Worksheet 5 (2.1)
Chapter 2 Equations and Inequalities
2.1 Solving First-Degree Equations
Important Terminology Used when Solving Equations
An equation is a statement that two symbols or groups of symbols are names for
An algebraic equation contains one or more variables. It is called an open
sentence because it is neither true nor false.
Solving an equation is the process used to find the number or numbers that
make
an algebraic equation a true numerical statement.
A solution or root of an equation is the number that satisfies a given equation.
The solution set is the set of all solutions of an equation.
First-degree equations in one variable are equations with one variable which has
an exponent of one.
Equivalent equations are equations that have the same solution.
Summary 1:
Warm-up 1. a) Is 5 a solution of 2x + 4 = 19 - x?
Verify this by showing that the numerical statement is true or false.
2( ) + 4 = 19 +4=
True or False?
14 =
Answer: _________
yes or no
Problem
1. Is 0 a solution of 3(x - 1) = 2x + 5?
Verify this by showing that the numerical statement is true or false.
Worksheet 5 (2.1)
17
the same nu
Further Properties of Equality Used to Solve Equations
1. Addition Property of Equality
For all real numbers a, b, and c, a = b if and only if a + c = b + c.
Note: Subtraction can be rewritten as addition. Therefore, this property
also allows
2. Multiplication Property of Equality
For all real numbers a, b, and c, where c  0, a = b if and only if ac = bc.
Note: Division can be rewritten as multiplication. Therefore, this property
Summary 2:
Warm-up 2. Solve:
a) - 11= 3x + 1
- 11 + = 3x + 1 +
- 12 = 3x
( )(-12 ) = ( ) ( 3 x)
=x
The solution set is
x=
.
b)
2(x - 1) = 5x + 7
2x - = 5x +7
2x - 2 + = 5x +7 +
- 3x - 2 = 7
- 3x - 2 + = 7 +
- 3x =
Worksheet 5 (2.1)
(
)(-3 x)= ( )( 9 )
x=
The solution set is
Problems - Solve:
2. -5x - 3 = 12
18
.
also allows
3. 2(2x - 3) = 5x - 11
Solving First-Degree Equations in One Variable
1. Simplify each side of the equation.
2. Apply the addition property of equality to isolate the variable term on one side
3. Apply the multiplication property of equality to obtain the coefficient of 1 for
4. Write an appropriate solution set and check the solution to verify that the
resulting numerical statement is true.
Summary 3:
Warm-up 3. Solve and check:
a)
5n - 4 - 2n = 4n + 12 + 3n
-4 =
+ 12
3n - 4 +
= 7n + 12 +
-4n - 4 = 12
-4n - 4 +
= 12 +
-4n =
( )(-4n) = ( )(16)
n =
The solution set is
.
Worksheet 5 (2.1)
Check:
b)
5( ) - 4 - 2( ) = 4( ) + 12 + 3( )
____ - 4 - ____ = ____ + 12 + ____
____ = ____ True or False?
4(m - 1) - (m + 6)
4m - 4 - m - ____
____ - 10
3m - 10 + ____
____ - 10
-2m - 10 + ____
-2m
19
= 5(m + 2)
= 5m + 10
= 5m + 10
= 5m + 10 + ____
= 10
= 10 + ____
=
and constant
the variable.
(
)(-2m) = ( )(20)
m = ____
The solution set is _______.
Problems - Solve and check:
4. 3n - 1 - 4n = 6n - 7 - 4n
5. -(n - 5) + 3(n + 2) = 4(n - 3) - 1
Using Equations to Solve Word Problems
1. Carefully read the word problem. Reread to get an overview of the situation.
Determine known facts and identify unknown quantities.
2. Declare the variable by choosing a letter to use as a variable and tell in words
3. Express other unknown quantities using this variable.
4. Translate the word problem or use a guideline from the word problem to
write an algebraic equation to solve.
5. Check results to determine whether or not they satisfy the conditions stated in
6. Express answer as a complete sentence.
Summary 4:
Worksheet 5 (2.1)
Warm-up 4. Set up and write an algebraic equation, then solve:
a) The sum of two numbers is 55. The larger number is eleven more
than the smaller number. Find the two numbers.
