CHAPTER
2
2.1
2.2
2.3
2.4
2.5
2.6
Solving Linear Equations and
Inequalities
Equations, Formulas, and the ProblemSolving Process
The Addition Principle of Equality
The Multiplication Principle of Equality
Applying the Principles to Formulas
Translating Word Sentences to Equations
Solving Linear Inequalities
2.1
Equations, Formulas, and the
Problem-Solving Process
1. Verify solutions to equations.
2. Use formulas to solve problems.
Equation: Two expressions set equal.
For example, 4x + 5 = 9 is an equation
made from the expressions 4x + 5 and 9.
Solution: A number that makes an equation true
when it replaces the variable in the equation.
For example, the number 5 is the solution
to the equation x + 2 = 7 because when 5
replaces x, it makes the equation true.
Slide 2- 3
Checking a Possible Solution
To determine whether a value is a solution to a
given equation, replace the variable in the equation
with the value. If the resulting equation is true, the
value is a solution.
Slide 2- 4
Example 1a
Check to see if –3 is a solution to 6x + 5 = –13.
Solution
6x + 5 = –13
x with –3 to see if
6(–3) + 5 = –13 Replace
the equation is true.
–18 + 5 = –13
–13 = –13
Since –3 makes the equation true, it is a solution to
6x + 5 = –13.
Slide 2- 5
Example 1b
Check to see if –0.2 is a solution to x2 + 0.7 = 1.4.
Solution
x2 + 0.7 = 1.4
Replace x with –0.2 to see if
2
(−0.2) + 0.7 = 1.4
the equation is true.
0.04 + 0.7 = 1.4
0.74 ≠ 1.4
Because −0.2 does not make the equation true, it is
not a solution to x2 + 0.7 = 1.4.
Slide 2- 6
Problem-Solving Outline
1. Understand the problem.
a. Read the question(s) (not the whole problem, just
the question at the end) and write a note to yourself
about what it is you are to find.
b. Read the whole problem, underlining the key
words.
c. If possible or useful, draw a picture, make a list or
table or organize what is known and unknown,
simulate the situation, or search for a related example
problem.
2. Plan your solution by searching for a formula or
using key words to translate to an equation.
Slide 2- 7
Problem-Solving Outline continued
3. Execute the plan by solving the equation/formula.
4. Answer the question. Look at the note about what
you were to find and make sure you answer that
question. Include appropriate units.
5. Check results.
a. Try finding the solution in a different way, reversing
the process, or estimating the answer and make sure
the estimate and actual answer are reasonably close.
b. Make sure the answer is reasonable.
Slide 2- 8
Formula: An equation that describes a
mathematical relationship.
Perimeter: The distance around a figure.
Area: The total number of square units that
fill a figure.
Volume: The total number of cubic units
that fill a space.
Circumference: The distance around a
circle.
Radius: The distance from the center of a
circle to any point on the circle.
Diameter: The distance across a circle
through its center.
Slide 2- 9
Using a Formula
To use a formula:
1. Replace the variables with the corresponding given
values.
2. Solve for the missing value.
Slide 2- 10
Geometric Formulas—Plane Figures
Square
Rectangle
Parallelogram Trapezoid
a
s
s
w
l
P = 4s
A = s2
P = 2l + 2w
A = lw
Triangle
Circle
h
h
b
b
A = bh
1
A  h( a  b)
2
r
d
1
A  bh
2
C   d or C  2 r
A   r2
Slide 2- 11
Geometric Formulas--Solids
Box
Pyramid
Cylinder
Cone
W
V  lwh
SA  2lw  2lh  2wh
L
1
V  lwh
3
V   r 2h
1 2
V  r h
3
Sphere
4
V   r3
3
Slide 2- 12
Example 2a
A kennel owner is enclosing a rectangular area with
fencing. The kennel is to be 60 feet in length and 25
feet wide. What is the total amount of fencing
needed, disregarding a gate?
Understand The fencing creates the perimeter of the
kennel.
Plan Because the shape is a rectangle, we can use the
formula P = 2l + 2w.
Execute Replace l with 60 feet and w with 25 feet,
then calculate.
Slide 2- 13
continued
P  2l  2 w
P  2  60  2  25
P  120  50
P  170 ft.
Answer The fencing needed for the kennel is 170 ft.
Check The solution can be verified using an
alternative method. Add all four side lengths.
P  60  60  25  25  170 ft.
Slide 2- 14
Example 2b
A circular window is to be covered with an applied
stencil. If the window has a radius of 3 feet, how
much of the window cover is needed?
Understand The amount of the window stencil is the
same as the area of the window.
Plan Because the shape is a circle, we can use the
2
A


