Download Chapter 2

Ch. 2 Linear Equations and Inequalities in One Variable
Index
Section
Pages
2.1 The Addition Property of Equality
2.2 The Multiplication Property of Equality
2.3 More on Solving Linear Equations
2.4 An Intro to Applications of Linear Equations
2.5 Formulas and Applications from Geometry
2.6 Ratios and Proportions
2.7 More about Problem Solving
2.8 Solving Linear Inequalities
2 - 10
11 - 12
13 - 20
21 - 28
29 - 35
36 - 46
47 - 59
60 - 70
1
§2.1 The Addition Property of Equality
Outline
Definitions
Linear Equation in One Variable
Equivalent Equations
Addition Property of Equality
Check
 (Therefore)
Simplification Step
Solving Algebraic Equations
Solution Set – i.e. Checking (is it a true statement)
Addition Property of Equality – Add the opposite to both sides
Simple – Just add the opposite
2 Step – Simplify (including distributive property), then add opposite
Complex – Add the opposite of variable term and constant terms
Difference between Expression & Equation
Solve Equation – One Step is Simplify, but continue to solve
Simplify Expression – Don’t Place Equal Sign, DO NOT SOLVE
Solution Set
Isolating the Variable
Homework p. 98-99 #2, 3, 4, #6-36mult.of3, #37, 38, #42-57mult.of3, #61-64all
Linear Equations in One Variable are equations that can be simplified to an
equation with one term involving a variable raised to the first power which is added
to a number and equivalent to a constant. Such an equation can be written as
follows:
ax + b = c
a, b & c are constants
a0
x is a variable
Our goal in this and the next section will be to rewrite a linear equation into an
equivalent equation (an equation with the same solution set) in order to derive a
solution set (the answer; note that I will simply call it a solution at this point). We will
do this by forming the equivalent equation:
x=#
or
#=x
x is a variable
# is any constant
Remember that only 1 value will make a linear equation true and that is the value that we
desire. This also means that we can check our value and make sure that the original
equation, when evaluated at the constant, makes a true statement.
Recall the Fundamental Theorem of Fractions:
a  c = a (multiply/divide the numerator & denominator by the same number
b  c
b and the resulting fraction is equivalent)
2
The method for which we will be creating equivalent equations is very similar to the
Fundamental Theorem of Fractions.
The Addition Property of Equality says that given an equation, if you add (subtract) the
same thing to both sides of the equal sign, then the new equation is equivalent to the
original.
A = B is the same as A + C = B + C (since c is added to both A & B)
They are equivalent equations
Example:
Form an equivalent equation by adding the opposite of
the constant term on the left.
9 + x = 12
Note: We’ll use the additive property of equality to move all variables to one side of the
equation (for simplicity, in the beginning, we will move variables to the left) and all
numbers to the other side (in the beginning to the right). This is known as isolating the
variable.
Isolating the variable is done by adding the opposite of the constant term to both sides
of the equation (the expression on each side of the equal sign must first be simplified of course.) . We
add the opposite of the constant term in order to yield zero! This makes the constant
disappear from the side of the equal sign with the variable! Don’t get to carried away
and forget that in order to make it disappear, you must add the opposite to both sides of
the equal sign!
Example:
Solve by using the addition property of equality and the
additive inverse. Note: In order to solve these, you must look
at the equation and ask yourself, “How will I make x stand
alone?” This is where the additive inverse comes in – You must
add the opposite of the constant term in order to get the variable to
“stand alone.”
a)
x + 9 = 11
3
b)
x  3 = 10
c)
x  7/8 = 3/8
d)
x + 3 = 0
e)
-7 = y + 3
Yes, you can probably look at each of the above equations and tell me what the variable
should be equivalent to, but the point here is to develop a method that will work when the
equation is more complicated!!
4
How do we know that we got the right solution for the above answers? We check them
using evaluation! When you are asked to see if a number is a solution to an equation, this
is the process you take.
Example:
a)
Check the solutions to a), c) & d) from the previous
example
{2} so x = 2 so we replace x with 2 and get the
following
?
(2) + 9 = 11
?
11 = 11 true statement, therefore() this checks
c)
d)
5
Of course, every problem will not be as simplistic as the ones used as examples so far,
and sometimes we will have to simplify a problem before we can solve it. This means
that will have to combine all like terms on both the left and the right sides of the equation
before we can apply the addition property of equality. Sometimes this will also require
the use of the distributive property before we can combine like terms. Do not neglect
the simplification step!!
Example:
a)
Solve the following problems.
6a  2  5a = -9 + 1
a)
5 + 6 = 9a + 8  8a
b)
3(a + 2)  2a = 1
6
/3 a  2(1/5 + 1/3 a) = 4/5
c)
5
d)
4.6a + 1.1  3.6a = 3.2  5.7
Not only do we have to move constants from one side of the equal sign to the other, some
times we will also have to move the variable terms from one side of the equal sign to the
other. We’ll start by practicing with the variable term only then we’ll do both variable
and constant term! Remember that the idea is always to get the variables all on one
side (usually the left at this point) and the constants on the other side (usually the right at this
point)!
Example:
a)
Solve.
5y = 4y  2
7
b)
5y + 6 = 4y  2
c)
7z + ½ = 8z  7/2
Note: The variable term has to move to the right in this example and the constants
therefore to the left because we need to keep the variable’s numeric coefficient a positive
one!
d)
1.1  2v = 3.1  3v
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You will also be asked to translate problems and then solve them in this section. Here are
a few examples for a brief review. Translation is SO important!
Example:
a)
b)
Translate and solve.
The difference of x and 7/5 is -13.
The sum of 1/3 y and 6 is equal to the difference of twice
2
/3 y and 5.