Declare the variable:
Let
n
= the smaller number
= the larger number
20
what it will
the original
Write an algebraic equation and solve:
n + _______
2n + ____
2n + 11 + ____
2n
( )(2n)
n
= 55
= 55
= 55 + ____
= ____
= ( )(44)
= ____ ; n + 11 = ____
The numbers are _______ and _______.
Problems - Set up and write an algebraic equation, then solve:
6. In a class containing 58 students, the number of women students is one more than
twice the number of men students. How many women students are in the class?
7. An air-conditioner repair bill is $87. This included $30 for parts and an amount for 3
hours of labor. What was the hourly rate that was charged for labor?
Worksheet 6 (2.2)
2.2 Equations Involving Fractional Forms
Using the LCD to Clear Fractions in an Equation
Summary 1:
21
The least common denominator, LCD, refers to the least common multiple of
a set of deno
Clearing the equation of all fractions is accomplished by applying the
multiplicatio
When an equation has more than one term on one or both sides of the equation, the
distributive property is used to multiply each term by the LCD.
Warm-up 1. Find the LCD and multiply both sides of the equation by the LCD
to clear the fractions. Give the resulting equivalent equation.
3 1
2
- + x=
a)
The LCD is _____.
4 2
3
 3 1 
2
- + x=  
 4 2 
3
 3
1 
2
- +  x=  
 4
2 
3
-9 + _____ = _____
The resulting equivalent equation is __________________________.
2
1
(4x - 1) + 1 = - (5x + 2)
b)
The LCD is _____.
5
4
2

 1



4x
1
+
1
=
 5

- 4 5x + 2 
2

1= - 1 (5x + 2)
 5 (4x - 1) +
 4

(4x - 1) + = (5x + 2)
The resulting equivalent equation is ___________________________.
Worksheet 6 (2.2)
Problems - Find the LCD and multiply both sides of the equation by the LCD to
clear the fractions. Give the resulting equivalent equation.
4x - 1 3x + 1
=
- x+4
1.
9
3
2.
1
1
(6x - 1) + = - 2x - 5
2
2
22
Solving Equations Involving Fractional Forms
1. Find the LCD in the equation.
2. Apply the multiplication property of equality to clear the equation of
all fractions. This is done by multiplying each term in the equation by the LCD.
(See summary 1 above.)
3. Solve the resulting equivalent equation.
4. Check when directed to do so.
Summary 2:
3 1
2
Warm-up 2. a) Solve and check: - + x =
4 2
3
 3 1 
2
- + x=  
 4 2 
3
 3
1 
2
- +  x=  
 4
2 
3
-9 + _____ = _____
_____ = 17
x = _____
The solution set is ______.
Worksheet 6 (2.2)
3 1
Check: - + (
4 2
3
- +
4
2
3
2
=
3
)=
23
=
b) Solve:
2
True or False?
3
4x - 1 3x + 1
=
- x+4
9
3
 4x - 1 
 3x + 1

- x+4

=

 9 
 3

 4x - 1 
 3x + 1 

= 
 - (x)+
 9 
 3 
4x - 1= (3x + 1) - 9x + 36
4x - 1= 9x + - 9x + 36
4x - 1=
4x =
x=
The solution set is ______.
(4)
Problems - Solve:
3a - 1 a - 2 21 a - 1
+
= +
3.
4
3
20
5
4.
1
1
(2x - 1) = 3 + (5x + 2)
2
3
Worksheet 6 (2.2)
Summary 3:
Suggestions for Solving Word Problems
24
1. Carefully read the word problem. Reread to get overview of given
situation. Determine known facts and identify unknown quantities.
2. Utilize a chart, figure, or diagram to clarify the problem.
3. Declare the variable and express unknown quantities in terms of the variable.
4. Find a guideline in the word problem.
5. Translate into an equation using the declared variable.
6. Solve the equation and use solution to determine all facts asked for in the
7. Check results in the original problem. Express answer in a complete
sentence.
Warm-up 3. Set up and write an algebraic equation, then solve:
a) Find a number such that one-half of the number is three more than
one-fourth of the number.
Declare the variable:
Let ____ = a number
Write an algebraic equation and solve:
4(
=
+3
) = 4( ) + 4(3)
=
+ 12
= 12
The number is _______.