r
.
formula
Execute Replace  with 3.14 and r with 3 ft.
and simplify.
Slide 2- 15
continued
A  r
2
A   3.14  3ft.
2
A   3.14   9ft.2 
A  28.26 ft 2
Answer About 28.26 square feet of window stencil is
needed to cover the window.
Check Verify the reasonableness of the answer using
estimation. If we round  to 3, the answer is
(3)(3 ft)(3 ft) = 27 ft2. Because  is a little more than
3 and our answer is a little more than 27 ft.2, our
answer is reasonable.
Slide 2- 16
Calculating the Area of Composite Figures
For composite figures:
1. To calculate the area of a figure composed of two
or more figures that are next to each other, add
the areas of the individual figures.
2. To calculate the area of a region defined by a
smaller figure within a larger figure, subtract the
area of the smaller figure from the area of the
larger figure.
Slide 2- 17
Example 3
Following is a drawing of a deck that is to have
brick flooring installed. Calculate the area of the
deck.
12 ft
6 ft
Slide 2- 18
continued
Understand The figure can be viewed as a rectangle
and a semi-circle. The diameter of the circle is the
same as the width of the rectangle.
Plan We find the area of both shapes and add them
together to get the total area.
A = Area of the rectangle + Area of the semi-circle.
A
lw
+
1
2
 r2
Slide 2- 19
continued
Execute Replace the variables with the corresponding
values and calculate.
A  lw+ 12  r 2
A  12 ft  6 ft  
 3.14  3 ft 
A  12 ft  6 ft   12  3.14   9 ft 2 
1
2
2
A  72 ft 2  14.13 ft 2
A  86.13 ft 2
Slide 2- 20
continued
Answer The total area is 86.13 square feet.
Check Verify the reasonableness of the answer using
estimation. Suppose the figure had been a rectangle
measuring 6 ft by 15 ft. We would expect the area to
be slightly larger than that of the actual figure.
A = (6 ft.)(15ft.) = 90 ft2, which indicates that 86.13
square feet is reasonable.
Slide 2- 21
Sometimes, a figure may be placed within
another figure and we must calculate the area of
the region between the outside and inside figure.
Slide 2- 22
Example 4
A family room is being carpeted. Determine the area
to be carpeted excluding the area of the semicircular
hearth of the fireplace.
3 ft.
14 ft.
3 ft.
12 ft.
Slide 2- 23
continued
Understand The entire room will be carpeted
excluding the shaded semi-circle.
Plan We can exclude the area of the fireplace by
finding the area of the entire room and
subtracting the area of the fireplace.
A = Area of the rectangle – area of the semi-circle
A=
lw
–
 r2
2
Slide 2- 24
3 ft.
continued
Execute
A  lw 
 r2
2 3.14  4 2
A  14 12  
2
A  168  25.12
14 ft.
3 ft.
12 ft.
A  142.88 ft 2
Answer The area to be carpeted is 142.88 square feet.
Check Estimate the area of the semi-circle by dividing
the room into six equal parts. The fireplace
would occupy approximately one sixth of the
room. Since168  25.12  6.9 , the answer is
reasonable.
Slide 2- 25
Nongeometric Formulas:
The distance, d, an object travels given its rate, r,
and the time of travel, t: d = rt
The average rate of travel, r, given the total
d
distance, d, and total time, t: r 
t
The voltage, V, in a circuit with a current, i, in
amperes (A), and a resistance, R, in ohms,  : V = iR
The temperature in degrees Celsius given degrees
Fahrenheit: C  5  F  32 
9
The temperature in degrees Fahrenheit given
degrees Celsius: F  9 C  32
5
Slide 2- 26
Example 5
A truck driver begins a delivery at noon and travels
120 miles before taking a 30 minute break. He then
travels 136 miles, arriving at his destination at 5 pm.
What was his average driving rate?
Understand We are given travel distances and times,
and we are to find the driving rate.
Plan To find the average rate, we first need the total
distance traveled and the total time spent driving.
d
We can use the formula r  .
t
Slide 2- 27
continued
Execute Total distance = 120 + 136 = 256 miles = d
Total time driving = 5 – 0.5 = 4.5 hours = t
Now we calculate the average rate.
256 miles
r
4.5 hours
r  56.8 mph
Answer His average driving rate was 56.8 mph.
Check We can use d = rt to verify.
d  56.8  4.5
d  255.6 miles
Slide 2- 28
Determine which of the following is a
solution for the equation.
6  x  4  2  4  x  3  6
a) –5
b) – 1
c) 2
d) 4
2.1
Determine which of the following is a
solution for the equation.
6  x  4  2  4  x  3  6
a) –5
b) – 1
c) 2
d) 4
2.1
A thermometer reads 30° C: what is this
temperature in degrees Fahrenheit?
9
F  C  32
5
a) 34F
b) 49F
c) 86F
d) 112F
2.1
A thermometer reads 30° C: what is this
temperature in degrees Fahrenheit?
9
F  C  32
5
a) 34F
b) 49F
c) 86F
d) 112F
2.1
2.2
The Addition Principle of Equality
1. Determine whether a given equation is linear.
2. Solve linear equations in one variable using the
addition principle.
3. Solve equations with variables on both sides of
the equal sign.
4. Solve identities and contradictions.
5. Solve application problems.
Linear equation: An equation in which each
variable term contains a single variable raised
to an exponent of 1.
Linear equation in one variable: An equation
that can be written in the form ax + b = c,
where a, b, and c are real numbers and a  0.
Slide 2- 34
Example 1
Determine whether the equation is linear or nonlinear.
a. 6y + 8 = 12
Answer This equation is linear because the variable y
has an exponent of 1.
b. 8x3 + 2x = 9
Answer This equation is nonlinear because the variable
in the term 8x3 has an exponent of 3.
Slide 2- 35
The Addition Principle of Equality
If a = b, then a + c = b + c is true for all real
numbers a, b, and c.
Using the Addition Principle of Equality
To use the addition principle of equality to clear a
term in an equation, add the additive inverse of that
term to both sides of the equation. (That is, add or
subtract appropriately so that the term you want to
