The Improving Mathematics Education in Schools (TIMES) Project
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Algebraic expressions
Number and Algebra : Module 23Year : 7
June
PDF Version of module
ASSUMED KNOWLEDGE
 Fluency with addition, subtraction, multiplication and division of whole numbers
and fractions.  Ability to apply the anyorder principle for multiplication and addition (commutative law and associative law) for whole numbers and fractions.
 Familiarity with the order of operation conventions for whole numbers.
MOTIVATION
Algebra is a fascinating and essential part of mathematics. It provides the written language in which mathematical ideas are described.
Many parts of mathematics are initiated by finding patterns and relating to different quantities. Before the introduction and development of algebra, these patterns and relationships had to be expressed in words. As these patterns and relationships became more complicated, their verbal descriptions became harder and harder to understand. Our modern algebraic notation greatly simplifies this task.
A wellknown formula, due to Einstein, states that E = mc2. This remarkable formula gives the relationship between energy, represented by the letter E, and mass, represented by letter m. The letter c represents the speed of light, a constant, which is about metres per second. The simple algebraic statement E = mc2 states that some matter is converted into energy (such as happens in a nuclear reaction), then the amount of energy produced is equal to the mass of the matter multiplied by the square of the speed of light. You can see how compact the formula is compared with the verbal description.
We know from arithmetic that 3 × 6 + 2 × 6 = 5 × 6. We could replace the number 6 in this statement by any other number we like and so we could write down infinitely many such statements. All of these can be captured by the algebraic statement 3x + 2x = 5x, for any number x. Thus algebra enables us to write down general statements clearly and concisely.
The development of mathematics was significantly restricted before the 17th century by the lack of efficient algebraic language and symbolism. How this notation evolved will be discussed in the History section.
CONTENT
Using pronumerals
In algebra we are doing arithmetic with just one new feature − we use letters to represent numbers. Because the letters are simply standins for numbers, arithmetic is carried out exactly as it is with numbers. In particular the laws of arithmetic (commutative, associative and distributive) hold.
For example, the identities
2 + x = x + 2  2 × x = x × 2  
(2 + x) + y = 2 + (x + y)  (2 × x) × y = 2 × (x × y)  
6(3x + 1) = 18x + 6 
hold when x and y are any numbers at all.
In this module we will use the word pronumeral for the letters used in algebra. We choose to use this word in school mathematics because of confusion that can arise from the words such as ‘variable’. For example, in the formula E = mc2, the pronumerals E and m are variables whereas c is a constant.
Pronumerals are used in many different ways. For example:
 Substitution: ‘Find the value of 2x + 3 if x = 4.’ In this case the pronumeral is given the value 4.
 Solving an equation: ‘Find x if 2x + 3 = 8.’ Here we are seeking the value of the pronumeral that makes the sentence true.
 Identity: ‘The statement of the commutative law: a + b = b + a.’ Here a and b can be any real numbers.
 Formula: ‘The area of a rectangle is A = lw.‘ Here the values of the pronumerals are connected by the formula.
 Equation of a line or curve: ‘The general equation of the straight line is y = mx + c.’
Here m and c are parameters. That is, for a particular straight line, m and c are fixed.
In some languages other than English, one distinguishes between ‘variables’ in functions and ‘unknown quantities’ in equations (‘incógnita’ in Portuguese/Spanish, ‘inconnue’ in French) but this does not completely clarify the situation. The terms such as variable and parameter cannot be precisely defined at this stage and are best left to be introduced later in the development of algebra.
An algebraic expression is an expression involving numbers, parentheses, operation signs and pronumerals that becomes a number when numbers are substituted for the pronumerals. For example 2x + 5 is an expression but +) ×is not.
Examples of algebraic expressions are:
3x + 1 and 5(x2 + 3x)
As discussed later in this module the multiplication sign is omitted between letters and between a number and a letter. Thus substituting x = 2 gives
3x + 1 = 3 × 2 + 1 = 7 and 5(x2 + 3x) = 5(22 + 3 × 2) =
In this module, the emphasis is on expressions, and on the connection to the arithmetic that students have already met with whole numbers and fractions. The values of the pronumerals will therefore be restricted to the whole numbers and nonnegative fractions.
Changing words to algebra
In algebra, pronumerals are used to stand for numbers. For example, if a boxcontains x stones and you put in five more stones, then there are x + 5 stones in the box. You may or may not know what the value of x is (although in this example we do know that x is a whole number).
 Joe has a pencil case that contains an unknown number of pencils. He has three
other pencils. Let x be the number of pencils in the pencil case. Then Joe has x + 3 pencils altogether.
 Theresa has a boxwith least 5 pencils in it, and 5 are removed. We do not know how many pencils there are in the pencil case, so let z be the number of pencils in the box. Then there are z − 5 pencils left in the box.
 There are three boxes, each containing x marbles in each box, then the total number of marbles is 3 × x = 3x.
 If n oranges are to be divided amongst 5 people, then each person receives oranges. (Here we assume that n is a whole number. If each person is to get a whole number of oranges, then n must be a multiple of 5.)