Worksheet 6 (2.2)
b) In triangle ABC, angle A is one-third angle C and angle B is 40
degrees more than two-thirds of angle A. Find the measures of the
three angles in the triangle.
Draw a figure:
25
problem.
Declare the variable:
Let
x = measure of angle C
_________ = measure of angle A
____________ = measure of angle B
Write an algebraic equation and solve:
x+
+
9( ) + 9(
) + 9(
9x + 3x +
= 180
) = 9(180)
= 1620
14x + 360 = 1620
14x = 1260
x=
x ( )
=
=
3
3
2x
2( )
+ 40 =
+ 40 =
9
9
The angle measures are _____, _____, and _____.
Worksheet 6 (2.2)
Problems - Set up and write an algebraic equation, then solve:
5. John is paid 2½ times his normal hourly rate for each hour he works on a holiday. Last
week he worked 45 hours, five hours were on a holiday. Find his normal hourly
rate if he earned $420.
26
6. Maria took three college algebra exams and had an average score of 84. Her
second exam was eight points better than her first and her third exam was eleven
points better than her second exam. Find the three exam scores.
Worksheet 7 (2.3)
2.3 Equations Involving Decimals and Problem Solving
Summary 1:
Using 10n, a Power of 10, to Clear Decimals in an Equation
27
Multiplying a decimal by 10n will result in moving the decimal point n places to
Clearing the equation of all decimals is accomplished by applying the
multiplication property of equality. Both sides of the equation are to be
multiplied by an appropriate power of 10 to produce an equivalent
equation. This power of 10 is equal to the most number of decimal
places for any given decimal in the equation.
When an equation has more than one term on one or both sides of the equation, the
distributive property is used to multiply each term by the power of 10.
Warm-up 1. Find an appropriate power of 10 to clear all decimals. Give the
resulting equivalent equation:
a)
2y = .6y + .21
The power of 10 is _______.
_____(2y) = _____(.6y + .21)
_____(2y) = _____(.6y) + _____(.21)
The resulting equivalent equation is ____________________.
Problem - Find an appropriate power of 10 to clear all decimals. Give the
resulting equivalent equation:
1. 0.05x + 0.08(10000 - x) = 620
Worksheet 7 (2.3)
Summary 2:
Solving Equations Involving Decimals
28
the right.
1. Find an appropriate power of 10 to clear all decimals using the multiplication
2. Solve the resulting equivalent equation.
3. Check when directed to do so.
Warm-up 2. a) Solve and check:
.07x + 160 = 152 + .08x
_____(.07x + 160) = _____(152 + .08x)
_____(.07x) + _____(160) = _____(152) + _____(.08x)
_____ + 16000 = _______ + 8x
7x + 16000 + _____ = 15200 + 8x + _____
_____ + 16000 = 15200
-1x + 16000 + _______ = 15200 + ________
-1x = _____
_____(-1x) = _____(-800)
x = _____
The solution set is _______.
Check:
.07( ) + 160 = 152 + .08( )
_____ + 160 = 152 + _____
_____ = _____ True or False?
b) Solve:
.11(7000 - x) - .12x
_____[.11(7000 - x) - .12x]
_____[.11(7000 - x)] - _____[.12x]
11(7000 - x) - _____
77000 - _____ - 12x
77000 - _____
77000 - 23x + ________
-23x
_____(-23x)
x
= 310
= _____[310]
= _____[310]
= ________
= 31000
= 31000
= 31000 + ________
= __________
= _____(-46000)
= __________
The solution set is __________.
Worksheet 7 (2.3)
Problems - Solve:
2. 0.222x = 0.2 - 0.22x , round to nearest hundredth.
29
property of e
3. 0.06x + 0.05(8000 - x) = 450
Suggestions for Solving Word Problems
1. Carefully read the word problem. Reread to get overview of given
situation. Determine known facts and identify unknown quantities.
2. Utilize a chart, figure, or diagram to clarify the problem.
3. Declare the variable and express unknown quantities in terms of the variable.
4. Find a guideline in the word problem.
5. Translate into an equation using the declared variable.
6. Solve the equation and use solution to determine all facts asked for in the
7. Check results in the original problem. Express answer as a complete sentence.
Summary 3:
Warm-up 3. Set up and write an algebraic equation, then solve:
a) Kayla bought an opal necklace at a 15% discount sale for $1150.