clear becomes 0).
Slide 2- 36
Example 2
Solve and check. x – 19 = –34
Solution
To isolate x, we need to clear –19.
x – 19 = –34
Add 19 to the lefthand side so that
+ 19 + 19
–19 + 19 = 0.
x + 0 = –15
x = 15
Since we added 19 to
the left side, we must
add 19 to the right side
as well.
Slide 2- 37
continued
Check Replace the x in the original equation with –15
and verify that the equation is true.
x  19  34
15  19  34
34  34
Replace x with –15 .
True, so –15 is the solution.
Slide 2- 38
Some equations have expressions that can be
simplified. If like terms are on the same side of the
equation, we combine the like terms before isolating
the variable.
If the equation to be solved contains parentheses,
we use the distributive property to clear the
parentheses before isolating the variable.
Slide 2- 39
Example 3
Solve and check. 2x – 1.2 + 6.7 – x = −3.4 + 5
Solution Simplify the expressions; then isolate the x.
2x – 1.2 + 6.7 – x = −3.4 + 5
x + 5.5 = 1.6
− 5.5 = − 5.5
x + 0 = −3.9
x = −3.9
Check Replace x in the original equation with −3.9 and
verify the equation is true. We will leave this to the viewer.
Slide 2- 40
Example 4
Solve and check. 6x – 7 = 5x – 19
Solution Use the addition principle to get the variable
terms together on the same side of the equal sign.
6x – 7 = 5x – 19
− 5x
− 5x
x – 7 = –19
+7
+7
x = –12
Slide 2- 41
continued
Check
6x – 7 = 5x – 19
6(−12) – 7 = 5(−12) – 19
−72 – 7 = −60 – 19
−79 = −79
True; so −12 is the solution.
Slide 2- 42
Example 5
Solve and check. n   4n  8  6  n  3  8n
Solution Simplify both sides of the equation. Then
isolate n.
Distribute minus sign.
n   4n  8  6  n  3  8n
n  4n  8  6n  18  8n
3n  8  2n  18
3n
3n
8  n  18
18
18
10  n
Distribute the 6.
Combine like terms.
Add 3n to both sides.
Add 18 to both sides to
isolate the n.
Slide 2- 43
continued
Check
n   4n  8  6  n  3  8n
10   4 10   8  6 10  3  8 10 
10   40  8  6 10  3  80
10  48  6  7   80
Replace n in the
original equation with
10 and verify that the
equation is true.
10  48  42  80
38  38
True, so the solution is 10.
Slide 2- 44
Solving Linear Equations
To solve linear equations requiring the addition
principle only:
1. Simplify both sides of the equation as needed.
a. Distribute to clear parentheses.
b. Combine like terms.
2. Use the addition principle so that all variable terms
are on one side of the equation and all constants are
on the other side. Then combine like terms.
Tip: Clear the variable term that has the lesser
coefficient to avoid negative coefficients.
Slide 2- 45
In general, a linear equation in one variable has only
one real-number solution. However, there are two
special cases: an equation in which every real number
is a solution and one that has no solution.
Identity: An equation that has every real number as a
solution (excluding any numbers that cause an
expression in the equation to be undefined).
Contradiction: An equation that has no real number
solution
Slide 2- 46
Recognizing an Identity
When solving a linear equation, if after simplifying
each side of the equation the expressions are
identical, the equation is an identity and every real
number for which the equation is defined is a
solution.
Slide 2- 47
Example 6
Solve and check. 2 + 14x – 9 = 7(2x + 1) – 14
Solution Simplify both sides of the equation. Then
isolate the variable.
2 + 14x – 9 = 7(2x + 1) – 14
2 + 14x – 9 = 14x + 7 – 14
14x – 7 = 14x – 7
Check Every real number is a solution for an identity,
so any number we chose will check in the original
equation.
Slide 2- 48
Example 7
Solve and check. 11x + 6 = 11x + 4
Solution
11x + 6 = 11x + 4
−11x
−11x
0+6=0+4
6=4
Because the equation is a contradiction, it has no solution.
Check Because the variable terms, 11x, are identical on
both sides of the equal sign, replacing x with any
number will yield an identical product. The equation
cannot be true.
Slide 2- 49
Example 8
Raul wants to buy a membership to a gym which
costs $395 for the year. He currently has saved
$149. How much more does he need?
Understand We are given the total required for
the membership and the amount he
currently has. We must find how
much more he needs.
Plan Let x represent the amount Raul needs.
Write the equation, then solve.
Slide 2- 50
continued
Execute Current amount + amount needed = 395
149
+
x
= 395
Subtract 149 from both sides to isolate x.
149 + x = 395
–149
–149
0 + x = 246
x = 246
Answer Raul needs $246 to buy the membership.
Does $149 plus the additional $246 equal
$395?
149 + 246 = 395
395 = 395
Slide 2- 51
Solve for t. 8t + 5 –14t = 5t + 3 – 9t
a) t = 5
b) t = 3
c) t = 1
d) t = –1
2.2
Solve for t. 8t + 5 –14t = 5t + 3 – 9t
a) t = 5
b) t = 3
c) t = 1
d) t = –1
2.2
An 8 ft board is cut into 3 pieces. If one is
3 ½ ft and the second is 2 ¼ ft, what is the
length of the third piece?
a) 1 ¾ ft
b) 2 ¼ ft
c) 2 ½ ft
d) 3 ¼ ft
2.2
An 8 ft board is cut into 3 pieces. If one is
3 ½ ft and the second is 2 ¼ ft, what is the
length of the third piece?
a) 1 ¾ ft
b) 2 ¼ ft
c) 2 ½ ft
d) 3 ¼ ft
2.2
2.3
The Multiplication Principle of Equality
1. Solve linear equations using the
multiplication principle.
2. Solve linear equations using both the
addition and the multiplication principles.
3. Use the multiplication principle to clear
fractions and decimals from equations.
4. Solve application problems.
The Multiplication Principle of Equality
If a = b, then ac = bc is true for all real numbers a,
b, and c, where c  0.
Using the Multiplication Principle of Equality
To use the multiplication principle of equality to
clear a coefficient in an equation, multiply both
sides of the equation by the multiplicative inverse
of that coefficient or divide both sides by the
coefficient.
Slide 2- 57
Example 1
5
3
 x
4
8
Solve and check.
Solution
5
3
 x
4
8
1
1
1
4