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Ok, after all that, let’s make a point about the difference between solving an equation and
simplifying an expression! Don’t forget that if the problem has an equal sign it is an
equation and you will solve it! However, if the problem has no equal sign it is an
expression and you will only simplify it!
Example:
Simplify
2x + 2(x  2)  5
Example:
Solve
2x  (x  2)  5 = 7
Note: Your authors want the solution set for each equation, meaning when you get
variable = #, you give the answer as {#}. This is a personal preference of the authors
that goes along with more advanced mathematics. It is important that you realize that
they are equivalent and you may give your answers to me in either form. You will notice
that I will most often use variable = #, when presenting the answers in class.
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§2.2 The Multiplication Property of Equality
Outline
Definitions
Multiplication Property of Equality
Isolate the Variable
Multiplication Property of Equality
Multiply both sides by reciprocal of numeric coefficient
Multiplication by a reciprocal is division – So it's the same to multiply by reciprocal as dividing
Choose the easiest multiply by reciprocal or divide
Divide by whole numbers or decimals
Multiply by reciprocal for fractions or –1, whose reciprocal is –1!
Homework p. 104-106 #1-6all, #8-22even, #24-66mult.of3, #69-72all
The Mulitiplication Property of Equality says that two equations are equivalent if the
second is the same as the first, where both sides have been multiplied by a nonzero
constant.
a = b
is equivalent to ac = bc
a, b & c are real number
c  0
We will be using this to solve such problems as:
4x = 24
 By inspection we see that if x = 6, both sides are equivalent!
 However, if we want to isolate x what must 4x be multiplied by? Think about it’s
reciprocal!
 Using the idea of isolating the variable and the multiplication property of equality,
we can arrive at a solution for x!
Example:
Solve (isolate the variable) using the Multiplication
Property of Equality [Multiply both sides by the reciprocal of
the numeric coefficient – for fractions and –1 (the reciprocal of –1 is –1
) this is the best way to think of it and for whole numbers and
decimals you may choose to think of the equivalent operation –
division by the numeric coefficient!]
a)
16x = 48
11
b)
x
c)
14x = 0
d)
0.5z = 25
e)
-p = 5
f)
-2/5 m = -2/3
/3 = 36
(Recall that x/3 is equivalent to 1/3 x)
Really, this section is a breeze. There is one fact and that is pretty basic. It is the next
section where we put it all together that we have to really have our thinking caps on!
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§2.3 More on Solving Linear Equations
Outline
Definitions
Clearing an Equations of Fractions
Identity
Contradiction
Null Set ()
All Real Numbers ()
Empty Set ()
Solving Linear Equations in 1 Variable
Putting it all together – Simplification, Addition Property and then Multiplication Property
Removing Fractions
Removing Decimals
Homework p. 112-114 #2-46even, #47-52all
Now, let’s get a little more complicated with some material from section four. We have
all the tools that we need to solve any problem, we just have to apply them in the correct
order!
Solving Linear Equations in One Variable
Step 1: Simplify both sides of the equation
Step 2: Move all variables to one side of the equation using addition prop.
Step 3: Move all constants to the opposite of equation from variables
using addition prop.
Step 4: Remove numeric coefficient of variable by multiplying by its
reciprocal in order to solve the equation (multiplication prop.)
Step 5: Check your solution
Example: Solve 12x  5 = 23  2x
Step 1: Both sides of equation are simplified
Step 2: Move all variables to one side (left)
Step 3: Move all numbers to the other side (right)
Step 4: Isolate the variable (use multiplicative prop. of equality)
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Step 5: Check your solution from step 4
Note: Steps 2 & 3 are interchangeable. These steps both use the additive property of
equality.
Example: Solve
2x + 10 + 14x = 2x  18
Step 1: Simplify each side of the equation
Step 2: Move all variables to one side(left)
Step 3: Move all #’s to the other side(right)
Step 4: Isolate x
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Step 5: Check
Example: Solve
5(2x  1)  6x = 4
Step 1: Simplify each side of the equation (this includes
distributive property)
Step 2: Move all variables to one side (left)
Step 3: Move all #’s to the other side (right)
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Step 4: Isolate x
Step 5: Check
Ok, now that we have practiced all the basic steps for solving a linear equation in one
variable, let’s complicate the situation with some decimals and fractions. There are 2
ways to deal with fractions and decimals. The first way is to go about your business of
solving and not worry about them (in other words, just work with them) or you can get rid of
them as part of your simplification step. Here’s the low down on how to get rid of each,
with a few examples following each.
Simplifying: Getting Rid of Fractions (Clearing the Equation of Fractions)
Step 1: Find LCD of all fractions (on both left and right)
Step 2: Multiply all terms by LCD
Step 3: Simplify left & right sides of the equation
Example:
Solve
1
/2 x  6/12 = 1
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Example:
Solve
4
Example:
Solve
x
/3 (5  w) = - w
/2  1 = x/5 + 2
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Simplifying: Getting the Decimals Out
Step 1:
Examine all of the decimals in the equation, and determine which has the most
decimal places. Count those decimal places.
Step 2:
Multiply every term in the entire equation by a factor of 10 with the same
number of zeros as the number counted in step 1.
Step 3:
Simplify both sides of the equation
Example:
2.7  x = 5.4
Example:
Solve
4(y + 2.81) = 2.81
Note: In a problem involving the distributive property, it may be less confusing for you
to simplify the distributive property before doing the first step.
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Example:
Solve
0.60(z  300) + 0.05z = 0.70z  0.41(500)
Finally, we must discuss equations that have no solution or infinitely many solutions. An
equation that has no solution yields an untrue statement, in other words, when you go
through the process of solving such an equation you will get a false statement such as
0 = 5. This is a contradiction (an equation with no solution) and its solution set is
written as a null set (symbolized with ). An equation with infinitely many solutions
yields a true statement, in other words, when you go through the process of solving such
an equation you will get a true statement such as 5 = 5. This is called an identity (an
equation with infinite solutions and its solution set is all real numbers. I will allow you to
write  rather than {all real numbers}, but note that the book indicates the set notation as
the correct answer.
