Download 1.2 ADDING WHOLE NUMBER EXPRESSIONS

1.2 ADDING WHOLE NUMBER EXPRESSIONS
Student Learning
Objectives
After studying this section, you will
be able to:
Use symbols and key words
for expressing addition.
Use properties of addition
to rewrite algebraic
expressions.
Evaluate algebraic expressions
involving addition.
Using Symbols and Key Words for Expressing Addition
What is addition? We perform addition when we group items together. Consider the
following illustration involving the sale of bikes.
is equal to
Bikes sold Saturday
Bikes sold Sunday
4
Find the perimeters of
geometric figures.
3
7 bikes
We see that the number 7 is the total of 4 and 3. That is, 4 + 3 = 7 is an addition fact. The numbers being added are called addends. The result is called the sum.
+
4
Add whole numbers when
carrying is needed.
Total bikes sold
ⴝ
ⴙ
addend
=
3
addend
7
sum
In mathematics we use symbols such as “ +” in place of the words sum or plus. The
English phrase “five plus two” written using symbols is “5 + 2.” Writing English
phrases using math symbols is like translating between languages such as Spanish
and French.
There are several English phrases that describe the operation of addition. The
following table gives some of them and their translated equivalents written using
mathematical symbols.
English Phrase
Translation into Symbols
Six more than nine
9 + 6
The sum of some number and seven
x + 7
Four increased by two
4 + 2
Three added to a number
n + 3
One plus a number
1 + x
When we do not know the value of a number, we use a letter, such as x, to represent
that number. A letter that represents a number is called a variable. Notice that the
variables used in the table above are different. We can choose any letter as a variable. Thus we can represent “a number plus seven” by x + 7, a + 7, n + 7, y + 7,
and so on. Combinations of variables and numbers such as x + 7 and a + 7 are
called algebraic expressions or variable expressions.
EXAMPLE 1
Translate each English phrase using numbers and symbols.
(a) The sum of six and eight
(b) A number increased by four
Solution
(a) The sum of six and eight
6
8
(b) A number increased by four
x
4
Although we used the variable x to represent the unknown quantity in part (b),
any letter could have been used.
NOTE TO STUDENT: Fully worked-out
solutions to all of the Practice Problems
can be found at the back of the text
starting at page SP-1
10
Practice Problem 1
Translate each English phrase using numbers and
symbols.
(a) Five added to some number
(b) Four more than five
Section 1.2 Adding Whole Number Expressions
Using Properties of Addition to Rewrite Algebraic Expressions
Most of us memorized some basic addition facts. Yet if we study these sums, we
observe that there are only a few addition facts for each one-digit number that we
must memorize. For example, we can easily see that when 0 items are added to any
number of items, we end up with the same number of items: 5 + 0 = 5, 0 + 8 = 8,
and so on. This illustrates the identity property of zero: a + 0 = a and 0 + a = a.
EXAMPLE 2 Express 4 as the sum of two whole numbers. Write all possibilities. How many addition facts must we memorize? Why?
Solution Starting with 4 + 0, we write all the sums equal to 4 and observe any
patterns.
The numbers in this column
decrease by 1.
The numbers in this column
increase by 1.
4 0 4
3 1 4
2 2 4
The last two rows of the pattern
are combinations of the same
numbers listed in the first
three rows.
1 3 4
0 4 4
We need to learn only two addition facts for the number four: 3 + 1 and 2 + 2.
The remaining facts are either a repeat of these or use the fact that when 0 is added
to any number, the sum is that number.
Practice Problem 2
Express 8 as the sum of two whole numbers. Write all
possibilities. How many addition facts must we memorize? Why?
In Example 2 we saw that the order in which we add numbers doesn’t affect
the sum. That is, 3 + 1 = 4 and 1 + 3 = 4. This is true for all numbers and leads us
to a property called the commutative property of addition.
COMMUTATIVE PROPERTY OF ADDITION
a + b = b + a
4 + 9 = 9 + 4
13 = 13
Two numbers can be added in either order with the same result.
EXAMPLE 3
(a) 8 + 2
Use the commutative property of addition to rewrite each sum.
(b) 7 + n
(c) x + 3
(b) 7 + n = n + 7
(c) x + 3 = 3 + x
Solution
(a) 8 + 2 = 2 + 8
Notice that we applied the commutative property of addition to the expressions with variables n and x. That is because variables represent numbers, even
though they are unknown numbers.
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12
Chapter 1 Whole Numbers and Introduction to Algebra
Practice Problem 3
Use the commutative property of addition to rewrite
each sum.