5
5
3
 x
4
8
1
1
2
4

5
5
Clear the coefficient  by
4
multiplying both sides by its
multiplicative inverse,  4 .
5
3
x
10
Slide 2- 58
continued
Check
5
3
 x
4
8
1
5 3  3
   
4  10  8
2
3 3

8 8
True, therefore
3

10
Replace x in the original
3
equation with  10 and
verify that the equation
is true.
is correct.
Slide 2- 59
Solving Linear Equations
To solve linear equations in one variable,
1. Simplify both sides of the equation as needed.
a. Distribute to clear parentheses.
b. Combine like terms.
2. Use the addition principle so that all variable terms
are on one side of the equation and all constants are
on the other side. (Clear the variable term with the
lesser coefficient to avoid negative coefficients.)
Then combine like terms.
3. Use the multiplication principle to clear any
remaining coefficient.
Slide 2- 60
Example 2
Solve and check. 8x  9  7
Solution
8x  9  7
9 9
8x  0  16
8x  16
8 = 8
x  2
The answer checks so x = −2.
Slide 2- 61
Example 3
Solve and check. 7 y 11  2 y  46
Solution
7 y 11  2 y  46
2 y
2 y
5 y 11  0  46
5 y 11  46
11 11
5 y  35
5 = 5
y  7
The answer checks so y = −7.
Slide 2- 62
Example 4
Solve and check. 2  5  y  5  3 y  2 1
Solution
2  5  y  5  3  y  2   1
2  5 y  25  3 y  6 1
5 y  23  3 y  7
5 y
5 y
0  23  8 y  7
Distribute to clear parentheses.
Combine like terms.
Add 5y to both sides (–5y has
the lesser coefficient).
Slide 2- 63
continued
0  23  8 y  7
7
7
16  8 y  0
16 8 y

