4 Ships use the speed of sound in water to help find the water’s depth. A sonar pulse from a ship is sent to the bottom of the ocean floor. The time taken for the pulse to hit the ocean floor and return to the ship is used to calculate the distance. If the sonar pulse returns in 1.5 seconds, what is the ocean depth? Assume that the speed of sound in water is 1470 metres per second. How could you set up a procedure to quickly calculate the ocean depth for any time measurement? This chapter looks at using pronumerals to represent quantities in different situations. You will learn how to form and use algebraic expressions and how to express them in simpler forms. Algebra
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4Ships use the speed of sound in water to help find the water’s depth. A sonar pulse from a ship is sent to the bottom of the ocean floor. The time taken for the pulse to hit the ocean floor and return to the ship is used to calculate the distance. If the sonar pulse returns in 1.5 seconds, what is the ocean depth? Assume that the speed of sound in water is 1470 metres per second.
How could you set up a procedure to quickly calculate the ocean depth for any time measurement?
This chapter looks at using pronumerals to represent quantities in different situations. You will learn how to form and use algebraic expressions and how to express them in simpler forms.
Using pronumeralsThe basic purpose of algebra is to solve mathematical problems involving an unknown.Equations where an unknown quantity is replaced with a letter, for example x, can beused to solve problems like:
At what speed should I ride my bicycle to arrive at school on time?How do I convert a recipe for different numbers of guests?What volume of cement is needed to build a path?
A pronumeral is a letter that is used in place of a number. In Year 7 we saw thatpronumerals could be used to make expressions and equations. Often a pronumeral isused to represent one particular number. For example, in the equation
x + 1 = 7the pronumeral x has the value 6.
Pronumerals can also be used to show a relationship between two or more numbers,for example
a + b = 10Can you find some different pairs of values for a and b which fit this rule?
Algebra allows us to show complex rules in a more simple way, and to solve problems involving unknown numbers.
1 Suppose x people are in attendance at the start of a football match.a If a further y people arrive during the first quarter, write an expression for the
number of people at the ground.b At half-time 170 people leave. Write an expression for the number of people at the
ground after they have left.
2 The canteen manager at Browning Industries orders m vanilla slices each day. Write aparagraph which could explain the table below:
3 Imagine that your cutlery drawer contains a knives, b forksand c spoons.a Write an expression for the total number of knives and
forks you have.b Write an expression for the total number of items in the
drawerc You put 4 more forks in the drawer. Write an expression
for the number of forks now.d Write an expression for the number of knives in the
drawer after 6 knives are removed.
4 If y represents a certain number, write expressions for thefollowing numbers.a A number 7 more than yb A number 8 less than yc A number which is equal to five times yd The number formed when y is subtracted from 14e The number formed when y is divided by 3.
5 Using a and b to represent numbers, write expressions for:a the sum of a and bb the difference between a and bc three times a subtracted from two times bd the product of a and be twice the product of a and bf the sum of 3a and 7bg a multiplied by itself.
C h a p t e r 4 A l g e b r a 1216 If tickets to a Brisbane Bullets/Melbourne Tigers basketball match cost $27 for adults
and $14 for children, write an expression for the cost of:a y adult ticketsb d child ticketsc r adult and h child tickets.
7 If Naomi is now t years old.a Write an expression for her age in 2 years’ time.b Write an expression for Steve’s age, if he is g years older than Naomi.c How old was Naomi 5 years ago?d Naomi’s father is twice her age. How old is he?
8 Charles places p coins into a poker machine. He plays the machine and counts hiscoins every 3 minutes. The table below shows how many coins he has.
a Write a paragraph explaining what happened.b When did Charles start to lose money?c If he used $1 coins, how much did Charles win or lose, overall?
1 Licia has bought her lunch from the school canteen for $3.00. It con-sisted of a roll, a carton of milk and a piece of fruit. She paid 60 centsmore for the milk than the fruit and 30 cents more for the roll than themilk. How much did the roll cost her?
2 Find at least two 2-digit numbers that areequal to 7 times the sum of their digits.
3 Find 5 consecutive numbers thatadd to 120.
4 I’m thinking of a number. If Imultiply it by 5 and subtract4, I get the same number aswhen I multiply it by 4 andadd 2. What is the number?
If this pattern continues, how manycubes will it take to make 10 layers?
5
9 A microbiologist places m bacteria onto anagar plate. She counts the number ofbacteria at approximately 3 hour intervals.The results are shown in the table below:
a Explain what happens to the number ofbacteria in the first 5 intervals.
b What might be causing the number ofbacteria to increase in this way?
c What is different about the last bacteriacount?
d What may have happened to cause this?
10 If n represents an even number: a is the number n + 1 odd or even?b is 3n odd or even? c Write expressions for:
i the next three even numbers which are greater than nii the even number which is 2 less than n.