(a) x + 3
EXAMPLE 4
(b) 9 + w
(c) 4 + 0
If 2566 + 159 = 2725, then
159 + 2566 = ?
Solution
159 + 2566 = 2725
NOTE TO STUDENT: Fully worked-out
solutions to all of the Practice Problems
can be found at the back of the text
starting at page SP-1
Practice Problem 4
Why? The commutative property states that the order
in which we add numbers doesn’t affect the sum.
If x + y = 6075, then
y + x = ?
To simplify an expression like 8 + 1 + x, we find the sum of 8 and 1.
8 + 1 + x = 9 + x or x + 9
Simplifying 8 + 1 + x is similar to rewriting the English phrase “8 plus 1 plus some
number” as the simpler phrase “9 plus some number.” Since addition is commutative, we can write this simplification as either 9 + x or x + 9. We choose to write
this sum as x + 9, since it is standard to write the variable in the expression first.
EXAMPLE 5
Simplify. 3 + 2 + n
Solution To simplify, we find the sum of the known numbers.
3 + 2 + n = 5 + n
or
= n + 5
We cannot add the variable n and the number 5 because n represents an unknown quantity; we have no way of knowing what quantity to add to the number 5.
Practice Problem 5
Simplify. 6 + 3 + x
Addition of more than two numbers may be performed in more than one
manner. To add 5 + 2 + 1 we can first add the 5 and 2, or we can add the 2 and 1
first. We indicate which sum we add first by using parentheses. We perform the operation inside the parentheses first.
5 + 2 + 1 = 15 + 22 + 1 = 7 + 1 = 8
5 + 2 + 1 = 5 + 12 + 12 = 5 + 3 = 8
In both cases the order of the numbers 5, 2, and 1 remains unchanged and the sums
are the same. This illustrates the associative property of addition.
ASSOCIATIVE PROPERTY OF ADDITION
1a + b2 + c = a + 1b + c2
14 + 92 + 1 = 4 + 19 + 12
13 + 1 = 4 + 10
14 = 14
When we add three or more numbers, the addition may be grouped in any way.
Section 1.2 Adding Whole Number Expressions
EXAMPLE 6 Use the associative property of addition to rewrite the sum
and then simplify. 1x + 32 + 6
Solution
1x + 32 + 6 = x + 13 + 62
= x + 9
The associative property allows us to regroup.
Simplify: 3 + 6 = 9.
Practice Problem 6 Use the associative property of addition to rewrite
the sum and then simplify. 1w + 12 + 4
Sometimes we must use both the associative and commutative properties of
addition to rewrite a sum and simplify. In other words, we can change the order in
which we add (commutative property) and regroup the addition (associative property)
to simplify an expression.
EXAMPLE 7
Use the associative and/or commutative property as necessary
to simplify the expression. 5 + 1n + 72
Solution
5 + 1n + 72 = 5 + 17 + n2
The commutative property allows us to change
the order of addition.
= 15 + 72 + n
= 12 + n
5 + 1n + 72 = n + 12
Regroup the sum using the associative property.
Simplify.
Write 12 + n as n + 12.
Practice Problem 7 Use the associative and/or commutative property as
necessary to simplify each expression.
(a) 12 + x2 + 8
(b) 14 + x + 32 + 1
Understanding the Concept
Addition Facts Made Simple
There are many methods that can be used to add one-digit numbers. For example, if you can’t remember that 7 + 8 = 15 but can remember that 7 + 7 = 14,
just add 1 to 14 to get 15.
7+8=7+(7+1)
=(7+7)+1
=14+1
=15
Another quick way to add is to use the sum 5 + 5 = 10, since it is easy to
remember. Let’s use this to add 7 + 5.
7+5=(2+5)+5
=2+(5+5)
=2+10
=12
Exercises
1. Use the fact that 5 + 5 = 10 to add 8 + 5.
2. Use the fact that 6 + 6 = 12 to add 6 + 8.
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14
Chapter 1 Whole Numbers and Introduction to Algebra
Evaluating Algebraic Expressions Involving Addition
We have already learned that when we do not know the value of a number, we designate the number by a letter. We call this letter a variable. We use a variable to represent an unknown number until such time as its value can be determined. For
example, if 6 is added to a number but we do not know the number, we could write
n + 6
where n is the unknown number.
If we were told that n has the value 9, we could replace n with 9 and then simplify.
n + 6
9 + 6
15
Replace n with 9.
Simplify by adding.