The following table gives us some meanings of some commonly occurring
algebraic expressions.
x + 3 

x − 3 

3 × x 

x ÷ 3 

2 × x − 3 

x ÷ 3 + 2 

Notations and laws
Expressions with zeroes and ones
Zeroes and ones can often be eliminated entirely. For example:
x + 0 = x (Adding zero does not change the number.)
1 × x = x (Multiplying by one does not change the number.)
Algebraic notation
In algebra there are conventional ways of writing multiplication, division and indices.
Notation for multiplication
The × sign between two pronumerals or between a pronumeral and a number is usually omitted. For example, 3 × x is written as 3x and a × b is written as ab. We have been following this convention earlier in this module.
It is conventional to write the number first. That is, the expression 3 × a is written as 3a and not as a3.
Notation for division
The division sign ÷ is rarely used in algebra. Instead the alternative fraction notation for division is used. We recall 24 ÷ 6 can be written as . Using this notation, x divided by 5
is written as , not as x ÷ 5.
Index notation
x × x is written as x2 and y × y × y is written as y3.
EXAMPLE
Write each of the following using concise algebraic notation.
 a
 A number x is multiplied by itself and then doubled.
 b
 A number x is squared and then multiplied by the square of a second number y.
 c
 A number x is multiplied by a number y and the result is squared.
Solution
ax × x × 2 = x2 × 2 = 2x2. bx2 × y2 = x2y2.
c (x × y)2 = (xy)2 which is equal to x2y2
Summary


Substitution
Assigning values to a pronumeral is called substitution.
EXAMPLE
If x = 6, what is the value of:
a  3x + 4  b  2x + 3  c  2x − 5  
d  − 2  e  + 2 
Solution
a  3x + 4 = 3 × 6 + 4 = 22  b  2x + 3 = 2 × 6 + 3 = 15  
c  2x − 5= 2 × 6 − 5 = 7  d  − 2 = − 2 = 1  
e  + 2 = + 2 = 4 
Adding and subtracting like terms
Like terms
If you have 3 pencil case with the same number x of pencils in each, you have 3x pencils altogether.
If there are 2 more pencil cases with x pencils in each, then you have 3x + 2x = 5x pencils altogether. This can be done as the number of pencils in each case is the same. The terms 3x and 2x are said to be like terms.
Consider another example. If Jane has x packets of chocolates each containing y chocolates, then she has x × y = xy chocolates.
If David has twice as many chocolates as Jane, he has 2 × xy = 2xy chocolates.
Together they have 2xy + xy = 3xy chocolates.
The terms 2xy and xy are like terms. Two terms are called like terms if they involve exactly the same pronumerals and each pronumeral has the same index.
EXAMPLE
Which of the following pairs consists of like terms:
a 3x, 5xb 4x2, 8xc 4x2y, 12x2y
Solution
 a
 5x and 3x are like terms.
 b
 4x2 and 8x are not like terms since the powers of x are different.
 c
 4x2y, 12x2y are like terms
Adding and subtracting like terms
The distributive law explains the addition and subtraction of like terms. For example:
2xy + xy = 2 × xy + 1 × xy =(2 + 1)xy = 3xy
The terms 2x and 3y are not like terms because the pronumerals are different. The terms 3x and 3x2 are not like terms because the indices are different. For the sum 6x + 2y + 3x, the terms 6x and 3x are like terms and can be added. There are no like terms for 2y, so by using the commutative law for addition the sum is
6x + 2y + 3x = 6x + 3x + 2y = 9x + 2y.
The anyorder principle for addition is used for the adding like terms.
Because of the commutative law and the associative law for multiplication (anyorder principle for multiplication) the order of the factors in each term does not matter.
5x × 3y = 15xy = 15yx
6ab × 3b2a = 18a2b3 = 18b3a2
Like terms can be added and subtracted as shown in the example below.
EXAMPLE
Simplify each of the following by adding or subtracting like terms:
Solution
a  2x + 3x + 5x = 10x  b  3xy + 2xy = 5xy  
c  4x2 − 3x2 = x2  d  2x2 + 3x + 4x = 2x2 + 7x  
e  4x2y − 3x2y + 3xy2 = x2y + 3xy2 
The Use of brackets
Brackets fulfill the same role in algebra as they do in arithmetic. Brackets are used in algebra in the following way.
‘Six is added to a number and the result is multiplied by 3.’
Let x be the number. Then the expression is (x + 6) × 3. We now follow the convention that the number is written at the front of the expression (x + 6) without a multiplication sign. The preferred form is thus 3(x + 6).
EXAMPLE
For a party, the host prepared 6 tins of chocolate balls, each containing n chocolate balls.
 a
 The host places two more chocolates in each tin. How many chocolates are there altogether in the tins now?
 b
 If n = 12, that is, if there were originally 12 chocolates in each box, how many chocolates are there altogether in the tins now?