Example:
Solve
2x + 5 = 8x  2(3x + 5)
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Example:
Solve
2(x  5) = 8x  2(3x + 5)
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§2.4 An Introduction to Applications of Linear Equations
Outline
Definitions
Consecutive Integers
Consecutive Even Integers
Consecutive Odd Integers
Degree
Right Angle
Straight Angle
Supplementary 
Complementary 
Degrees in a Triangle()
Sum of  's in a 
Number Problems
More Complex Number Problems
Deeper Translation
Precursor to Mixture Problems (Really more number problems)
Still translation with multiplication
Consecutive Integer Problems
Definition
Consecutive Integers
x and x + 1
Even or Odd Integers Defined
x and x + 2
Angle Problems
Straight Angle
Relation to Supplementary Angles
Right Angle
Relation to Complementary Angles
Sum of Angles in a Triangle
1 + 2 + 3 = 180
Homework p.122-128 #3, #4-32even, #33-36all, #38-40all, #42-58even
Number Problems
Number problems are no more than the translation problems that we have already been
doing, except that the variable will not be predefined! You now need to be sure that you
define the variable!
Unknown # = x (or whatever you choose)
Equation  You will get this from translation
You will solve the equation and find the unknown number using your skills for solving
algebraic equations.
Example:
Four times some number is the same as that number
added to 15. Find the number.
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Example:
One number is twice another less 3. If their sum is 27
find each number.
More Complex Number Problems
These number problems involve not one but two unknowns that are defined in terms of
one another. This definition of one unknown in terms of the other will involve a
translation of a sum. Finally, the two unknowns will sum to a given total, completing the
similarity to the simple number problem.
First Unknown = x
Second Unknown = First Unknown + #
Total = First Unknown + Second Unknown
Example:
The Ringers play 5 more games than the Setters during a
regular season. If together they play a total of 51 games
how many does each team play?
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Example:
My husband and I took our 2 children to watch a movie.
It cost a total of $23, to view the movie. If the price one
child's ticket is $3 less than that of one adult ticket, how
much was the price of a child's ticket?
Precursor to Mixture Problems
The problems in this section are really a form of a more complex mixture problem that
we will encounter later, but they also have similarities with the number problems. For
instance like the more complex number problems they involve two unknown things that
are defined in terms of one another and they also involve a sum of these two unknowns.
However, what makes them different and is the most important thing to remember about
any mixture problem is that they involve a sum of products. The products will involve
an unknown and some portion OF that unknown. In this first very simple case the
unknown is one of the products and some portion of that is the other (since "of" means
multiplication, this is really not much more than a 2 step number problem, with the special translation step
to arrive at the second number in the sum).
First Unknown = x
Second Unknown = Some Portion of the First unknown
= (#, Fraction, Percent, Decimal, Etc.)  x
Total = First Unknown + Second Unknown
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Example:
For every cup of margarine in a chocolate chip cookie
recipe, there are twice that many cups of flour. If there
are a total of 9 cups of flour and margarine in the
restaurant style recipe, how many cups of each are there
in this recipe?
Example:
In my tamale pie recipe there is ½ as much cumin as
there is salt. If I make a recipe using the basic ingredient
ratio there is a total of ¼ teaspoon (tsp.) of both cumin
and salt. How much salt must I use?
24
Consecutive Integer Problems
These are of the same form as number problems, but with the twist that you must
correctly define the integers using a variable and an expression(s) for the consecutive
integers. Here is the trick.
Consecutive Integers – Define 1st as x and the next as x + 1, next as x + 2, etc.
Consecutive Even/Odd Integers – Define 1st as x and next as x + 2, next as x + 4,
and so on.
Example:
Two consecutive odd integers sum to 56. What
are the integers?
Complementary and Supplementary Angle Problems
A degree is the basic unit of measure in angles. It is defined as 1/360th of a rotation in a
whole circle. A degree can be denoted by a tiny circle above and to the right of a number
(). A right angle (denoted with a square inside the angle; see below) is a 90 angle and a
straight angle is a 180 angle. Complementary and supplementary angles are related to
right and straight angles respectively. Just one more note – the word angle can be written
short hand with the symbol .
Complementary angles sum to 90. (I remember complementary is 90 because it comes before
180 and "C" is comes before than "S" in the alphabet.)
Example:
The following angles are complementary.
x = 30
y = 60
x + y = 90;
30 + 60 = 90
25
Supplementary angles sum to 180. (Likewise, I remember supplementary is 180 because it
comes after 90 and "S" is comes after than "C" in the alphabet.)
Example:
The following angles are supplementary.
x + y = 180;
53 + 127 = 180
x = 53
y = 127
Sometimes we will be given the information that two angles are either supplementary or
complementary, but we will only be given the measure of one angle and we will have to
use algebra to find the missing angle.
Example:
Find the missing angle:
x = 73
y=?
26
Example:
If the sun is at a 45 angle from the eastern horizon, what
is its angle from the western horizon (assuming that there is
a straight line between the two).
Example:
If one angle is 15 more than twice is complement, find
the other.
27
While we are on the topic of angles let's briefly discuss a topic that comes up in some
exercises – the number of degrees in a triangle. We have discussed the three-sided
polygon before in terms of its perimeter, but we have not discussed any other properties
of this figure. Since there are three sides that make up a triangle, each pair meet at an
angle, and although each angle in any given triangle is unique there are some basic rules
that they must adhere to. The most basic is that the sum of all angles in a triangle
(degrees in a triangle) is always 180. This means that no angle can ever measure 180 or
more.