Thus n + 6 has the value 15 when n is replaced by 9. This is called evaluating the
expression n + 6 if n is equal to 9.
To evaluate an algebraic expression, we replace the variables in the expression
with their corresponding values and simplify.
An algebraic expression has different values depending on the values we use
to replace the variable.
EXAMPLE 8
Evaluate x + y + 3 for the given values of x and y.
(a) x is equal to 6 and y is equal to 1
(b) x is equal to 4 and y is equal to 2
Solution
(a) x + y + 3 Replace x with
6 + 1 + 3 6 and y with 1.
Simplify.
10
When x is equal to 6 and y is equal
to 1, x + y + 3 is equal to 10.
NOTE TO STUDENT: Fully worked-out
solutions to all of the Practice Problems
can be found at the back of the text
starting at page SP-1
Practice Problem 8
(b) x + y + 3 Replace x with
4 + 2 + 3 4 and y with 2.
Simplify.
9
When x is equal to 4 and y is equal
to 2, x + y + 3 is equal to 9.
Evaluate x + y + 6 for the given values of x and y.
(a) x is equal to 9 and y is equal to 3
(b) x is equal to 1 and y is equal to 7
Adding Whole Numbers When Carrying Is Needed
Of course, we are often required to add numbers that have more than a single digit.
In such cases we must:
1. Arrange the numbers vertically, lining up the digits according to place value.
2. Add first the digits in the ones column, then the digits in the tens column, then
those in the hundreds column, and so on, moving from right to left.
Sometimes the sum of a column is a multidigit number—that is, a number
larger than 9. When this happens we evaluate the place values of the digits to find
the sum.
15
Section 1.2 Adding Whole Number Expressions
EXAMPLE 9
Add. 68 + 25
Solution We arrange numbers vertically and add the digits in the ones column
first, then the digits in the tens column.
68
+ 25
6 tens
2 tens
8 tens
8 ones
5 ones
13 ones
We cannot have two digits in the ones column,
so we must rename 13 as 1 ten and 3 ones.
8 tens+1 ten+3 ones
9 tens+3 ones=93
A shorter way to do this problem involves a process called “carrying.” Instead
of rewriting 13 ones as 1 ten and 3 ones we would carry the 1 ten to the tens column
by placing a 1 above the 6 and writing the 3 in the ones column of the sum.
1
68
8 ones+5 ones=13 ones
+ 25
3
1
68
+ 25
93 Add 1 ten+6 tens+2 tens.
Practice Problem 9
Add. 247 + 38
Often you must carry several times, by bringing the left digit into the next
column to the left.
EXAMPLE 10 A market research company surveyed 1870 people to determine the type of beverage they order most often at a restaurant. The results of the
survey are shown in the table. Find the total number of people whose responses
were iced tea, soda, or coffee.
Solution We add whenever we must find the “total” amount.
21
357
577
+ 84
1018
Iced tea
Soda
Coffee
Type of
Beverage
We add 7+7+4=18. Since 18 equals 1 ten and 8 ones,
we carry 1 ten placing a 1 at the top of the tens column.
We add 1+5+7+8=21. Since 21 tens equals 2
hundreds and 1 ten, we carry 2 hundreds placing a 2 at the
top of the hundreds column.
We add 2+3+5=10. Since 10 hundreds equals 1
thousand and zero hundreds, we write 0 in the hundreds
column and 1 in the thousands column.
357 + 577 + 84 = 1018
A total of 1018 people responded ice tea, soda, or coffee.
Number of
Responses
Soda
577
Orange juice
475
Coffee
84
Iced tea
357
Milk
286
Other
91
16
Chapter 1 Whole Numbers and Introduction to Algebra
NOTE TO STUDENT: Fully worked-out
solutions to all of the Practice Problems
can be found at the back of the text
starting at page SP-1
Practice Problem 10 Use the survey results from Example 10 to answer the
following: Find the total number of people whose responses were milk, orange
juice, or other.
Finding the Perimeters of Geometric Figures
Geometry has a visual aspect that many students find helpful to their learning.
Numbers and abstract quantities may be hard to visualize, but we can take pen in
hand and draw a picture of a rectangle that represents a room with certain dimensions. We can easily visualize problems such as “what is the distance around the outside edges of the room (perimeter)?” In this section we study rectangles, squares,
triangles, and other complex shapes that are made up of these figures.
A rectangle is a four-sided figure like the ones shown here.