Solution
 a
 The number of chocolates in each tin is now n + 2. There are 6 tins, and therefore there are 6 (n + 2) chocolates in total.
 b
 If n = 12, the total number of chocolates is 6 (n + 2) = 6 (12 + 2) = 6 × 14 =
EXERCISE 1
Each crate of bananas contains n bananas. Three bananas are removed from each crate.
 a
 How many bananas are now in each crate?
 b
 If there are 7 crates, how many bananas are there now in total?
EXERCISE 2
Five extra seats are added to each row of seats in a theatre. There were s seats in each row and there are 20 rows of seats. How many seats are there now in total?
Use of brackets and powers
The following example shows the importance of following the conventions of order of operations when working with powers and brackets.
 2x2 means 2 × x2
 (2x)2 = 2x × 2x = 4x2
EXAMPLE
For x = 3, evaluate each of the following:
a 2x2b (2x)2
Solution
a 2x2 = 2 × x × x = 2 × 3 × 3 = 18 b (2x)2 = (2 × x)2 = (2 × 3)2 = 36
EXERCISE 3
Evaluate each expression for x = 3.
a 10x3b (10x)3
Multiplying terms
Multiplying algebraic terms involves the anyorder property of multiplication discussed for whole numbers.
The following shows how the anyorder property can be used
3x × 2y × 2xy  = 3 × x × 2 × y × 2 × x × y 
= 3 × 2 × 2 × x × x × y × y  
= 12x2y2 
EXAMPLE
Simplify each of the following:
a 5 × 2ab 3a × 2ac 5xy × 2xy
Solution
 a
 5 × 2a = 10a
 b
 3a × 2a = 3 × a × 2 × a = 6a2
 c
 5xy × 2xy = 5 × x × y × 2 × x × y = 10x2y2
With practice no middle steps are necessary.
For example, 2a × 3a × 2a2 = 12a4
Arrays and areas
Arrays
Arrays of dots have been used to represent products in the module, Multiplication of Whole Numbers.
For example 2 × 6 = 12 can be represented by 2 rows of 6 dots.
The diagram below represents two rows with some number of dots.
Let n be the number of dots in each row. Then there are 2 × n = 2n dots.
If an array is m dots by n dots then there are mn dots. If the array is m × n then by convention we have m rows and n columns. We can represent the product m × n = mn by such an array.
EXAMPLE
The pattern goes on forever.
How many dots are there in the nth diagram?
Solution
In the 1st diagram there are 1 × 1 = 12 dots.
In the 2nd diagram there are 2 × 2 = 22 dots.
In the 3rd diagram there are 3 × 3 = 32 dots.
In the nth diagram there will be n × n = n2 dots.
Areas
We can also represent a product like 3 × 4 by a rectangle.
The area of a 3cm by 4cm rectangle is 12cm2.
The area of a x cm by y cm rectangle is x × y = xy cm2.
(x and y can be any positive numbers.)
The area of a x cm by x cm square is x2 cm2.
(x can be any positive number.)
EXAMPLE
Find the total area of the two rectangles in terms of x and y.
Solution
The area of the rectangle to the left is xy cm2 and the area of the rectangle to the right is 2xy cm2. Hence the total area is xy + 2xy = 3xy cm2.
Number patterns
Some simple statements with numbers demonstrate the convenience of algebra.
EXAMPLE
The nth positive even number is 2n.
 a
 What is the square of the nth positive even number?
 b
 If the nth positive even number is doubled what is the result?
Solution
 a
 The square of the nth positive even number is (2n)2 = 4n2
 b
 The double of the nth positive even number is 2 × 2n = 4n.
EXERCISE 4
a If b is even, what is the next even number?
b If a is a multiple of three, what are the next two multiples of 3?
c If n is odd and n ≥ 3, what is the previous odd number?
EXERCISE 5
Think of a ‘number’. Let this number be x.
Write the following using algebra to see what you get.
 Multiply the number you thought of by 2 and subtract 5.
 Multiply the result by 3.
 Add
 Subtract 5 times the number you first thought of.
EXERCISE 6
Show that the sum of the first n odd numbers is n2.
Algebraic fractions
Quotients of expressions involving pronumerals often occur. We call them algebraic fractions we will meet this again in the modules, Special Expansions and Algebraic Fractions.
EXAMPLE
Write each of the following in algebraic notation.
a A number is divided by 5, and 6 is added to the result.
b Five is added to a number, and the result is divided by 3.
Solution
a  Let x be the number. 
Dividing by 5 gives .  
Adding 6 to this result gives + 6.  
b  Let the number be x. 
Adding 5 gives x + 5.  
Dividing this by 3 gives . 
Notice that the vinculum acts as a bracket.
EXAMPLE
A vat contains n litres of oil. Forty litres of oil are then added to the vat.
a How many litres of oil are there now in the vat?
b The oil is divided into 50 containers. How much oil is there in each container?
Solution
a There is a total of n + 40 litres of oil in the vat.
b There are litres of oil in each container.
EXERCISE 7
A shed contains n tonnes of coal. An extra tonnes are then added.
 a
 How many tonnes of coal are there in the shed now?
 b
 It is decided to ship the coal in 10 equal loads. How many tonnes of coal are there in each load?