Example:
In a triangle, the second angle is 5 less than the twice the
smallest and the largest is 10 more than four times the
smallest. Find the largest angle's measure.
28
§2.5 Formulas and Applications from Geometry
Outline
Definitions
Trapezoid
Rectangular Solid
Sphere
Pyramid
Circumference
Area
Perimeter
Volume
Pi
Base of Triangle
Radius
Diameter
Big Base
Little Base
Height of Triangle
Vertical  's
Opposite  's
Adjacent  's
Basic Formulas
Rectangles
P = 2L + 2W
A = LW
V = LWH
Circle
C = 2r = d
A = r2
V = 4/3 r3
Triangle
P = S 1 + S2 + S3
A = ½ bh
V = 1/3 BH
Trapezoid
A = ½ (B + b)h
Temperature Conversion
F C
C = 5/9 (F  32)
CF
F = 9/5 C + 32
Solving Formulas for a Variable
Step 1: Simplify (by either multiplying out or multiplying both sides by a reciprocal)
Step 2: Addition Property of Equality
Step 3: Multiplication Property of Equality
Opposite(Vertical) & Adjacent Angles
Opposite  's are equal
Adjacent  's sum to 180
Homework p. 133-136 #3,4,5,8,9,12, #15-33mult.of3, #38, #45-50all, #51-72mult.of3, &
#73
This section is a specialized section dealing with Geometry formulas and formulas in
general, their algebraic manipulation and word problems containing them. A formula is
an algebraic equation that contains 2 or more variables. Geometry formulas in this
section will refer to area, perimeter, volume. The area of a geometric figure is the
amount of surface that it covers measured in square units. If you break this down into a
very basic visual exercise you would need to count the number of 1-unit squares that
could be fit into a figure. The perimeter of a figure is the distance around the outside
edge. In the case of a circle the perimeter is called the circumference. The volume of a
figure is the amount of substance it can contain measured in cubic units. Following are
some of the basic formulas needed for this section. Each contains special variables which
I will briefly define. For more formulas refer to the rear inside cover of your text.
29
Rectangle
L = length
W = width
H = height
Area
A = LW
Perimeter
P = 2L + 2W
Volume of a Rectangular Solid
Circle
V = LWH
r = radius (the distance from the center to any side)
d = diameter (the distance from one side to another through the
 = pi
middle; twice the radius)
(a Greek letter representing the constant that defines the ratio of
the circumference to the diameter;  3.14 or 22/7)
Circumference
C = 2r = d
Area
A = r2
Volume of a Sphere V = 4/3 r3
Triangle
b = base (the side to which the height meets at a right angle)
h = height (the perpendicular distance from a vertex to an opposite
side)
B = base of Pyramid (the area of the base rectangle)
H = height of Pyramid (the  height from the vertex to rectangle)
Perimeter
P = S1 + S2 + S3
Area
A = ½ bh
Volume of a Pyramid V = 1/3 BH
Trapezoid
Area
B = big base [the longest perpendicular () side]
b = little base (the smaller  side)
h = height (the  distance from B to b or vice versa)
A = ½(B + b)h
Temperature Conversion
C = Degrees Celsius (the scale based upon 0 freezing & 100
boiling)
F = Degrees Fahrenheit (the scale most commonly used in the US.
32 is freezing and 212 is boiling)
9
F = /5 C + 32
C = 5/9(F  32)
Now let's practice with a few of the less basic ones.
30
Example:
Find the area of the following trapezoid.
2 ft.
1.5 ft.
3 ft.
Example:
Find the temperature in Celsius, to the nearest degree, if
it is 113 F.
Example:
Find the temperature in Fahrenheit, to the nearest degree,
if it is 14 C.
31
Example:
Find the volume of a sphere with the diameter of 2 m.
The next item of importance in this section is solving a formula for one of its variables.
This process seems difficult, but it uses the same steps as solving any algebraic equation.
(Addition Property to Isolate Variable, Multiplication Property to Further Isolate Variable) I find that it
makes it more manageable to think of all other variables as numbers until I have
manipulated them away from the variable I want to solve for. Another trick to make it
more manageable is to use a colored pen or pencil to denote the variable that you wish to
solve for.
Example:
Solve the following for z.
G = az
Example:
Solve the following for x.
x + by = c
32
Example:
Solve the above (x + by = c ) for y.
Example:
Solve the following for g.
½(g + a) = 9
Note: There are several ways to solve this problem. In keeping with the method used to
solve any equation, you can simplify it by multiplying it out on the left and proceeding
from there. You may also choose to multiply both sides by the reciprocal of the ½ to free
the variable. Either method will yield equivalent solutions although they may look
different.
33
Finally, we must discuss some more facts about angles. We already know that
supplementary angles sum to 180 and complementary angles sum to 90. We must now
discuss when two lines intersect (cross) and form opposite (vertical) angles and adjacent
angles. Opposite (vertical) angles are equal. Adjacent angles are supplementary –
they sum to 180. This can be proven via methods from geometry, but you will have to
take my word. The following diagrams show opposite (vertical) and adjacent angles.
A
B
C
C
D
B & C are Opposite  's
Example:
A
A & C are Adjacent  's
What other angles are opposite? Adjacent?
34
Example:
In the following diagram find the measure of the missing
angles.
A
(7x  3)
(3x + 5)
Example:
What is the measure of A in the above problem?
