200
Opposite: -200
1
Reciprocal:
200
Properties of Real Numbers
Property
Addition
Multiplication
Closure
ab
ab
Commutative
ab  ba
ab  ba
Associative
a  b   c  a  b  c 
 ab  c  a  bc 
Identity
a  0  a,0  a  a
a  1  a,1 a  a
a   a   0
1
a 1
a
Inverse
Distributive
a  b  c   ab  bc
Example – 5, Page 9, #42, 46, 48, 50
Name the property of real numbers illustrated by each
equation.
42  92.5(1)  92.5
Identity Prop. of Mult.
46  29    29
Comm. Prop. of Mult.
48  ( 8)   ( 8)  0
Inverse Prop. of Add
50  25(2 x  5 y )  50 x  125 y
Dist. Prop.
Absolute Value
The absolute value of a real number is the distance from
zero on the number.
Example 6 – Page 9, #54, 60
Simplify each expression.
54 
0.06  0.06
60 
5  7  5  7  2
Lesson 1-2
Algebraic Expressions
Definitions



Variable is a symbol, usually a letter that
represents one or more numbers.
An expression that contains one or more
variables is an algebraic expression or
variable expression.
When you substitute numbers for the
variables in an expression and follow the
order of operations, you evaluate the
expression.
Examples 2 – Page 15, #4
Evaluate the expression for the given values of the variables.
k 2   3k  5n   4n; k  1 and n  2
  1   3  1  5  2    4  2  
2
1   3  10   8 
1  (7)  8 
16
Example 3 – Page 15, #10
The expression 16t 2 models the distance in feet that an
object falls during t seconds after being dropped. Find
the distance an object falls during 0.5 seconds.
16t 2 where t  0.5
16  0.5   16(0.25) 
2
4 feet
Example 4 – Page 15, #30
Simplifying by combining like terms.
3(2x  1)  8
6x  3  8 
6x  5
Example 5 – Page 15, #36
Find the perimeter of the figure.
ab
P abbbabaa
 4a  2b  2b
 4a
b
b
ab
a
a
Lesson 1-3
Solving Equations
Example 2 – Page 21, #16
Solve the equation. Check your answers.
10 1  2y   5  2y  1
10  20 y  10 y  5
20 y  10 y  5  10
10 y  5
y
1
2
Example 3 – Page 21, #18
Solve the formula for the indicated variable.
1 2
s  gt ; for g
2
s gt 2

1
2
2s  t 2g
2s t 2g
 2
2
t
t
2s
g 2
t
Example 4 - Page 21, #28
Solve the equation for x. Find any restrictions.
2
 x  1  g
5
2( x  1) g

5
1
2  x  1  5g
2  x  1
2
5g

2
5g
x 1
2
5g
x
1
2
Example 5, Page 22, #32
One side of a triangle is 1 in. longer than the shortest side
and is 1 in. shorter than the longest side. The perimeter
is 17 in. Find the dimensions of the triangle.
x  1 x  1 x  17
3 x  17
x  5.66
S  5.66  1  4.66in
M  5.66in
L  5.66  1  6.66in
x
x  1S
M
L
x 1
Example 6 – Page 22, #34
The sides of a triangle are in the ratio 3 : 4 : 5. What is the
length of each side if the perimeter of the triangle is 30 cm?
3 x  4 x  5 x  30
12x  30
x  2.5
4x
5x
3 x  3(2.5)  7.5cm
4 x  4(2.5)  10cm
5 x  5(2.5)  12.5cm
3x
Examples 7 – Page 22, #30
Two planes left an airport at noon. One flew east at a
certain speed and the other flew west at twice the speed.
The planes were 2700 mi apart in 3 h. How fast was each
plane?
distance east + distance west = 2700 d  r  t
East
West
r  t  d
x
3 3x
3 6x
2x
East: 300 mi/hr and West: 600 mi/hr
3 x  6 x  2700
9 x  2700
9 x 2700