Expanding brackets and collecting like terms
Expanding brackets
Numbers obey the distributive laws for multiplication over addition and subtraction. For example:
3 × (4 + 5) = 3 × 4 + 3 × 5 7 × (6 − 3) = 7 × 6 − 7 × 3
The distributive laws for division over addition and subtraction also hold as shown.
For example:
(8 + 6) ÷ 2 = 8 ÷ 2 + 6 ÷ 2 and = −
As with adding like terms and multiplying terms, the laws that apply to arithmetic can be extended to algebra. This process of rewriting an expression to remove brackets is usually referred to as expanding brackets.
EXAMPLE
Use the distributive law to rewrite these expressions without brackets.
a 5(x − 4) b 4(3x + 2) c 6(4 − 2x)
Solution
a 5(x − 4) = 5x − 20 b 4(3x + 2) = 12x + 8
c 6(4 − 2x) =24 − 12x
EXAMPLE
Use the distributive law to rewrite these expressions without brackets.
a  b 
Solution
a  = x +  b  = 1 − 
Collecting like terms
After brackets have been expanded like terms can be collected.
EXAMPLE
Expand the brackets and collect like terms:
a 2(x − 6) + 5xb 3 + 3(x − 1) c 3(2x + 4) + 6(x − 1)
Solution
a 2(x − 6) + 5x = 2x − 12 + 5x = 7x − 12 b 3 + 3(x − 1) = 3 + 3x − 3 = 3x
c 3(2x + 4) + 6(x − 1) = 6x + 12 + 6x − 6 = 12x + 6
EXERCISE 8
Expand the brackets and collect like terms:
a 5(x + 2) + 2(x − 3) b 2(7 + 5x) + 4(x + 6)
c 3(2x + 7) + 2(x − 5)
links forward
A sound understanding of algebra is essential for virtually all areas of mathematics.
The introduction to algebra is continued in the modules, Negatives and the Index Laws in Algebra, Special Expansions and Algebraic Fractions and Fractions and the Index Laws in Algebra.
HISTORY
It was only in the 17th century that algebraic notation similar to that used today was introduced. For example, the notation used by Descartes (La Geometrie, ) and Wallis () was very close to modern notation. However, algebra has a very long history.
There are examples of the ancient Egyptians working with algebra. About BC, the Egyptian scribe Ahmes made a transcript of even more ancient mathematical scriptures dating to the reign of the Pharaoh Amenemhat III. In the Scottish antiquarian, Henry Rhind, came into possession of Ahmes's papyrus. The papyrus is a scroll 33 cm wide and about m long filled with mathematical problems. One of the problems is as follows:
measures of corn must be divided among 5 workers, so that the second worker gets as many measures more than the first worker, as the third gets more than the second, as the fourth gets more than the third, and as the fifth gets more than the fourth. The first two workers shall get seven times fewer measures of corn than the three others. How many measures of corn shall each worker get? (The answer involves fractional measures of corn. Answer: 1 , 10 , 20, 29 and 38 measures.)
Euclid (circa BC) dealt with algebra in a geometric way and algebraic problems are solved without using algebraic notation of any form. Diophantus (circa AD ) who is sometimes called the father of algebra, produced a work largely devoted to the ideas of the subject.
Chinese and Indian authors wrote extensively on algebraic ideas and achieved a great deal in the solution of equations. The earliest Chinese text with algebraic ideas is The Nine Chapters on the Mathematical Art, written around BC. A later text is Shushu chiuchang, or Mathematical Treatise in Nine Sections, which was written by the wealthy governor and minister Ch'in Chiushao (circa − circa AD).
In approximately BC an Indian mathematician Baudhayana, in his Baudhayana Sulba Sutra, discovered Pythagorean triples algebraically, and found geometric solutions of linear equations and quadratic equations of the forms, ax2 = c and of ax2 + bx = c.
In AD the Indian mathematician Brahmagupta, in his treatise Brahma Sputa Siddhanta, worked with quadratic equations and determined rules for solving linear and quadratic equations. He discovered that quadratic equations can have two roots, including both negative and irrational roots.
Indian mathematician Aryabhata, in his treatise Aryabhatiya, obtains wholenumber solutions to linear equations by a method equivalent to the modern one.
Arab and Persian mathematicians had an interest in algebra and their ideas flowed into Europe. One algebraist of special prominence was alKhwarizmi, whose aljabr w’al muqabalah (circa AD) gave the discipline its name and gave the first systematic treatment of algebra.
The works of alKhwarizmi were translated into European languages by several different translators during the twelfth century and so his and other Arab writers’ work were well known in Europe.
Fibonacci was the greatest European writer on algebra during the middle ages and his work Liber Quadratorum (circa AD) includes many different ingenious ways of solving equations.