35
§2.6 Ratios and Proportions
Outline
Definitions
Ratio
Proportion
Means
Extremes
Cross Multiply
Rate
Cross Product Unit Price
Similar 
Means-Extremes Property\
Ratios
Definition
Writing
a to b, a:b, a/b
Always lowest terms
Never Mixed Number
Always one # to another
Rates (Unit Prices)
Similarity to Ratios
One # to another
Never a mixed #
Differences
The denominator is always 1
Divide numerator by denominator to simplify and make denominator 1
Proportions
Mean & Extremes
Property
Cross Products
Solving
Using
Similar Triangles
Proportionality of
Using Proportions to find missing sides
Homework p. 141-144 #3, #3-12mult.of3, #14,20,24,25, #27-48mult.of3, #56, #57-63all
A ratio is a quotient of two numbers where the divisor isn't zero. A ratio is stated as:
a to b or
a:b
or
a
b
where a & b are whole numbers and b0
A ratio is always written as one whole number to another although it may not start out
being written in this form. A ratio is really a special fraction and we should always
simplify it by putting it in lowest terms (recall that this means that all common factors have been
divided out). The only difference between a ratio and a fraction is that a ratio is never
simplified to a whole number or a mixed number. A ratio must always be the quotient of
two whole numbers. Another important thing to remember is that ratios do not have
units, because both numerator and denominator have the same units so they cancel and a
ratio appears unitless.
36
Writing a Ratio Correctly
Step 1: Write one number divided by another number (order will be dependent upon your
problem, pay attention to the word to or the word quotient or its equivalent to help you out)
Step 2: Make sure that both numbers are in the same units. Convert if necessary.
Step 3: Divide out (cancel out) all common factors
Step 4: Rewrite the ratio making sure that it is written as one whole number to another
Example:
a)
b)
Example:
a)
b)
Write the following correctly, as ratios
8 to 32
50 to 150
What is the ratio of the following?
20 min. to 1 hour
$1.50 to 25¢
37
A special case of ratios is called rates. Rates are sometimes, as in the case of your book,
referred to as unit pricing. Rates (unit prices) are ratios in the sense that they are one
number to another, but a rate is written as any number to one. We find a rate using the
same principle as finding a correct ratio, except that we divide out the number that is in
the denominator. This can create a number that is not a whole number, which is fine in a
rate. A rate is not written as a whole number to another, it is simply written as an integer.
The ratio form of a rate is shown in its units.
Writing a Rate (Unit Price)
Step 1: Write as one number to another, also writing the units in this form
Step 2: Divide the numerator by the denominator (round only when told)
Step 3: Write the rate with its ratio like units
The following three examples are true examples of rates. The fourth and fifth examples
are unit pricing. Unit pricing can help us in shopping at the grocery store.
Example:
Write 28 miles to 6 gallons as a rate.
Example:
Find the average speed of a car that travels 292 miles in 4
hours.
Note: Average speed is a rate, found by the rate equation r = d / t, which can also be
written as d = r t, which is know as the distance equation. r = rate, d = distance & t =
time.
38
Example:
A Ferris wheel completes one revolution in 20 seconds.
If its circumference is 240 feet, what is the rate that a
person travels when riding on the ride?
Example:
A large box of Tidal Wave detergent (263 oz.) costs
$19.95 while the smaller box (96 oz.) costs $13.95.
Which box is the better buy? (Calculator OK)
39
A proportion is a mathematical statement that two ratios are equal.
2 = 4
3
6
It is read as:
is a proportion
2 is to 3 as 4 is to 6
The numbers on the diagonal from left to right, 2 and 6, are called the extremes
The numbers on the diagonal from right to left, 4 and 3, are called the means
If we have a true proportion, then the product of the means equals the product of the
extremes. This is called the means-extremes property. It actually comes from the fact
that when we multiply both sides of an equation by the same number (in this case the LCD)
we get an equivalent equation. It is a very useful property for finding a missing term (one
of the four numbers in a proportion 2 – 1st term, 3 – 2nd term, 4 – 3rd term, 6 – 4th term, in the example
above) in a proportion.
The products of the means and extremes are also called the cross products and finding
the product of the means and extremes is called cross multiplying.
Example:
Find the cross products of the following to show that this
is a true proportion
24 = 3
64
8
40
Solving a Proportion
Step 1: Find the cross products and set them equal.
Step 2: Simplify the expressions on both the left and the right side of the equal sign.
Step 3: Proceed as a linear equation by isolating the variable using addition property of
equality.
Step 4: Further isolate the variable by multiplying by the reciprocal of the numeric
coefficient.
Example:
a)
b)
Solve each of the following proportions
a = 21
4
28
z = 7
3
8
41
c)
d)
x + 2 = 2
3
5
m  8 = m
18
2
42
e)
m + 1 = 5  m
5
9
Truly, the most useful thing about ratios and proportions are their usefulness in word
problems. We can solve distance problems, rate of pay problems, gas mileage problems,
unit price problems, equal concentration mixture problems, etc. using this method.
The key is to set up equal ratios of one thing to another and write it out in words 1st.
Example:
a)
Using proportions, solve each of the following problems.
If Alexis gets 24 mpg in her car how many miles can she
drive on 7 gallons of gas?
43
b)
Joe gets a check for $225 for working 5 hours on
Tuesday, if he works 3 hours on Wednesday, how much
should he expect his check to be?
d)
A car travels 251 miles in 3 hours. If it continues at the
same speed, how long would you expect it to take in
order to travel 362 miles? (Calculator OK. Round up to
nearest whole hour.)
44
One more way to apply our knowledge of proportions is in finding the lengths of missing
sides in similar triangles. Similar triangles are triangles that have exactly the same angles
with corresponding sides that are proportional (not the same length, but all sides between
equivalent angles will form an equivalent fractions).
Given A = D, B = E, C = F in the following 2 triangles
E
B
A
Then AB : DE,
Meaning that
C
BC : EF,
AB = BC ,
DE
EF
AC : DF
AB = AC ,
DE
DF
F
D
BC = AC
EF
DF
(read AB is to DE as
BC is to EF, etc.)
The only thing left to recall, is that in a proportion (which is what the above “meaning that”
statements are called) the cross products are equal. Cross products are the diagonal numbers,
across the equal sign, multiplied.