8
8
2  y
Check
Add 7 to both sides.
Divide both sides by 8 to clear
the 8 coefficient.
2  5  y  5  3  y  2   1
2  5  2  5  3 2  2 1
2  5 3  3 4 1
2  15  12  1
13  13 True
Replace y in the original
equation with –2 and verify that
the equation is true.
Therefore 2 is correct.
Slide 2- 64
If the equation contains fractions, we multiply
both sides by a number that will clear all the
denominators. We could multiply both sides by
any multiple of the denominators; however,
using the LCD (least common denominator)
results in the simplest equations.
The multiplication principle can also be used to
clear decimals in an equation. We can clear
decimals by multiplying both sides of the
equation by an appropriate power of 10. The
power of 10 we use depends on the decimal
number with the most decimal places.
Slide 2- 65
Example 5
7
1 3
1
x  x  x
8
4 4
16
Solve and check.
Solution
1 3  1
7

16  x   x     x 16
4 4   16
8

2
4
4
1
16 7
16 1 16 3
16 1
 x     x    16 x
1 8
1 4 1 4
1 16
1
1
1
1
14 x  4  12 x  1  16 x
26 x  4  1  16 x
10 x  5
1
x
2
Slide 2- 66
Example 6
Solve and check. 16.3  7.2n  26.9
Solution
16.3  7.2n  26.9
10 16.3  7.2n    26.910
163  72n  269
163
163
0  72n  432
72n 432

72
72
n6
Check
16.3  7.2n  26.9
16.3  7.2  6  26.9
16.3  43.2  26.9
26.9  26.9
Slide 2- 67
Solving Linear Equations
To solve linear equations in one variable,
1. Simplify both sides of the equation as needed.
a. Distribute to clear parentheses.
b. Clear fractions or decimals by multiplying through
by the LCD. In the case of decimals, the LCD is the
power of 10 with the same number of zero digits as
decimal places in the number with the most decimal
places. (Clearing fractions and decimals is optional.)
c. Combine like terms
Slide 2- 68
continued Solving Linear Equations
2. Use the addition principle so that all variable terms
are on one side of the equation and all constants are
on the other side. (Clear the variable term with the
lesser coefficient.) Then combine like terms.
3. Use the multiplication principle to clear any
remaining coefficient.
Slide 2- 69
Example 7
The perimeter of the figure shown is 72 inches. Find
the width and the length.
x
x +11
Understand The width is represented by x and
the length is represented by x + 11.
We are given the perimeter, so we
can use the formula P = 2l + 2w.
Slide 2- 70
continued
Plan In the perimeter formula, replace P with 72,
l with x + 11, w with x, and solve for x.
Execute
P  2l  2 w
72  2  x  11  2 x
72  2 x  22  2 x
72  4 x  22
22
22
50  4 x  0
50 4 x

4
4
12.5  x
Distribute.
Combine like terms.
Subtract 22 from both sides.
Divide both sides by 4.
Slide 2- 71
continued
Answer The width is 12.5 inches. To find the length,
we evaluate the expression that represents the
length, x + 11, with x = 12.5 inches
Length = 12.5 + 11 = 23.5 inches
Check
P  2l  2w
P  2  23.5  2 12.5 
P  47  25
P  72
It checks.
Slide 2- 72
Solve. 8m + 6 = 3(12 + 2m)
a) 3
b) 5
c) 8
d) 15
2.3
Solve. 8m + 6 = 3(12 + 2m)
a) 3
b) 5
c) 8
d) 15
2.3
Solve. 0.8 – 4(a – 1) = 0.2 + 3(4 – a)
a)  10
b) 7.4
c) 8.6
d) 20.6
2.3
Solve. 0.8 – 4(a – 1) = 0.2 + 3(4 – a)
a)  10
b) 7.4
c) 8.6
d) 20.6
2.3
2.4
Applying the Principles to Formulas
1. Isolate a variable in a formula using the
addition and multiplication principles.
Isolating a Variable in a Formula
To isolate a particular variable in a formula, treat
all other variables like constants and isolate the
desired variable using the outline for solving
equations.
Slide 2- 78
Example 1
Isolate W in the formula Z = W – Y.
Solution
Z=W–Y
+Y
+Y
Z+Y=W+0
Z+Y=W
To isolate W we must clear Y.
Because Y is subtracted from W,
we add Y to both sides.
Slide 2- 79
Example 2
Isolate l in the formula for the volume of a
box, V = lwh.
Solution
V = lwh
To isolate l we must clear w and h.
Because w and h are multiplying l,
we divide both sides by w and h.
V
lwh