Cardano, Tartaglia (16th century), Vieta (16th − 17th century) and others developed the ideas and notation of algebra. The Ars Magna (Latin: “The Great Art”) is an important book on Algebra written by Gerolamo Cardano. It was first published in under the title Artis Magnæ, Sive de Regulis Algebraicis Liber Unus (Book number one about The Great Art, or The Rules of Algebra). There was a second edition in Cardano’s lifetime, published in It is considered one of the three greatest scientific treatises of the Renaissance. The book included the solutions to the cubic and quartic equations. The solution to one particular case of the cubic, x3 + ax = b (in modern notation), was communicated to him by Niccolò Fontana Tartaglia, and the quartic was solved by Cardano’s student Lodovico Ferrari.
Development of algebraic notation
Here are some of the different notations used from the Middle Ages onwards together with their modern form. They reveal how useful modern algebraic notation is.
Trouamen0.che gito al suo drat0 facia 

4 Se. −51 Pri. −30 N. dit is ghelijc 45 

cub p: 6 reb ælis 

1 Q C −15 Q Q + 85 C − Q + N æquator 

aaa − 3.bba ======= + 2.ccc 

References
A History of Mathematics: An Introduction, 3rd Edition, Victor J. Katz, AddisonWesley, ()
ANSWERS TO EXERCISES
Exercise 1
an − 3 bananas in each crate b 7(n − 3) bananas in total
Exercise 2
a 20(s + 5) seats in total
Exercise 3
a b 27
Exercise 4
ab + 2 ba + 3, a + 6 cn − 2
Exercise 5
x 2x − 5 3(2x − 5) = 6x − 15 6x 6x − 5x = x
Exercise 6
Sum of the first n odd numbers is: 1 + 3 + 5 + …. + (2n − 5) + (2n − 3) + (2n − 1)
Reverse the sum: (2n − 1) + (2n − 3) + (2n − 5) + … + 5 + 3 + 1
Add to the original sum pairing terms yields
(2n − 1 + 1) + (2n − 3 + 3) + … + (3 + 2n − 3) + (1 + 2n − 1) = 2n × n =2n2
The sum is n2
Exercise 7
an + tonnes b tonnes
Exercise 8
a 7x + 4 b 14x + 38 c 8x + 11
The Improving Mathematics Education in Schools (TIMES) Project was funded by the Australian Government Department of Education, Employment and Workplace Relations.
The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations.
© The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICEEM), the education division of the Australian Mathematical Sciences Institute (AMSI), (except where otherwise indicated). This work is licensed under the Creative Commons AttributionNonCommercialNoDerivs Unported License.
https://creativecommons.org/licenses/byncnd//
HOW TO TAKE HALF
OF ANY NUMBER
Lesson 16 Section 2
Back to Section 1
7.  Which natural numbers are called the even numbers? 
Numbers that can be divided into two equal parts.  
8 is an even number because we can divide it "evenly" into two 4's.
4 is half of 8.
9 is not an even number. There is an "odd" 1:
Therefore we call 9 an odd number. We cannot take half of a natural number that is odd. We cannot take half of 9 people.
The even numbers, then, are the numbers divisible by 2: 2, 4, 6, 8, and so on. The odd numbers are 1, 3, 5, 7, and so on.
If you do not know half of some even number, you can find it by decomposing it mentally into two even numbers whose half you do know. We saw how to do that in Section 1, Example 8.
Example 1. How much is half of 54?
Answer. Half of 54  =  Half of 50 + Half of 4 
=  25 + 2  
= 
As for odd numbers, note that every odd number is equal to an even number plus 1.
3 = 2 + 1.
5 = 4 + 1.
7 = 6 + 1.
9 = 8 + 1.
And so on.
Again, if a natural number is odd, then we cannot take half of it. Therefore, we will now assume that we are not necessarily dealing with natural numbers, and that 1 is now a continuous unit  a unit we use to measure  and that has a half: 1 minute, 1 centimeter, 1 pound, and so on. (Lesson )
8.  What number is half of 1? 
½ ("Onehalf")  
½ must be one half of something. It is one half of 1.
Half of $1, of course, is $.
9.  How can we take half of an odd number? 
Half of 7  
Take half of its even part, and add half of 1.  
Example 2. How much is half of 7?
Answer. 7 = 6 + 1. Half of 6 is 3; half of 1 is ½; therefore half of 7
is 3½.
Half of $7.00 is $3.
Example 3. A scarf that normally sells for $17 is being sold at half price. What do you pay?
Answer. 17 = 16 + 1. Half of 16 is 8. Therefore, half of $17 is $8.
Example 4. How much is half of 25%?
Answer. 12½%.
25 = 24 + 1. Half of 24 is Half of 1 is ½.
And so if you knew that 25% means a quarter, then you now know that 12½% means an eighth.
(See Lesson 27, Example 7.)
Example 5 How much is half of $27.40?
Answer. We will take half of $27 and add it to half of $.
Since 27 = 26 + 1, then
Half of $27 is $13.
Half of $.40 is $.
Therefore, half of $ is $13.
Example 6. How much is half of 90?