Find the original price of the necklace, rounded to the nearest cent.
Declare the variable:
Let
x = original price
_______ = amount of discount
Write an algebraic equation and solve:
x - _______ = _______
_____(x - .15x) = _____(1150)
_____(x) - _____(.15x) = _____(1150)
100x - _____ = 115000
_____ = 115000
x = _______ The original price was______.
Worksheet 7 (2.3)
b) A total of $12,000 was invested, part at 6% and the remainder at
8%. Find how much was invested at each rate, if the total yearly
interest earned was $820.
Declare the variable:
30
problem.
Let
x
___________
___________
___________
= amount invested at 6%
= amount invested at 8%
= interest earned at 6%
= interest earned at 8%
Write an algebraic equation and solve:
.06x + .08(__________)
_____[.06x + .08(12000 - x)]
_____[.06x] + _____[.08(12000 - x)]
6x + _____(12000 - x)
6x + ________ - 8x
_____ + 96000
-2x
x
12000 - x
= _____
= _____(820)
= _____(820)
= 82000
= 82000
= 82000
= ________
= ________
= ________
There was __________ invested at 6% and __________
invested at 8%.
Problems - Set and write an algebraic equation, then solve:
4. Higinio bought a car with 6.5% sales tax included for $17,944. Find the price of
the car, rounded to the nearest cent, before tax.
5. Xurry invests a certain amount of money at 7% interest and $1500 less than that
amount at 4.5%. His yearly interest is $392.50, find how much he invested at each
rate.
Worksheet 8 (2.4)
2.4 Formulas
Summary 1:
31
Literal equations are equations that contain more than one variable.
Formulas are usually literal equations that state a rule in symbolic form.
Warm-up 1. a) Solve i = Prt for i, given that P = $800, r = 8 21 % ,
and t = 3 years.
i = Prt
i = (800)(
)(3)
i = _______
b) Solve 3(x - 2y) = 4 for y, given that x = -1.
3(x - 2y)
3( - 2y)
3( ) - 3( )
-3 - _____
-3 - 6y + _____
-6y
_____(-6y)
=4
=4
=4
=4
= 4 + _____
= _____
= _____(7)
y = _____
Problems
1. Solve A = P + Prt for r, given that A = $1080, P = $800, and t = 7 years.
Express r as a percent.
2. Solve 2x - 5y = -10 for x, given that y = 0.
Worksheet 8 (2.4)
Changing the Form of a Literal Equation or Formula
Summary 2:
32
1. If necessary, clear all fractions using the LCD and multiplication property of
equality.
2. Apply distributive property to clear parentheses when appropriate.
3. Use the addition property of equality to collect terms containing the desired
variable on one side and all other terms on the opposite side.
4. If two or more terms contain the desired variable on one side, use the
distributive property to rewrite the expression.
5. Apply the multiplication property of equality to obtain a coefficient of 1 for
the desired variable.
6. If needed, apply the symmetric property of equality so the desired variable is on
the left side of the equation.
Warm-up 2. a) Solve A = 21 h( b1 + b2 ) for b2.
_____[A] = _____ 21 h( b1 + b2 )
_____ = h(b1 + b2)
2A = _____ + _____
2A + _____ = hb1 + hb2 + _____
2A - hb1 = _____
_____(2A - hb1) = _____(hb2)
h
= b2
b2 =
h
b) Solve -3x - y = 5 + 4y for y.
-3x - y + _____
-3x - _____
-3x - 5y + _____
-5y
_____(-5y)
y=
-5
= 5 + 4y + _____
=5
= 5 + _____
= _____ + _____
= _____(3x + 5)
or
y=
Worksheet 8 (2.4)
Problems - Solve:
3. A= 2lh + 2hw + 2lw , for w.
4.
x-c y-c
=
, for x.
b
a
33
5
Suggestions for Solving Word Problems Using Formulas
1. Carefully read the word problem. Reread to get overview of given
situation. Determine known facts and identify unknown quantities.