wh wh
V
l
wh
Slide 2- 80
Example 3
Isolate m in the formula jm + c = n.
Solution
jm + c = n
 c c
jm = n – c
jm n  c

j
j
nc
m
j
Isolate jm by subtracting c from
both sides.
Isolate m by dividing both sides by j.
Slide 2- 81
Isolate a:
a)
b)
W
a
N
N
a
W
c) a  NW
d) a  N  W
2.4
W
N
a
Isolate a:
a)
b)
W
a
N
N
a
W
c) a  NW
d) a  N  W
2.4
W
N
a
Isolate a: a  b  c  180
180  c
a) a 
b
180
b) a 
bc
c) a  180  b  c
d) a  180  b  c
2.4
Isolate a: a  b  c  180
180  c
a) a 
b
180
b) a 
bc
c) a  180  b  c
d) a  180  b  c
2.4
2.5
Translating Word Sentences to Equations
1. Translate sentences to equations using key
words, then solve.
Key Words and Their Translations
Addition
Translation
Subtraction
Translation
The sum of x
and 3
x+3
The difference of x
and 3
x–3
h plus k
h+k
h minus k
h–k
7 added to t
t+7
7 subtracted from t
t–7
3 more than a
number
n+3
3 less than a
number
n–3
y increased by 2
y+2
y decreased by 2
y–2
Note: Since addition is a
commutative operation, it does not
matter in what order we write the
translation.
Note: Subtraction is not a
commutative operation; therefore,
the way we write the translation
matters.
Slide 2- 87
Key Words and Their Translations
Multiplication
Translation
Division
Translation
The product of
x and 3
3x
The quotient of x
and 3
x  3, x/3
h times k
hk
h divided by k
h  k, h/k
Twice a number
2n
h divided into k
k  h, k/h
Triple the
number
3n
The ratio of a to b
a  b, a/b
Two-thirds of a
number
2
n
3
Note: Like addition, multiplication
is a commutative operation, it does
not matter in what order we write
the translation.
Note: Division is like subtraction in
that it is not a commutative
operation; therefore, the way we
write the translation matters.
Slide 2- 88
Key words for an equal sign:
is equal to
is
yields
is the same as
produces
results in
Translating Word Sentences
To translate a word sentence to an equation,
identify the unknown(s), constants, and key words;
then write the corresponding symbolic form.
Slide 2- 89
Example 1
The sum of thirty-five and a number is equal to
eighteen. Translate to an equation, then solve for the
number.
Understand The key word sum indicates addition,
is equal to indicates an equal sign, and a number
indicates a variable.
Plan Translate the key words to an equation, and then
solve the equation. We will use n as the variable.
Slide 2- 90
continued
Execute
Translate:
The sum of thirty-five and a number is equal to 18.
35
Solve:
Answer
+
35 + n = 18
35
35
0 + n = 17
n = 17
n
=
18
Check
35 + n = 18
35 + (–17) = 18
18 = 18
Slide 2- 91
Example 2
Two-thirds of a number is negative seven-eighths.
Translate to an equation and then solve.
Understand When of is preceded by a fraction, it
means multiply. The word is means an equal sign.
Plan Translate the key words and then solve.
Execute Translate:
Two-thirds of a number is negative seven-eighths.
2
3

n

7

8
Slide 2- 92
continued
Solve:
2
7
n  
3
8
3 2
7 3
 n   
2 3
8 2
21
n
Answer
16
Check 2  n   7
3
1
8
7
Clear the coefficient
2 21 2/3 by
7
  sides
 by
 its
multiplying both
8
reciprocal 33/2. 16
1
8
7
7
 