Answer. 90 = 80 + Half of 80 is 40; half of 10 is 5; therefore half of 90 is
See Problem
Example 7. How much is half of $9.70?
Answer. Half of $9.00 (8 + 1) is $4. Half of $.70 is $. Therefore half of $ is $4.
Example 8. The famous number π ("pi") is approximately 3. How
much is half of π; that is,  π 2  ? 
Answer. Ignore the decimal point in 3. Then half of is ; half of 14 is 7; therefore half of is On replacing the decimal point,
π 2  1. 
How can we take a fourth or 25% of a number?  
Take half of half.  
To divide a number in Half means to divide it into two equal parts. To divide it into Fourths, or quarters, means to divide it into four equal parts. And we can do that by taking half of each Half.
A quarter is half of a half. Compare Lesson
Example 9. How much is a fourth, or a quarter, of 60?
Answer. Half of 60 is Half of 30 is
Example How much is 25% of ?
Answer. 25% means a fourth. (Lesson ) Half of is Half of 90 is
Example How much is 25% of ?
Answer. Half of = Half of + Half of 12 =
Answer. Half of 56 = Half of 50 + Half of 6 = 25 + 3 =
Example How much is 25% of $?
Answer. Half of $ = $ + $ = $ Half of $ = $2.40
How can we multiply by 5?  
5 ×  
Take half of the multiplication by  
Since 5 is half of 10, then 5 times a number will be half of 10 times that number.
Example 5 × = Half of 10 ×
Example 13 5 × = Half of (Lesson 4, Question 1.)
Example 5 × =
Example 5 × $1.50 = Half of $15
Example 5 × $1.50 = $7.
Example 5 × $46.80 = Half of $
Example 5 × $46.80 = $
How can we take 5% of a number?  
Take half of 10%.  
Example How much is 5% of $?
Answer. 10% is $16. (Lesson 4, Question 7.) Therefore, 5% is $8.
To write and multiply, is a written method for those who do not understand percent.
Example How much is 5% of $?
Answer. 10% is $47. Therefore, 5% is $23.50 + $.25 = $23.
How can we multiply by 15?  
15 × 8  
Multiply by 10, and add half.  
For, 15 = 10 + Half of
Example 15 × 8 = 80 + Half of 80
Example 15 × 8 = 80 + 40
Example 15 × 8 =
Example 15 × 42 = +
Example 15 × 42 =
How can we take 15% of a number?  
Take 10% and add half.  
Example How much is 15% of $70??
Answer. 15%  =  10% + Half of 10% 
=  $7.00 + $3.50  
=  $10. 
Example If you tip at the rate of 15%, and the bill is $40, how much do you leave?
Answer. 15%  =  10% + Half of 10% 
=  $4.00 + $2.00  
=  $6. 
At this point, please "turn" the page and do some Problems.
or
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by Marilyn Burns
In this lesson, Marilyn Burns shows fifthgrade students a fraction and they decide if it’s more or less than onehalf and then explain their reasoning. This focus on onehalf helps establish it as an important and useful benchmark. The lesson also provides practice with mental computation of whole numbers as students compare numerators and denominators. More or Less Than OneHalf will appear in Marilyn’s forthcoming book Teaching Arithmetic: Lessons for Introducing Fractions, Grades 4–5, to be published in fall by Math Solutions Publications.
I used this activity when I had some time left at the end of a periodor as a warmup at the beginning of class. I’d write a fraction on theboard and ask if it was more or less than onehalf. Students who answered also had to explain their thinking. After each student’s response, I’d ask, “Does anyone have another way to explain that?” In this way, students focused not only on answering and explaining their reasoning but also on trying to think of different ways to explain answers. I’d continue discussing the fraction until all students who wanted had had a chance to explain.
Here’s how the activity went with a class of fifth graders in the fall of the year. I wrote on the board:
2/3
“Is this more or less than onehalf?” I asked.
Davy began by saying, “There are three thirds in a whole, and twothirds is more than halfway to the whole.”
“How do you know it’s more than halfway?” I probed. Davy wasn’t sure.
“Listen to other ideas,” I said, “and see if they can help you explain more.” I called on Ramon next.
“On a measuring cup, the line for twothirds is above the onehalf line,” Ramon said. “It’s like halfway to a whole cup after half a cup.”
Leslie asked to come to the board. She drew a circle and divided it into three equalsize wedges. She said, “If you had a cookie cut into thirds like this, you can see that onethird is less than onehalf. If there were two people and you each took onethird, then you’d have to share some more to get onehalf each. So onehalf is onethird plus some more.”
Rachel’s explanation was more abstract. “If twothirds was the same as onehalf, then two would have to be half of three. But it’s more, so twothirds has to be more.”
One day, I chose one and onequarter for the fraction. When I wrote it on the board, some students laughed and others blurted out comments: “Simple!” “That’s easy.” “Nobrainer.” Practically every student raised a hand.
“More or less?” I asked. “Let’s all say the answer softly together.” That took care of the answer.
“Who can explain why one and a fourth is more than onehalf?” I then asked. Again, there were lots of volunteers.