SubstitutionWhen a pronumeral is replaced by a number, we say that the number is substituted forthe pronumeral. If the value of the pronumeral (or pronumerals) is known, it is possibleto evaluate (work out the value of) an expression.
For example, if we know that x = 2 and y = 3, the expression x + y can be evaluatedas shown:
x + y = 2 + 3= 5
When writing expressions with pronumerals:1. We leave out the multiplication sign.
For example: 8n means 8 × n and 12ab means 12 × a × b. 2. The division sign is rarely used.
For example, y ÷ 6 is shown as .
When substituting pronumerals, replace the multiplication signs, as shown in theworked example below.
The same methods are used when substituting into a formula or rule.
y6---
Find the value of the following expressions if a = 3 and b = 15.
a 6a b
THINK WRITE
a Substitute the pronumeral (a) with its correct value and replace the multiplication sign.
a 6a= 6 × 3
Multiply. = 18
b Substitute each pronumeral with its correct value and replace the multiplication signs.
b
=
Do the first multiplication. =
Do the next multiplication. =
Do the division. = 21 − 10Do the subtraction. = 11
C h a p t e r 4 A l g e b r a 1253 Evaluate the following expressions, if d = 5 and m = 2.
a d + m b m + d c m − d d d − m
e 2m f md g 5dm h
i −3d j −2m k 6m + 5d l
m 25m − 2d n o 4dm − 21 p
4 The formula for finding theperimeter (P) of a rectangleof length l and width w isP = 2l + 2w. Use this for-mula to find the perimeterof the rectangular swim-ming pool at right.
5 The formula F = 2c + 30 is used to convert temperatures measured in degrees Celsiusto an approximate Fahrenheit value. F represents the temperature in degrees Fahrenheitand c the temperature in degrees Celsius.a Find F when c = 100.b Convert 28° Celsius to Fahrenheit.c Water freezes at 0° Celsius. What is the freezing temperature of water in Fahrenheit?
6 The formula for the perimeter (P) of a square of side length l is P = 4l.Use this formula to find the perimeter of a square of length 2.5 cm.
7 The formula C = 0.1a + 42 is used to calculate the cost in dollars (C) of renting a carfor one day from Poole’s Car Hire Ltd, where a is the number of kilometres travelledon that day. Find the cost of renting a car for one day if the distance travelled is220 kilometres.
8 Distances in the USA and Canada are often expressed in both miles and kilometres.The formula D = 0.6T can be used to convert distances in kilometres (T) to the approxi-mate equivalent in miles (D). Use this rule to convert the following distances to miles:a 100 kilometresb 248 kilometresc 12.5 kilometres.
9 The area (A) of a rectangle of length l and width w can be found using the formulaA = lw. Find the area of the rectangles below:a length 12 cm, width 4 cmb length 200 m, width 42 mc length 4.3 m, width 104 cm.
Working with bracketsBrackets are ‘grouping’ symbols. For example, the expression 3(a + 5) can be thoughtof as ‘three groups of (a + 5)’, or (a + 5) + (a + 5) + (a + 5).
When substituting into an expression with brackets, remember to place a multipli-cation (×) sign next to the brackets.For example, 3(t + 2) means 3 × (t + 2)6(h − 4) means 6 × (h − 4)g(2 + 3k) means g × (2 + 3k)(3 + 2k) 4 means (3 + 2k) × 4(x + y) (6 − 2p) means (x + y) × (6 − 2p).
We evaluate expressions inside a bracket first, then multiply by the value outside the bracket.
a Substitute r = 4 and s = 5 into the expression 5(s + r) and evaluate.b Substitute t = 4, x = 3 and y = 5 into the expression 2x(3t − y) and evaluate.
THINK WRITE
a Put the multiplication sign back into the expression.
a 5(s + r)= 5 × (s + r)
Substitute the pronumerals with their correct values.
= 5 × (5 + 4)
Work out the bracket first. = 5 × 9Complete the multiplication. = 45
b Put the multiplication signs back into the expression.
b 2x(3t − y)= 2 × x × (3 × t − y)
Substitute the pronumerals with their correct values.
= 2 × 3 × (3 × 4 − 5)
Do the multiplication inside the brackets.
= 2 × 3 × (12 − 5)
Do the subtraction inside the brackets.
= 2 × 3 × 7
Do the final multiplication. = 42
1
2
3
4
1
2
3
4
5
4WORKEDExample
remember1. Brackets are ‘grouping’ symbols.2. When substituting into an expression with brackets, remember to place a
multiplication (×) sign next to the brackets.3. Work out the brackets first.