8
8
Slide 2- 93
Example 3
Eight less than five times a number is equal to thirtyseven. Translate to an equation, and then solve.
Understand
Less than indicates subtraction in reverse order,
times indicates multiplication,
is equal to indicates an equal sign.
Plan Translate to an equation using the key words
and then solve the equation.
Slide 2- 94
continued
Eight less than five times a number is equal to thirty-seven.
5n
Solve:
Answer
 8
5n – 8 = 37
+8 +8
5n + 0 = 45
5n = 45
5
5
n=9
=
37
Check:
5n – 8 = 37
5(9) – 8 = 37
45 – 8 = 37
37 = 37
Slide 2- 95
Example 4
Nine times the sum of a number and seven subtracted
from three times the number results in negative
twenty-seven.
Understand
Times means multiply,
subtracted from indicates subtraction,
sum means addition,
result in indicates an equal sign.
Plan Translate the key words to an equation and then
solve.
Slide 2- 96
continued
Nine times the sum of a number and seven subtracted
from three times the number results in negative
twenty-seven.
Execute:
3x  9(x + 7) = 27 Distribute to clear the
3x – 9x – 63 = 27 parentheses.
6x – 63 = 27 Simplify.
+ 63 +63 Add 63 to both sides.
6x + 0 = 36
Check:
6x = 36
Divide both sides by 6.
3x – 9(x + 7) = 27
6 6
3(6) – 9(6 + 7) = 27
x
=
6
Answer
18 – 9(1) = 27
27 = 27
Slide 2- 97
Translate: Thirty-two less than three times
a number is fifteen.
a) 32 – 3x = 15
b) 3x – 32 = 15
c) 32 + 3x = 15
d) 3x  32 = 15
2.5
Translate: Thirty-two less than three times
a number is fifteen.
a) 32 – 3x = 15
b) 3x – 32 = 15
c) 32 + 3x = 15
d) 3x  32 = 15
2.5
Translate: The quotient of four less than a
number and five is the same as the number
divided by eight.
a) 4  n  8n
5
b) 5(n  4)  8n
5
8
c)

n4 n
d) n  4  n
5
2.5
8
Translate: The quotient of four less than a
number and five is the same as the number
divided by eight.
a) 4  n  8n
5
b) 5(n  4)  8n
5
8
c)

n4 n
d) n  4  n
5
2.5
8
2.6
Solving Linear Inequalities
1. Represent solutions to inequalities
graphically and using set notation.
2. Solve linear inequalities.
3. Solve problems involving linear inequalities.
Not all problems translate to equations. Sometimes a
problem can have a range of values as solutions.
In mathematics we write inequalities to describe
situations where a range of solutions is possible.
<
is less than
>
is greater than

is less than or equal to

is greater than or equal to
Slide 2- 103
Linear inequality: An inequality containing
expressions in which each variable term contains
a single variable with an exponent of 1.
Examples of linear inequalities:
x>5
n+2<6
2(y – 3)  5y – 9
Set builder notation
{ x | x  5}
The set of all x such that x is greater than or equal to 5.
Slide 2- 104
We can graph solution sets for inequalities on a
number line.
Since the solution set for x  5 contains 5 and every
real number to the right of 5, we draw a dot (or solid
circle) at 5 and shade to the right of 5.
[
x < 2: Open circle on 2 and shade to the left of 2
)
Parentheses and brackets can also be used on
graphs instead of open and solid circles.
Slide 2- 105
Graphing Inequalities
To graph an inequality on a number line,
1. If the symbol is  or , draw a bracket (or solid
circle) on the number line at the indicated number
open to the left for  and to the right for . If the
symbol is < or >, draw a parenthesis (or open
circle) on the number line at the indicated number
open to the left for < and to the right for >.
2. If the variable is greater than the indicated
number, shade to the right of the indicated
number. If the variable is less than the indicated
number, shade to the left of the indicated number.
Slide 2- 106
Example 1
Write the solution set in set-builder notation and
interval notation, then graph the solution set.
a. x  2
b. n > 3
Solution
a. x  2
Set-builder notation: {x|x  2}
Interval notation: (, 2]
Graph:
]
Slide 2- 107
continued
b. n > 3
Set-builder notation: {n|n > 3}
Interval notation: (3, )
Graph:
(
Slide 2- 108
Inequalities containing two inequality symbols are
called compound inequalities. Compound inequalities
are useful in writing a range of values between two
numbers.
For example, 1 < x < 6
x can be any number greater than 1 and less than 6.
The solution set contains every real number between 1
and 6, but not 1 and 6.
Set-builder notation: {x|1 < x < 6}
Interval notation: (1, 6)
Graph:
(
)
Slide 2- 109
Example 2
Write the solution set for 7 < x  3 in set-builder
notation and interval notation, then graph the solution
set.
Solution
Set-builder notation: {x|7 < x  3}
Interval notation: (7, 3]
(
]
Graph:
Slide 2- 110
To solve inequalities, we will follow essentially the
same process as for solving equations.
The Addition Principle of Inequality
If a < b, then a + c < b + c is true for all real
numbers a, b, and c. The principle also holds
true when < is replaced with >, , or .
Slide 2- 111
Example 3
Solve x + 8  7 and write the solution set in set-builder
notation and interval notation; then graph the solution
set.
Solution
x+87
−8 −8
x  −1
Set-builder notation: {x| x  −1}
Interval notation: (∞, −1]
Graph:
]
Slide 2- 112
The multiplication principle, on the other hand, does
not work as neatly as it did for equations.
The Multiplication Principle of Inequality
If a and b are real numbers, where a < b, then
ac < bc is true if c is a positive real number.
If a and b are real numbers, where a < b, then
ac > bc is true if c is a negative real number.
The principle also holds true for >, , and .
Slide 2- 113
Example 4
Solve and write the solution set in set-builder notation
and interval notation. Then graph the solution set.
7 x  28
Solution
7 x  28
7 x 28