9
9
x  300
Application
The formula d  5000c  525,000 models the relationship between
the annual number of deaths (d) in the United States from heart
disease and average adult cholesterol level (c, in milligrams per
deciliter of blood). In 1990, 500,000 American died from heart
disease. What was the average cholesterol level at the time? If
the United States could reduce its average cholesterol level to 180,
how many lives could be saved compared to 1990?
Let d = 500,000 deaths
d  5000c  525,000
500,000  5000c  525,000
1,025,000  5000c
205  c
In, 1990, the average
cholesterol level was 205
Application
Let c = 180 cholesterol level
d  5000c  525,000
d  5000(180)  525,000
 375,000
500,000  375,000  125,000
125,000 lives could be saved
Lesson 1-4
Solving Inequalities
Example 1 – Page 29, #6
Solve the inequality. Graph the solutions
14  4y  38
4 y  38  14
4 y  24
4 y 24

4
4
y  6
-6
0
Example 2 – Page 29, #10
Solve the inequality. Graph the solutions
9  x  2  9  x  3 
9 x  18  9 x  27
9 x  9 x  18  27
18  27
True
All real numbers are solutions
0
Example 3 – Page 29, #14
Solve the problem by writing the inequality.
The length of a picture frame is 3 in. greater than the
width. The perimeter is less than 52 in. Describe the
dimensions of the frame.
2( x  3)  2( x )  52
l  x 3
w x
P  2l  2w  52
2 x  6  2 x  52
4 x  6  52
4 x  52  6
4 x  46
4 x 46

4
4
x  11.5
Example 4, Page 29, #20
Solve each compound inequality. Graph the solution.
6 x  24 and 9 x  54
x  4
x6
4
0
6
4  x  6
Example 5 – Page 30, #22
Solve the compound inequality. Graph the solution.
4 x  16 or 12x  144
x4
x  12
0
4
12
x  4 or
x  12
Example 6 – Page 30, #26
A baker needs between 40 lb and 50 lb of a flour-sugar
mixture that contains ten times as much flour as sugar.
What are the possible weights of flour the baker can use?
40  x  50
40  f  0.10 f  50
40  1.10 f  50
40  1.10 f
36.36  f
f  36.36
1.10 f  50
f  45.45
Lesson 1-5
Absolute Value Equations and
Inequalities
Absolute Value Equations
x 4
4 4
x  4 or
4  4
x  4
Examples 2 – Page 36, #4
Solve the equation. Check your answers.
2 3 x  2  14
2 3x  2
2
14

2
3x  2  7
3x  2  7 or
3x  2  7
3x  7  2
3 x  7  2
3x  9
3 x  5
3x 9

3
3
3 x 5

3
3
x 3
or
5
x
3
Examples 2 – Page 36, #6
Solve the equation. Check your answers.
2 x  3  1
2 x  3  1
No Solutions
Extraneous Solution

An extraneous solution is a solution
of an equation derived from an original
equation that is not a solution of the
original equation
Example 3, Page 36, #12
Solve the equation. Check for extraneous solutions
3x  5  5x  2
3 x  5  5 x  2 or
3x  5   5x  2
3 x  5x  2  5
3 x  5  5 x  2
2x  3
3 x  5 x  2  5
2 x 3

2
2
3
x
2
8 x  7
8 x 7

8
8
7
x
8
Absolute Value Inequalities
x 4
x 4
x  4
x4
-4
0
x  4 or
4
x  4
x4
x  4
-4
0
4
x  4 and x  4
4  x  4
Example 4 – Page 36, #16
Solve the inequality. Graph the solution.
x 3 9
x  3  9 or
x  3  9
x  93
x  9  3
x6
x  12
-12
x  6 or
0
6
x  12
Example 5 – Page 36, #26
Solve the inequality. Graph the solution.
4 2w  3  7  9
2w 3  4 and 2w  3  4
4 2w  3  9  7
2w  4  3
2w  4  3
4 2w  3  16
2w  1
2w  7
4 2w  3
2w 1

2
2
1
w
2
16

4
4
2w  3  4
2w 7

2
2
1
w  3
2
Example 5 – Page 36, #26
1
w
2
-3.5
and
0
1
w  3
2
0.5
3.5  w  0.5