2. Utilize a chart, figure, or diagram to clarify the problem.
3. Declare the variable and express unknown quantities in terms of the
variable.
4. Use a formula as a guideline in the word problem.
5. Translate into an equation using the declared variable.
6. Solve the equation and use the solution to determine all facts asked for in
7. Check results in the original problem. Express answer as a complete
Summary 3:
Warm-up 3. Set up and write an algebraic equation, then solve:
a) The length of a rectangular lot is 16 feet. If the total distance around
the lot is 56 feet, find the width of the lot.
Draw a figure:
Worksheet 8 (2.4)
Declare the variable:
Let x = width
Write an algebraic equation and solve:
P = 2l + 2w
_____ = 2( ) + 2( )
_____ = 32 + _____
_____ = 2x
_____ = x
34
the problem.
sentence.
The width is __________.
b) After 3 hours, two cars traveling in opposite directions on I-24 are
360 miles apart. One car is traveling 10 mph faster than the other
car. Find the rate of speed for both cars.
Declare the variable:
Let
x = speed of slower car
_______= speed of faster car
Chart:
Diagram:
d=rt
r
t
-----------------------------------fast
x + 10 3
3x 3(
)
------------------------------------ <-------+---------->
slow
x
3
distance traveled
-----------------------------------360 miles
Write an algebraic equation and solve:
3x + ___________
3x + 3x + _____
_____ + 30
6x
x
= _____
= 360
= 360
= _____
= _____ ; x + 10 = _____
The speed of the slower car is _______ and the speed of the
faster car is _______.
Worksheet 8 (2.4)
Problems - Set up and write an algebraic equation, then solve:
5. How many milliliters of 15%-salt solution must be added to 200 ml of 25%-salt
solution to obtain a 20%-salt solution?
35
6. In a given triangle, the largest angle is three times the smallest angle. The third
angle is ten more than the smallest angle. Find the measures of all three angles.
Worksheet 9 (2.5)
2.5 Inequalities
Summary 1:
Statements of inequality express one of the following:
36
1.
2.
3.
4.
a <b
ab
a >b
ab
means a is less than b.
means a is less than or equal to b.
means a is greater than b.
means a is greater than or equal to b.
Algebraic inequalities contain one or more variables. They are open
sentences which are neither true nor false.
Solving an inequality is the process used to find the number or numbers
that
make an algebraic inequality a true numerical statement. These
numbersare
solutions of the inequality.
Warm-up 1. a) Is -3 a solution of 2x + 5  2 - x ?
Verify this by showing that the numerical statement is true or
false.
2( ) + 5  2 - ( )
+5  2+
 5 True or False?
Answer: _________
yes or no
Problem
1. Is 2 a solution of 2x + 5  2 - x ?
Verify this by showing that the numerical statement is true or false.
Worksheet 9 (2.5)
Properties of Inequalities
Summary 2:
37
1. Addition Property of Inequality
For all real numbers a, b, and c, a > b if and only if a + c > b + c.
Note: When adding the same number on both sides of the inequality,
the s
2. Multiplication Property of Inequality
a) For all real numbers a, b, and c with c > 0,
a > b if and only if ac > bc.
Note: When multiplying the same positive number on both sides of
b) For all real numbers a, b, and c with c < 0,
a > b if and only if ac < bc.
Note: When multiplying the same negative number on both sides of
the inequality, the sense of the inequality changes; therefore,
In general, the process for solving inequalities closely parallels that for
Warm-up 2.
Solve:
a)
b)
2x - 5 > 7
2x - 5 + _____
> 7 + _____
2x ___ 12
_____(2x)
> _____(12)
x ___ 6
x + 2  5x - 14
x + 2 +  5x - 14 +
x  5x x +  5x - 16 +
Worksheet 9 (2.5)
 - 16
(-4x)  (-16)
x 4
Problems - Solve:
2. - 3x + 8  2
38
the i
solving equ
3. 7x + 8 > 2(x - 6)
Summary 3:
Appropriate Ways to Express the Solution of an Inequality
SET
{ x| x > a }
{ x| x  a }
{ x| x <b }
{ x| x  b }
GRAPH
(===>
a
<
[===>
a
<====)
>
b
<====]
>
b
INTERVAL NOTATION
( a, infinity )
<
[ a, infinity )
( - infinity, b )
( - infinity, b ]
Notes:
1. Set notation like { x | x < - 4 } is translated as "the set of all x such that
2. The symbol ( or ) is used to exclude an endpoint or with the
3. The symbol [ or ] is used to include an endpoint.
Worksheet 9 (2.5)
Warm-up 3. a) Sketch a graph of x < 3 and express in interval notation.