“It’s obvious,” Daniel said, “because onehalf is less than one and one and one fourth is more than one, so it has to be more than onehalf.”
“It’s more than twice as big,” Sadie said, “because onehalf and onehalf are one and one and a fourth is even more than that.”
“On the number line, they’re on different sides of one,” Emma said. “That shows onehalf is less.”
Even for an obvious solution, students may think in different ways. Listening to their thinking often gives me new insights into students’ understanding. Also, sometimes students think in ways I hadn’t thought of, giving me new ways to look at the mathematics.
For this activity, I sometimes selected fractions with larger numerators and denominators, at times choosing fractions that were easy to analyze, such as 61/ or /, but at other times choosing fractions that also offered a mental computation challenge, such as /, 89/, even / The students seemed to like stretching their thinking to decide if fractions like these were more or less than onehalf.
From Online Newsletter Issue Number 2, Summer
Related Publication:
Teaching Arithmetic: Lessons for Introducing Fractions, Grades 4–5
by Marilyn Burns
HOW TO TAKE HALF
OF ANY NUMBER
Lesson 16 Section 2
Back to Section 1
7.  Which natural numbers are called the even numbers? 
Numbers that can be divided into two equal parts.  
8 is an even number because we can divide it "evenly" into two 4's.
4 is half of 8.
9 is not an even number. There is an "odd" 1:
Therefore we call 9 an odd number. We cannot take half of a natural number that is odd. We cannot take half of 9 people.
The even numbers, then, are the numbers divisible by 2: 2, 4, 6, 8, and so on. The odd numbers are 1, 3, 5, 7, and so on.
If you do not know half of some even number, you can find it by decomposing it mentally into two even numbers whose half you do know. We saw how to do that in Section 1, Example 8.
Example 1. How much is half of 54?
Answer. Half of 54  =  Half of 50 + Half of 4 
=  25 + 2  
=  27. 
As for odd numbers, note that every odd number is equal to an even number plus 1.
3 = 2 + 1.
5 = 4 + 1.
7 = 6 + 1.
9 = 8 + 1.
And so on.
Again, if a natural number is odd, then we cannot take half of it. Therefore, we will now assume that we are not necessarily dealing with natural numbers, and that 1 is now a continuous unit  a unit we use to measure  and that has a half: 1 minute, 1 centimeter, 1 pound, and so on. (Lesson 20.)
8.  What number is half of 1? 
½ ("Onehalf")  
½ must be one half of something. It is one half of 1.
Half of $1, of course, is $.50.
9.  How can we take half of an odd number? 
Half of 7  
Take half of its even part, and add half of 1.  
Example 2. How much is half of 7?
Answer. 7 = 6 + 1. Half of 6 is 3; half of 1 is ½; therefore half of 7
is 3½.
Half of $7.00 is $3.50.
Example 3. A scarf that normally sells for $17 is being sold at half price. What do you pay?
Answer. 17 = 16 + 1. Half of 16 is 8. Therefore, half of $17 is $8.50.
Example 4. How much is half of 25%?
Answer. 12½%.
25 = 24 + 1. Half of 24 is 12. Half of 1 is ½.
And so if you knew that 25% means a quarter, then you now know that 12½% means an eighth.
(See Lesson 27, Example 7.)
Example 5 How much is half of $27.40?
Answer. We will take half of $27 and add it to half of $.40.
Since 27 = 26 + 1, then
Half of $27 is $13.50.
Half of $.40 is $.20.
Therefore, half of $27.40 is $13.70.
Example 6. How much is half of 90?
Answer. 90 = 80 + 10. Half of 80 is 40; half of 10 is 5; therefore half of 90 is 45.
See Problem 21.
Example 7. How much is half of $9.70?
Answer. Half of $9.00 (8 + 1) is $4.50. Half of $.70 is $.35. Therefore half of $9.70 is $4.85.
Example 8. The famous number π ("pi") is approximately 3.14. How
much is half of π; that is,  π 2  ? 
Answer. Ignore the decimal point in 3.14. Then half of 300 is 150; half of 14 is 7; therefore half of 314 is 157. On replacing the decimal point,
π 2  1.57. 
10.  How can we take a fourth or 25% of a number? 
Take half of half.  
To divide a number in Half means to divide it into two equal parts. To divide it into Fourths, or quarters, means to divide it into four equal parts. And we can do that by taking half of each Half.
A quarter is half of a half. Compare Lesson 15.
Example 9. How much is a fourth, or a quarter, of 60?
Answer. Half of 60 is 30. Half of 30 is 15.
Example 10. How much is 25% of 180?
Answer. 25% means a fourth. (Lesson 15.) Half of 180 is 90. Half of 90 is 45.
Example 11. How much is 25% of 112?
Answer. Half of 112 = Half of 100 + Half of 12 = 56.
Answer. Half of 56 = Half of 50 + Half of 6 = 25 + 3 = 28.
Example 12. How much is 25% of $9.60?