Cross Products are equivalent in any true proportion and this allows us to set up an
equation to solve.
AB  EF = BC  DE
Example:
Find the missing side using your new knowledge.
B
E
A
F
C
AB = ?, BC = 9, AC = 13 1/3 While AE = 9, EF = ? and AF = 10
45
Example:
Find the missing side of these congruent triangles.
B
E
A
Example:
C
D
F
AB = 4.5 in.
DE = 3 in.
BC = 7.5 in.
EF = 5 in.
AC = ?
DF = 4 in.
A tree, standing beside a building, is known to be 27 feet
tall. If it casts a shadow that is 1 city block long and the
building casts a shadow that is 3 1/3 blocks long, how tall
is the building?
46
§2.7 More About Problem Solving
Outline
Definitions
Simple Interest
Percent
Equivalent Forms
Review of Percentages
Memorizing Fraction to Decimal Conversions (and vice versa)
Definition of percent
Solving Percentage problems
Mixture Problems
Table
Interest Problems
Table
Money Problems
Table
Distance Problems
Formula
Table
Homework p. 152-157 #1-6all, #16-34even, #46-54even
This section is about mixture problems. I mentioned before that we had started the baby
steps for mixture problems. They are problems that involve multiplication and addition.
There are three main types of mixture problems and we will go over a few examples of
each. The three main types are mixtures of solutions, simple interest (I = PRT), money
problems and rate (D = RT) problems. The mixture of solutions can look like mixtures of
any type of things, simple interest and rate problems always looks like what they are and
money problems can also look like value of an item times the number of the items!
Because I gave you percentage problems as homework but never covered them in class, I
would like to do a review at this time. It is useful to memorize some decimal to fraction
conversions. The following table is a list of very helpful conversions to memorize.
Recall that you can calculate these at any time you wish by simply dividing the
numerator by the denominator, but it saves a lot of time if you have them memorized.
Here are some tricks that I use to remember the fraction to decimal conversions.
4ths –
Money! 1 quarter is $0.25, 2 quarters is half a dollar, which is $0.50 and 3
quarters is $0.75
3rds –
1
/3 is 0.333, 1st in 3rds so repeats 3’s
/3 is 0.666, 2nd in 3rds so repeats 6’s (23)
5ths – 1/5, 10  5 is 2 so 0.2
2
/5, 20  5 is 4 so 0.4
3
/5, 30  5 is 6 so 0.6
4
/5, 40  5 is 8 so 0.8
2
47
6ths –
Think of the pattern is 3rds and remember that 6ths and 3rds are related
1
/6 is half of 1/3 so 0.1666
2
/6 is equal to 1/3 so 0.333
/6 is equal to ½ so 0.5
3
4
/6 is equal to 2/3 so 0.666
5
8
ths
/6 is 0.8333 (it’s following the repeating 3’s pattern and must be bigger than 2/3’s)
– The pattern is in last 2 digits, and is the most difficult
1
/8 is 0.125 (recall that 1/8 is ½ of ¼ )
3
/8 is 0.375 (I always have to divide 30 by 8 to get the 3 and then I know that 75 is the
5
/8 is 0.625
7
/8 is 0.875
ending)
(I always have to divide 50 by 8 to get the 6 and then I know that 25 is the
ending)
(I always have to divide 70 by 8 to get the 8 and then I know that 75 is the
ending)
9ths – It’s a repeating decimal of whatever the numerator is
At no time is it ever acceptable to write the * or any similar decimals with fractions
in them! The bar is the only acceptable method of showing repeat!!!!!!!
Fraction
½
1
/3
2
/3
¼
¾
1
/5
2
/5
3
/5
4
/5
1
/6
5
/6
1
/8
3
/8
5
/8
7
/8
1
/9
2
/9
4
/9
5
/9
7
/9
8
/9
Decimal
0.5
*0.333
*0.666
0.25
0.75
0.2
0.4
0.6
0.8
0.1666
0.8333
0.125
0.375
0.625
0.875
0.111
0.222
0.444
0.555
0.777
0.888
Note: You must know any repeating decimal’s equivalent fraction because there is no method for
conversion back to its equivalent fraction!
48
A percent is a part of a hundred.
There are 3 equivalent ways to write a percentage. They are as follows:
As a fraction – A part of one hundred (this must be reduced, of course, if possible)
As a decimal – Convert the fractional form to a decimal (see §5.6)
As a percentage – Multiply the decimal form by 100, since a percentage is just the
numerator portion of the fractional representation when the
denominator is 100!
We can write percentages as parts of 100, but since a percentage is a fraction in fractional
form it needs to be reduced.
Example:
75% = 75 = 3 = 0.75
100
4
All these are called equivalent forms, because they indicate the same thing.
The following is the only way that I will be teaching you to solve percentage problems. I
do not like the other method of creating a proportion. I find that the other method is
unreliable at best, not to mention confusing.
We will be seeing percentage problems in word problems. The following will be the
forms that we will be seeing and their corresponding algebraic equation. The portion in
italics is the portion that is unknown and thus represented by the x in the algebraic
equation.
Forms
Percent of a whole is what part
What percent of the whole is the part
Percent of what whole is the part
Algebra Equation
(%)(whole) = x
(x%)(whole) = part
(%)(x) = part
When approaching any percentage problem, you should write out the following form and
use the key words “of” and “%” to fill in the appropriate blanks in the form. The form
directly translates into an algebra problem since the word “of” means multiply and the
word “is” means equals. The only two things that you must remember are the
following:
1. Change the percentage to a decimal to solve the algebra problem
2. If the answer that you are getting is the percentage, then the algebra problem
will give you the equivalent decimal, and you must change it to a percentage
for your final answer.
Here is the form that you should always write down and fill in.