Graph:
<───┬──────┬──────┬──────┬──────┬──────┬──────┬─>
-1
0
1
2
3
4
5
Interval notation:
b) Express (-, 8] in set-builder notation.
Set: { x |
}
Problems
39
x is less than
infini
4. Express the graph below both in set notation and interval notation:
<
-6 -5
[========>
-4
5. Sketch a graph for the solution expressed as (3 , ).
Summary 4:
Key Factors to Consider when Solving Inequalities
1. The solving process for inequalities closely parallels that of solving
equations.
2. The inequality sign reverses when multiplying both sides of the inequality
by a negative
3. Use an appropriate format to express the solution of an inequality - set,
graph, or int
4. One solution can be checked to possibly catch a mistake. It is not
possible to c
Worksheet 9 (2.5)
Warm-up 4.
a) Solve and graph the solution set on a number line:
2(3 - x) < 14
6 - _____ < 14
6 - 2x + _____ < 14 + _____
-2x < _____
_____(-2x) < _____(8)
x ___ -4
Graph:
b) Solve and express the solution set using interval notation:
2(x + 4) - (x - 2)  - 3(x - 1) -3(x - 1)
2x + 8 - _____ + _____  -3x + _____
x + _____  -3x + 3
40
x + 10 + _____
_____ + 10
4x + 10 + _____
4x
_____(4x)
x






-3x + 3 + _____
3
3 + _____
_____
_____(-7)
_____
Interval notation:
Problems
6. Solve and graph the solution set on a number line:
-3 + 6x > -15
7. Solve and express the solution set using interval notation:
-3(2x + 1) - 2  -2(x + 5)
Worksheet 10 (2.6)
2.6 More on Inequalities and Problem Solving
Compound Inequalities
Summary 1:
41
Compound statements are formed when mathematical statements are joined
the words and or or.
by
Conjunctions are compound statements that use and. The solution set for a
given
conjunction is the intersection of the solution sets for the
inequalities
joined by the word and.
Expressing the Solution of a Conjunction:
SET
{ x|a < x <b }
GRAPH
(===) >
a b
< [===) >
a b
< (===] >
a b
< [===] >
a b
<
{ x|a  x <b }
{ x | a < x  b}
{ x | a  x  b}
INTERVAL NOTATION
(a, b)
[a, b)
(a, b]
[a, b]
Note: a < x < b is the compact form for a < x and x < b.
The compact form can only be used for conjunctions where the
Disjunctions are compound statements that use or. The solution set for a
given
disjunction is the union of the solution sets for the inequalities
joinedbythe
word or.
Expressing the Solution of a Disjunction:
SET
{ x | x < a or x > b }
GRAPH
INTERVAL NOTATION
( - infinity, a) union ( b, infinity )
<===)
(===>
a b
Note: The endpoints a and b will be included in the solution for , .
In this case, use [ or ].
Worksheet 10 (2.6)
Warm-up 1. Graph the compound inequality and express in interval notation:
a) x > -2 and x < 3
Give the compact form of this statement: __________
Graph:
42
solution is be
<─┬──────┬──────┬──────┬──────┬──────┬──────┬──────┬─>
Interval notation:
b) x  0 or x > 4
Graph:
<─┬──────┬──────┬──────┬──────┬──────┬──────┬──────┬─>
Interval notation:
c) x > -2 or x > 3
Graph:
<─┬──────┬──────┬──────┬──────┬──────┬──────┬──────┬─>
Interval notation:
Problems - Graph the compound inequality and express in interval notation:
1. x > 1 and x < 3 (What is the compact form of this statement?)
2. x < -5 or x > 1
3. x  -3 or x < 2
Worksheet 10 (2.6)
Summary 2:
Key Factors to Consider when Solving Compound Statements
43
1. For disjunctions, solve each inequality separately in the compound
sentence. The solution set is the union of these solutions.