7 7
x  4
Because we divided both sides by
a negative number, we reversed
the direction of the inequality
symbol.
Set-builder notation: {x|x  4}
Interval notation: (, 4]
Graph:
]
Slide 2- 114
Solving Linear Inequalities
To solve linear inequalities,
1. Simplify both sides of the inequality as needed.
a. Distribute to clear parentheses.
b. Clear fractions or decimals by multiplying through
by the LCD just as we did for equations. (Clearing
fractions and decimals is optional.)
c. Combine like terms.
2. Use the addition principle so that all variable terms are on
one side of the inequality and all constants are on the other
side. Then combine like terms.
3. Use the multiplication principle to clear any remaining
coefficient. If you multiply (or divide) both sides by a
negative number, reverse the direction of the inequality
symbol.
Slide 2- 115
Example 5
Solve 8x + 13 > 3x – 12.
Solution
8x + 13 > 3x – 12
3x
3x
5x + 13 > 0  12
5x + 13 > 12
13 13
5x + 0 > 25
5x > 25
5
5
x > 5
Subtract 3x from
both sides.
Subtract 13 from
both sides.
Divide both sides
by 5 to isolate x.
({x|x
Graph:
Set
builder
notation:(5,
Interval
notation:
)> 5}
Slide 2- 116
Key Words and Their Translations
Less Than:
Greater Than:
A number is less
than seven.
n<7
A number is greater
than two.
n>2
A number must be
smaller than five.
n<5
A number must be
greater than three.
n>3
A number must be
more than negative
six.
n > 6
Less Than or
Equal to:
Greater Than or
Equal to:
A number is at
most nine.
n9
A number is at least
two.
n2
The maximum is
fourteen.
n  14
The minimum is
eighteen.
n  18
Slide 2- 117
Example 6
Seven-eighths of a number is at least twenty-one.
Translate to an inequality, then solve.
Understand Because the word of is preceded by a
fraction, it means multiplication. The key words at
least indicate a greater-than or equal-to symbol.
Plan Translate the key words, then solve. Use n for
the variable.
Execute Translate.
Seven-eighths of a number is at least twenty-one.
7
 n  21
8
Slide 2- 118
continued
Solve:
1
7
 n  21
8
1
3
8 7
8 21
 n   
7 8
7 1
1
Answer
Check
1
Multiply both sides by 8/7 to isolate n.
1
n  24
Verify that 24 and any number greater
than 24 satisfy the original sentence.
Slide 2- 119
continued
Test 24:
Test a number greater than 24:
7
of 24 at least 21?
8
3
7 24
  21
8 1
1
21  21
Is
True
Is
7
of 32 at least 21?
8
4
7 32
  21
8 1
1
28  21
True
Slide 2- 120
Example 7
A park board is planning to enclose a new playground
area with fence. Cost restricts the board to a total of
320 feet of fencing materials. The board wants the
playground to be 75 feet wide. What is the maximum
length of the playground?
Understand The fence surrounds the playground, so
that 320 feet is the maximum perimeter of the
playground. The width is to be 75 feet, we need to find
the length.
Slide 2- 121
continued
Plan Formula for perimeter of a rectangle: P = 2l + 2w
320 is the maximum perimeter so we write an
inequality so the expression used to calculate
perimeter is less than or equal to 320. Because the
width is to be 75 feet we replace w with 75.
Execute Translate:
The perimeter must be less than or equal to 320.
2l + 2(75)  320
Slide 2- 122
continued
2l + 2(75)  320
2l + 150  320 Subtract 150 from both sides.
 150 150
2l  170
Divide both sides by 2.
2
2
l  85
Answer The length must be less than or equal to 85
feet, which means the maximum length is 85 feet.
Solve:
Slide 2- 123
Solve 9x + 12 > 3x – 18.
a) x > 5
b) x > 5
c) x < 5
d) x < 1
2.6
Solve 9x + 12 > 3x – 18.
a) x > 5
b) x > 5
c) x < 5
d) x < 1
2.6
Solve. Eight times a number less thirty-two
is at least seventy-two.
a) x  5
b) x  5
c) x  13
d) x  13
2.6
Solve. Eight times a number less thirty-two
is at least seventy-two.
a) x  5
b) x  5
c) x  13
d) x  13
2.6