A rectangle has the following two properties:
1. Any two adjoining sides are perpendicular.
2. Opposite sides are equal.
When we say that any two adjoining sides are perpendicular we mean that any two
sides that join at a corner form an angle that measures 90 degrees (called a right
angle) and thus forms one of the following shapes.
When we say that opposite sides are equal we mean that the measure of a side is
equal to the measure of the side across from it. When all sides of a rectangle are the
same length, we call the rectangle a square.
7
11
4
4
11
Opposite
sides are
equal.
7
7
All sides of a
square are
equal.
7
A triangle is a three-sided figure with three angles.
The distance around an object (such as a rectangle or triangle) is called the
perimeter. To find the perimeter of an object, add the lengths of all its sides.
Section 1.2 Adding Whole Number Expressions
EXAMPLE 11
Find the perimeter of the triangle. (The abbreviation “ft”
means feet).
5 ft
5 ft
7 ft
Solution We add the lengths of the sides to find the perimeter.
5 ft
5 ft
5 ft + 5 ft + 7 ft = 17 ft
7 ft
The perimeter is 17 ft.
Practice Problem 11 Find the perimeter of the square.
15 ft
If you are unfamilar with the value, meaning, and abbreviations for the metric
and U.S. units of measure, refer to Appendix A, which contains a brief summary of
this information.
EXAMPLE 12
Find the perimeter of the shape consisting of a rectangle and a
square.
150 ft
65 ft
50 ft
65 ft
215 ft
Solution We want to find the distance around the figure. We look only at the
outside edges since dashed lines indicate inside lengths.
150 ft
65 ft
We cross off 65 ft
since inside lengths
are not included
? ft in the perimeter.
50 ft
65 ft
? ft
215 ft
Now we must find the lengths of the unlabeled sides. The shaded figure is a
square since the length and width have the same measure. Thus each side of the
shaded figure has a measure of 65 ft.
150 ft
65 ft
50 ft
115 ft
65 ft
This side is 65 ft
because the shaded
figure is a square.
65 ft
215 ft
This side equals 50 65
or 115 ft because opposite
sides of a rectangle have
the same length.
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18
Chapter 1 Whole Numbers and Introduction to Algebra
Next, we add the length of the six sides to find the perimeter.
150 ft + 115 ft + 215 ft + 65 ft + 65 ft + 50 ft = 660 ft
The perimeter is 660 ft.
NOTE TO STUDENT: Fully worked-out
solutions to all of the Practice Problems
can be found at the back of the text
starting at page SP-1
Practice Problem 12 Find the perimeter of the shape consisting of a rectangle and a square.
125 ft
40 ft
30 ft
30 ft
155 ft
Understanding the Concept
Using Inductive Reasoning to Reach a Conclusion
When we reach a conclusion based on specific observations, we are using
inductive reasoning. Much of our early learning is based on simple cases of inductive reasoning. If a child touches a hot stove or other appliance several times and
each time he gets burned, he is likely to conclude, “If I touch something that is
hot, I will get burned.” This is inductive reasoning. The child has thought about
several actions and their outcomes and has made a conclusion or generalization.
The following is an illustration of how we use inductive reasoning in mathematics.
Find the next number in the sequence 10, 13, 16, 19, 22, 25, 28, . . .
We observe a pattern that each number is 3 more than the preceding number:
10 + 3 = 13; 13 + 3 = 16, and so on. Therefore, if we add 3 to 28, we conclude
that the next number in the sequence is 31.
Exercise
1. For each of the following find the next number by identifying the pattern.
(a) 8, 14, 20, 26, 32, 38, . . .
(b) 17, 28, 39, 50, 61, . . .
For more practice, complete exercises 89–94 on page 22.
Preparing to Learn Algebra
Many people have learned arithmetic by memorizing facts and
properties without understanding why the facts are true or
what the properties mean. Learning strictly by memorization
can cause problems. For example:
• Many of the shortcuts in arithmetic do not work in
algebra.
• Memorizing does not help one develop reasoning and
logic skills, which are essential to understanding algebra
concepts.
• Memorization can eventually cause memory overload.
Trying to remember a collection of unrelated facts can
cause you to become anxious and discouraged.
In this book you will see familiar arithmetic topics. Do not skip
them, even if you feel that you have mastered them. The
explanations will probably be different from those you have
already seen because they emphasize the underlying concepts.
If you don’t understand a concept the first time, be patient and
keep trying. Sometimes, by working through the material you
will see why it works. Read all the Understanding the Concept
boxes in the book since they will help you learn mathematics.
Exercise
1. Write in words why the commutative property of addition
reduces the amount of memorization necessary to learn
addition facts.