1 Substitute r = 5 and s = 7 into the following expressions and evaluate.a 3(r + s) b 2(s − r) c 7(r + s) d 9(s − r)e s(r + 3) f s(2r − 5) g 3r(r + 1) h rs(3 + s)i 11r(s − 6) j 2r(s − r) k s(4 + 3r) l 7s(r − 2)m s(3rs + 7) n 5r(24 − 2s) o 5sr(sr + 3s) p 8r(12 − s)
2 Evaluate each of the expressions below, if x = 3, y = 5 and z = 9.
a xy(z − 3) b c
d (x + y) (z − y) e (z − 3)4x f zy(17 − xy)
g h (8 − y) (z + x) i
j k l 2x(xyz − 105)
m 12(y − 1) (z + 3) n
3 The formula for the perimeter (P) of a rectangleof length l and width w is P = 2l + 2w. Thisrule can also be written as P = 2(l + w). Usethe rule to find the perimeter of rectangularcomic covers with the following measure-ments.a l = 20 cm, w = 11 cmb l = 27.5 cm, w = 21.4 cm
4 A rule for finding the sum of the interiorangles in a many-sided figure such as apentagon is S = 180(n − 2) where S represents thesum of the angles inside the figure and n representsthe number of sides. The diagram at right shows theinterior angles in a pentagon.
Use the rule to find the sum of the interior angles forthe following figures:a a hexagon (6 sides) b a pentagonc a triangle d a quadrilateral (4 sides)e a 20-sided figure.
1 Substitute m = 6 and n = −3 into the following expressions and evaluate.
a m + n b m − n c n − m d n + me 3n f −2m g 2n − m h n + 5
i 2m + n − 4 j 11n + 20 k −5n − m l
m n o p
q r 6mn − 1 s t
2 Substitute x = 8 and y = −3 into the following expressions and evaluate.
a 3(x − 2) b x(7 + y) c 5y(x − 7)d 2(3 − y) e (y + 5)x f xy(7 − x)
g (3 + x) (5 + y) h 5(7 − xy) i
j k l
3 Substitute a = −4 and b = −5 into the following expressions and evaluate.
a a + b b a − b c b − 2a d 2abe 12 − ab f −2(b − a) g a − b − 4 h 3a(b + 4)
i j k l
m 45 + 4ab n 8ab − 3b o p 2.5b
q 11a + 6b r (a − 5)(8 − b) s (9 − a)(b − 3) t 1.5b + 2a
rememberWhen substituting, if the pronumeral you are replacing has a negative value, simply remember the rules for directed numbers:1. For addition and subtraction, signs that occur together can be combined.
Same signs positive for example, 7 + +3 = 7 + 3 and 7 − −3 = 7 + 3
Different signs negative for example, 7 − +3 = 7 − 3 and 7 + −3 = 7 − 32. For multiplication and division.
Same signs positive for example, +7 × +3 = +21 and −7 × −3 = +21
Different signs negative for example, +7 × −3 = −21 and −7 × +3 = −21
1 If a kilogram of oranges cost $0.89 and a kilogram of carrots cost $0.99, what isthe cost of p kg of oranges and q kg of carrots.
2 If d represents a certain number, write an expression for the number formed whend is divided by 5.
3 True or false? If y = 4 and z = 1 then .
4 The area of a circle is p × r2 where p = 3.14 and r = radius of the circle. Find thearea of the circle when r = 0.5 cm.
5 If p = 1, what is the value of q, when pq(5p − 2) = 9?
6 Evaluate if r = 4 and s = 6.
7When m = 7 and n = 4 are substituted into the expression , the value is:
8 Substitute p = 7 and q = −2 into .
9 From the list −2, 1, 3, 4 choose the value of a and b when .
10 Substitute x = −3 and y = −5 into the expression and evaluate.
‘Rules of thumb’A ‘rule of thumb’ is a rule or pattern which people use to estimate things. They obtain this rule by observing a pattern.
1 Write an algebraic expression for each of the following ‘rules of thumb’. Explain what each pronumeral represents in your expressions.a Your adult height will be twice your height when you were 2.b To estimate the number of kilometres you are from a thunderstorm,
count the number of seconds between the lightning and the thunder and divide by 3.
c To convert temperature in degrees Celsius to degrees Fahrenheit, double it and add 30.
2 Write a question that could be solved for each of the algebraic expressions found and clearly show how you would solve it.
3 How would you go about verifying the accuracy of these ‘rules of thumb’?
4 If the accurate expression for converting temperature in degrees Celsius (C) to
degrees Fahrenheit (F) is F = C + 32, investigate at which temperatures the
‘rule of thumb’ expression gives the best results.
Simplifying expressionsExpressions can often be written in a more simple form.
For example, the expression 3x + 4x can be written more simply as 7x.Notice that the expression was simplified (put into a more simple form) even though
we did not know the value of the pronumeral (x).When simplifying expressions, we can collect (add or subtract) only like terms.