2. For conjunctions, solve each inequality separately in the compound
sentence. The solution set is the intersection of these solutions.
Note: If the conjunction is in compact form, it can be solved by isolating
Warm-up 2. Solve, then express the solution set as a graph and in interval
notation:
a) x - 5 < -3 or x - 3 > 3
x - 5 + _____ < -3 + _____
or
x < _____
x - 3 + _____ > 3 + _____
x > _____
The solution set is {x________or________}.
Graph:
<─┬──────┬──────┬──────┬──────┬──────┬──────┬─>
Interval notation:
__________ union __________
b) - 8  3x - 2  7
-8+
 3x - 2 +
7 +
 3x 
(-6 ) 
(3 x) 
-2
(9 )
3
Worksheet 10 (2.6)
The solution set is {x_______________}.
Graph:
<─┬──────┬──────┬──────┬──────┬──────┬──────┬──────┬─>
Interval notation:
44
the v
Problems - Solve, then express the solution set as a graph and in interval
notation:
4. x - 3 > -2 or x + 1 < 3
5. 0  4x + 2 < 8
Summary 3:
Solving Word Problems Involving Inequalities
1. The strategies previously outlined for equations hold true for inequalities as wel.
2. Analyze the situation to determine which inequality symbol is appropriate.
Worksheet 10 (2.6)
Warm-up 3. Set up and write an algebraic inequality, then solve:
a) In general chemistry, Kate scored 95, 83, 89, and 85 on four exams
this semester. If she is to earn an A, 90 or higher, for the semester,
what must she score on the fifth and final test for the semester?
Declare the variable:
Let x = score on the fifth test
45
Write an algebraic inequality and solve:
95 + 83 + 89 + 85 + ( )
 90
( )
 95 + 83 + 89 + 85 + x 
_____ 
  _____(90)
5


95 + 83 + 89 + 85 + x  _____
_____ + x  450
352 + x + _____  450 + _____
x  _____
She must score at least ______.
Problem - Set up and write an algebraic inequality, then solve:
6. Darla has $2000 to invest. If she invests $1000 at 9% interest, at what rate must she
invest the remaining amount so that the two investments earn more than $130
of combined yearly interest?
Worksheet 11 (2.7)
2.7 Equations and Inequalities Involving Absolute Value
Solving Absolute Value Statements when k > 0
Summary 1:
46
Absolute value equations and inequalities must be converted to their
corresponding equivalent compound statements when k > 0. The compound
1. | ax + b |= k is equivalent to ax + b = - k or ax + b = k
2. | ax + b | < k is equivalent to ax + b > - k and ax + b < k
or - k < ax + b < k
3. | ax + b | > k is equivalent to ax + b < - k or ax + b > k
Warm-up 1. a) Solve and give the solution set in set notation:
| 2 - m |= 4
or 2 - m =
2 - m=
2 - m+ = - 4 +
2 - m+ = 4 +
- m=
- m=
(-m) = (-6)
(-m)= (2)
or
m=
m= 6
The solution set is {
,
}
b) Solve and graph the solution set:
| 2x + 1 |  11
 2x + 1 
- 11+  2x + 1+  11+
Worksheet 11 (2.7)
- 12  2x 
(-12)  (2x)  (10)
 x 
Graph:
Problem - Solve and write solution in interval notation:
1. | 6 - 7x | > 22
47
statement ca
Solving Absolute Value Statements when k  0
Absolute value equations and inequalities are solved by inspection
when k < 0.
1. For | ax + b |= k , the solution set is .
2. For | ax + b | < k , the solution set is .
3. For | ax + b | > k , the solution set is (-, ).
Absolute value equations have exactly one solution when k = 0.
1. | ax + b |= 0 is equivalent to ax + b = 0.
Summary 2:
Worksheet 11 (2.7)
Warm-up 2. Solve:
a) | 2x + 1| < - 5
The solution set is _____.
b) | 2x + 1| > - 5
The solution set is _____.
c) | 2x + 1|= 0
2x + 1 = 0
x = _____
The solution set is _____.
Problems - Solve:
2. | y - 5 | > - 2
48
3. | 3m + 4 | < - 10
49