_____% of _____ is _____
Now let’s practice!
49
Example:
a)
Solve the following problems showing form, algebraic
equation and solution.
25% of 180 is what?
b)
16% of 100 is what?
50
c)
What percent of 10 is 3?
d)
18% of what is 125?
51
e)
8% of what is 62?
f)
What is 120% of 120?
Note: When the percentage is more than 100% the part will be larger than the whole! In
every case where the percentage is less than 100% the part will never exceed the whole
and you can use this as a quick check to see if you got an appropriate answer.
52
g)
What percentage of 25 is 95?
Note: The whole is less than the part so it is quite appropriate that the percentage is
greater than 100%
The key to doing word problems with percentages is to remember the form and to put
each problem in that form!
______% of ________ (whole) is ______ (part)
Example:
It took 12% of the Miller’s monthly income to pay for
daycare of their child. If the Miller’s income is $2500
per month, what is the cost of daycare per month?
53
Mixture Problems
Keys to solving are remembering that these are percentage problems and building a table
like the following!

Amount Solution
Unknown
Relation to unknown or given
Given or Sum of 1st two rows
Example:
Strength
=
Given %
Given % (Different from other)
Given % (Different from
others)
Amount of Pure
Calculate by multiplying across
Calculate by multiplying across
Calculate by multiplying across
Equal to sum of 1st two rows
Sum
Equals
How many cubic centimeters (cc) of a 25% anitbiotic
solution should be added to a 10 cc of a 60% antibiotic
solution in order to get a 30% antibiotic solution?
54
Example:
How much water should be added to 30 gallons of a
solution that is 70% antifreeze in order to get a mixture
that is 60% antifreeze?
Interest Problems
Again the table set up with multiplication across and addition down works well. It is
important to note that simple interest (I) is calculated by multiplying principle (P) by rate
(R) by time (T). The equation looks like P  R  T = I. In our work, time will most
often be 1 year, but we will put it into our table just in case at some point it is not!
P

x
x + Some amount
R

Given Rate
2nd Given Rate
T
=
Usually 1 year
Usually 1 year
I
Given  x
+
2nd Given (x + amt.)
= Given Total Interest Made
55
Example:
Joe invested some of his inheritance in a bond that yields
7% annually. He invested $900 more than he invested on
the bond in a high risk investment that yields 12%
annually. What was his total inheritance if he earned
$165 on both in one year?
Money Problems
The key to these type of problems is to realize that there must be a multiplication of the
value of the coin and the number of coins to find the total value of each type of coin.
These total values, of each type of coin, then usually sum to your total value or will have
some relationship to one another. You will never be given the value of each type of coin
of course because that is common knowledge – you must simply remember that you must
use this common knowledge.
Coin Type
Nickels
Quarters
Value of Coin
0.05 or 5
0.25 or 25
# of Coins
x
Relationship
between # of
Nickels & Quarters
is given
Total Value
5x
25(expression in #
of coins)
5x + 25(?) = Total
Value Given
56
Example:
I have 28 coins in my pocket. There are quarters and
dimes. If the total value is $4.60, how many of each type
do I have?
Example:
In my change jar I have $3.85. I have 5 more quarters
than nickels and 2 less dimes than nickels. How many of
each type of coin do I have?
57
Distance Problems
This type of problems can be tricky because there are so many scenarios. We will be
looking at distances and their sums. Sometimes the time will be unknown and sometimes
the rate will be unknown. Really, the most important thing is to read the problem
carefully and think logically about what is happening. I find a diagram of the scenario as
helpful as the table (which by the way looks just like the table for all other mixture problems) .
Important keys to remember for this type of problem are the formula
D=RT
Distance
=
Rate
#1  t
#2  t
#1t + #2t = Given D
Second Scenario
Distance
=
T = D/R

Time
#1
#2
2R
2(R + #)
2R + 2(R + #) = Given D
Example:
also
Unknown = t
Same Unknown = t
Rate

Unknown = R
Unknown Relation to 1st = R + #
Time
Known = 2
Same Known = 2
Two cars are traveling toward one another at a rate of 55
mph and 70 mph. If they begin 500 miles apart, how
long before they meet?
58
Example:
Two hikers are 14 miles apart and walking toward each
other. They meet in 2 hours. Find the rate of each hiker
if one hiker walks 1 mile per hour faster than the other.
59
§2.8 Solving Linear Inequalities
Outline
Definitions
Inequalities
Simple Inequality
Standard Form (of an inequality)
Endpoint
Compound Inequality
Interval Notation
Sense of the Inequality
Simple Inequalities
Definition
Review of Inequality Symbols
< less than &  less than or equal to
> greater than &  greater than or equal to
Graphing
< or > Open Circle (or Parentheses)
 or  Solid Dot (or Bracket)
Standard Notation – Line goes in direction of arrow
Interval Notation
Relation to parentheses and brackets in graphing
Addition Property & Solving Using
Multiplication Property & Solving Using
CAUTION: Reverse sense of inequality when mult. or dividing by a negative
Compound Linear Inequality (Three Part Linear Inequality)
Definition
Reading
Solving
Addition Property
Multiplication Property
CAUTION – Don't forget to reverse the sense of the inequality when multiplying by a negative
Writing solutions using interval notation
Applications
Translation Type
Mixture Type
Can Be Any Previously Studied Type
Consecutive Integers
Geometry Problems
Etc.
Homework p. 166-169 #3-15mult.of3, #25, #34-88even
Inequalities are algebraic expressions that contain inequality symbols. The following are
the inequality symbols:




60
Graphing of simple inequalities (contain only one inequality)
 -- Graph with an open circle and a line to the left, if in standard
form; Lial uses a parentheses – we won't!