Answer. Half of $9.60 = $4.50 + $.30 = $4.80. Half of $4.80 = $2.40
11.  How can we multiply by 5? 
5 × 123  
Take half of the multiplication by 10.  
Since 5 is half of 10, then 5 times a number will be half of 10 times that number.
Example 13. 5 × 123 = Half of 10 × 123
Example 13 5 × 123 = Half of 1230 (Lesson 4, Question 1.)
Example 13. 5 × 123 = 615.
Example 14. 5 × $1.50 = Half of $15
Example 14. 5 × $1.50 = $7.50.
Example 15. 5 × $46.80 = Half of $468
Example 15. 5 × $46.80 = $234.
12.  How can we take 5% of a number? 
Take half of 10%.  
Example 16. How much is 5% of $162?
Answer. 10% is $16.20. (Lesson 4, Question 7.) Therefore, 5% is $8.10.
To write .05 and multiply, is a written method for those who do not understand percent.
Example 17. How much is 5% of $475?
Answer. 10% is $47.50. Therefore, 5% is $23.50 + $.25 = $23.75.
13.  How can we multiply by 15? 
15 × 8  
Multiply by 10, and add half.  
For, 15 = 10 + Half of 10.
Example 18. 15 × 8 = 80 + Half of 80
Example 18. 15 × 8 = 80 + 40
Example 18. 15 × 8 = 120.
Example 19. 15 × 42 = 420 + 210.
Example 19. 15 × 42 = 630.
14.  How can we take 15% of a number? 
Take 10% and add half.  
Example 20. How much is 15% of $70??
Answer. 15%  =  10% + Half of 10% 
=  $7.00 + $3.50  
=  $10.50. 
Example 21. If you tip at the rate of 15%, and the bill is $40, how much do you leave?
Answer. 15%  =  10% + Half of 10% 
=  $4.00 + $2.00  
=  $6.00. 
At this point, please "turn" the page and do some Problems.
or
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Half some number of one
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Writing Expressions
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Algebraic Expressions
The following table gives some common words or phrases that are usually imply one of the four operations: add, subtract, multiply, divide. However, depending on the word problem, there are exceptions. Scroll down the page for more examples and explanations.
Phrase To Algebraic Expression
To write an expression, we often have to interpret a written phrase.
For example, the phrase “6 added to some number” can be written as the expression x + 6, where the variable x represents the unknown number.
Some examples of common phrases and corresponding expressions that involve addition are:
Phrase  Expression 
4 more than some number  x + 4 
a number increased by 10  y + 10 
8 plus some number  t + 8 
the sum of a number and 12  w + 12 
Some examples of common phrases and corresponding expressions that involve subtraction are:
Phrase  Expression 
4 less than some number  x – 4 
a number decreased by 10  y – 10 
8 minus some number  8 – t 
the difference between a number and 12  w – 12 
Some examples of common phrases and corresponding expressions that involve multiplication are:
Phrase  Expression 
4 times some number  4x 
twice a number  2y 
onethird of some number  
the product of a number and 12  12w 
Some examples of common phrases and corresponding expressions that involve division are:
Some examples of common phrases and corresponding expressions that involve two operations are:
Phrase  Expression 
6 more than 5 times a number  5x + 6 
4 times the sum of a number and 7  4(y + 7) 
5 less than the product of 3 and a number  3w – 5 
twice the difference between a number and 9  2(z – 9) 
How Do You Write Mathematical Expressions From Word Problems?
There are so many words that you come across when you’re working on algebra problems, and these words are really code for very specific mathematical symbols.
How To Write Algebraic Expressions For Word Phrases, By Analyzing The Language Used?
An algebraic expression is a mathematical phrase that contains a combination of numbers, variables and operational symbols.
A variable is a letter that can represent one or more numbers.
How to write expressions with variables?
Examples:
Write the algebraic expressions to represent the statements.
a) The sum of 7 and the quantity 8 times x
b) Take the quantity 3 times x and then add 1.
c) 6 plus the product of 1 and x.
How to write algebraic expressions with parentheses?
Examples:
First consider the expression for
5 plus the quantity of 4 times x
Now take the product of 8 and that expression and then add 6.First consider the expression for
the sum of 7 and the product of 2 and x
What expression would be:
4 plus the quantity of 2 times that expression.
How to Write Equations from Word Problems?
This video teaches how to dissect a word problem in order to define a variable and write an equation.
Examples:
Half of a number is Write an equation to represent the situation. Define your variable and solve.
Mrs. Gaddie has two dogs. Her friend AnnaMarie has three less than twice as many as Mtr. Gaddie. How many dogs does AnnaMarie have?
A recycling plant recycles 2 tons of cans yesterday. This is a third of their usual amount. How much does the plant usually recycle?
How to Write Algebraic Expressions for Situations?
Example:
Translate the following phrases into algebraic or numeric expressions.
a) less than a number ‘b’
b) Quotient of and ‘b’
c) ‘b’ times
d) more than ‘b’
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the stepbystep explanations.
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But isn't it.  she did not finish. Marisa looked at Cernunnos as if he were a phantom, a ghost from another world.