 -- Graph with an open circle and a line to the right, if in standard
form
 -- Graph with a closed circle and line to the right, if in standard
form; Lial uses a bracket – we won't!
 -- Graph with a closed circle and a line to the left, if in standard
form
Just a note about the brackets and parentheses that we won't be using – their use comes
from the interval notation, a method for writing a solution set that tells us which
endpoints are included. An included endpoint is always enclosed by a bracket ("[" or "]")
and an excluded endpoint is enclosed by a parenthesis ( "(" or ")" ). If an interval
continues on to infinity, we use either  for infinity or – for negative infinity. Either
infinity symbol is enclosed by a parenthesis. An endpoint is exactly what it sounds like;
it is the end of the solution set.
Standard form (of an inequality) means that the variable is on the left and the constant
on the right. In order to achieve this simply place the variable on the left and constant on
the right, being careful to keep the point of the inequality facing toward the number or
variable as it was before the swap.
Examples: Put the following in standard form and write them using
interval notation.
a)
5  x
b)
–2  y
c)
5002  m
61
Examples: Graph each of the following (on a number line)
a)
x3
b)
2y
c)
-2  v
d)
t 5
Note: Parts b and c in the example are examples of inequalities written in nonstandard
form. To read an inequality written in nonstandard form, we must read right to left,
instead of our usual left to right. Thus it is easiest to always put inequalities in standard
form before reading and trying to graph them.
A compound linear inequality (three part linear inequality) contains 2 inequalities
connected by the word “and”. When joined by “and” it means the intersection of the sets.
This means that we will be graphing where the two sets overlap. If there was no overlap
then there would be no solution. The “and” is usually not explicitly written, and is
usually written in the shorthand form
#<x<#
or
# > x > #.
Example:
15  x  20 is read as:
x is greater than 15 and x is less than 20
62
In order to graph a compound inequality written in this form we must first graph the
endpoints and then draw a line in between the endpoints as this is the intersection of a
compound inequality joined by “and”. Make sure that you graph the endpoints correctly
as either an open circle (parenthesis) for a strict inequality or a closed circle (bracket) for
greater than/less than or equal to.
Graphing a Compound Linear Inequality (and)
Step 1: Graph each endpoint. Open circle (parenthesis) for < or >. Closed circle (bracket)
for  or .
Step 2: Join endpoints with a line, after making sure that there is an intersection.
Example:
a)
Graph the following compound inequalities and put them
in interval notation.
8 x9
b)
–2  x  5
c)
5  x  -2
63
Solving Linear Inequalities
If all we must do to solve a linear inequality is to add or subtract then they are solved in
the exact same manner as a linear equation. Recall the steps to solving a linear equation
covered in §2.1.
Example:
a)
Solve and graph the linear inequality. Put the solution in
interval notation.
x + 2  7
b)
2x + 7  9 + x
c)
2x  15 + x  15 + 2(x  1)
64
Solving a compound linear inequality is a little more complex because there are
actually two problems. The first problem is from the middle to the left and the second if
from the middle to the right. Remember that it is the intersection of these solution sets.
Now this could be quite complicated if you solved every one as two separate problems,
but thank heavens we have a shortcut. The shortcut is to treat the problem as having 3
parts instead of 2. Translation: What you do to the middle, do to both the left and
the right! Equivalence in solving equations: What you do to one side do to the other!
In terms of the addition property, add or subtract the same things from both the left and
the right that you add or subtract from the middle to isolate the variable. Note: The
variable must stay in the middle!
Example:
a)
Solve the following compound linear inequalities and
answer in interval notation. Also graph each answer.
5 < x  1  9
b)
-2  5 + x < 0
65
Multiplication Property of Equality & Linear Inequalities
If we must multiply or divide by a positive number to get the solution, then the steps for
finding the solution are still the same!
Example:
Solve and graph the linear inequality. Give the answer in
interval notation.
5x  15 + 2  15 + 3x
Problem Area!!
The problem occurs when I must multiply or divide by a negative number! Whenever I
multiply or divide by a negative number, the sense of the inequality must reverse. The
sense of the inequality means that the direction the inequality symbol faces changes.
Example:
a)
Solve and graph. Give the answer in interval notation.
3
/4x  1  2x + ¼
66
b)
-2x + 5  9
Why Must Sense of Inequality Reverse?
To see why it is true that the sense of the inequality must reverse, let’s rework the last
problem keeping the numeric coefficient of the variable positive. Notice that you must
read the inequality in nonstandard form (right to left) and the answer is the same as
above.
Example:
Solve by keeping the variable positive.
-2x + 5  9
67
Multiplication Property and Compound Inequalities
Just as with addition property of equality, we treat a compound inequality as having 3
parts and what we do to the middle we do both the left and right. Thus when we wish to
remove the numeric coefficient by multiplying by its reciprocal we must multiply both
the left and the right sides by the reciprocal as well! Since there is no way to get around
reversing the sense of the inequality when multiplying by a negative –be careful!
Example:
a)
Solve for the variable. Give the answer using interval
notation.
5 < 2x  5  9
b)
-0.5  0.1y + 0.25 < 0.23
68
Applications
The most important comment that I have for you when solving word problems that
involve linear inequalities is to point out 2 very important phrases and their mathematical
interpretation!
At Most means 
At Least means 
Example:
Three times some number less five is greater than or
equal to the result of five subtracted from the number.
Find all such numbers.
69
Example:
John's phone has a monthly charge of $34.25 and each
long distance call costs 5¢ per minute. If he has $45.50
in his budget per month for his phone bill, how many
minutes of long distance can his use at most?
Example:
The length of a rectangle is three less than twice its
width. If the perimeter is at most 53 cm, what are the
possible values of the width?
Of course there are many other problems that can come into play here, but they all rely
upon the skills developed in §2.4, & §2.7, so we won’t cover any more at this time.
70