Like terms are terms that contain the same pronumeral parts.
For example:3x and 4x are like terms. 3x and 3y are not like terms.3ab and 7ab are like terms. 7ab and 8a are not like terms.2bc and 4cb are like terms. 8a and 3a2 are not like terms.3g2 and 45g2 are like terms.
Simplify the following expressions.a 3a + 5ab 7ab − 3a − 4abc 2c − 6 + 4c + 15
THINK WRITE
a Write down the expression and check that the pronumeral parts of the 2 terms are the same. They are.
a 3a + 5a
Add the 2 terms. = 8a
b Write down the expression. b 7ab − 3a − 4ab
Rearrange the terms so that the like terms are together. Remember to keep the correct sign in front of each term.
= 7ab − 4ab − 3a
Simplify by subtracting the like terms.
= 3ab − 3a
c Write down the expression. c 2c − 6 + 4c + 15
Rearrange the terms so that the like terms are together. Remember to keep the correct sign in front of each term.
4 Simplify the following expressions.a 4c + 2c b 2c − 5c c 3a + 5a − 4ad 6q − 5q e −h − 2h f 7x − 5xg 3a − 7a − 2a h −3f + 7f i 4p − 7pj −3h + 4h k 11b + 2b + 5b l 7t − 8t + 4tm 9m + 5m − m n x − 2x o 7z + 13zp 5p + 3p + 2p q 9g + 12g − 4g r 18b − 4b − 11bs 13t − 4t + 5t t −11j + 4j u −12l + 2l − 5lv 13m − 2m − 4m + m w m + 3m − 4m x t + 2t − t + 8t
5 Simplify the following expressions.a 3x + 7x − 2y b 3x + 4x − 12c 11 + 5f − 7f d 3u − 4u + 6e 2m + 3p + 5m f −3h + 4r − 2hg 11a − 5b + 6a h 9t − 7 + 5i 12 − 3g + 5 j 6m + 4m − 3n + nk 5k − 5 + 2k − 7 l 3n − 4 + n − 5m 2b − 6 − 4b + 18 n 11 − 12h + 9o 12y − 3y – 7g + 5g − 6 p 8h − 6 + 3h − 2q 11s − 6t + 4t − 7s r 2m + 13l − 7m + ls 3h + 4k − 16h − k + 7 t 13 + 5t − 9t − 8u 2g + 5 + 5g − 7 v 17f − 3k + 2f − 7k
A 12 B 12a C 6a D 12a2 E The expression cannot be simplified.
A 4 B 4x2 C 4x D 2x E The expression cannot be simplified.
A 12ab B 6ab C 36ab D 12a E The expression cannot be simplified.
remember1. When simplifying expressions, we can collect (add or subtract) only like terms.2. Like terms are terms that contain the same pronumeral parts.
1 Simplify the following.a 4 × 3g b 7 × 3h c 4d × 6d 3z × 5 e 6 × 5r f 5t × 7g 4 × 3u h 7 × 6p i 7gy × 3j 2 × 11ht k 4x × 6g l 10a × 7h m 9m × 4d n 3c × 5h o 9g × 2xp 2.5t × 5b q 13m × 12n r 6a × 12ds 2ab × 3c t 4f × 3gh u 2 × 8w × 3xv 11ab × 3d × 7 w 16xy × 1.5 x 3.5x × 3y y 11q × 4s × 3 z 4a × 3b × 2c
2 Simplify the following.a 3 × −5f b −6 × −2dc 11a × −3g d −9t × −3ge −5t × −4dh f 6 × −3stg −3 × −2w × 7d h −4a × −3b × 2c × ei 11ab × −3f j 3as × −3b × −2xk −5h × −5t × −3q l 4 × −3w × −2 × 6pm −7a × 3b × g n 17ab × −3gho −3.5g × 2h × 7 p 5h × 8j × −kq 75x × 1.5y r 12rt × −3z × 4ps 2ab × 3c × 5 t −4w × 34x × 3
Sonar measurementsAt the start of the chapter, we introduced the situation where a sonar pulse took 1.5 s to travel from the ship to the ocean floor and back again. (The speed of sound in water is assumed to be 1470 m/s.) Let us look at this problem again.1 Draw a diagram to show this situation.2 How far does the sonar pulse travel in:
a 1 second? b 2 seconds? c 1.5 seconds?
3 Calculate the ocean depth when the pulse took 1.5 seconds to return.4 Write a rule to find the ocean depth for any time measurement. Explain what
each pronumeral represents. 5 Use the rule found in part 4 to calculate the ocean depth for the following
pulse-return time measurements.a 1.8 secondsb 4.22 secondsc 0.64 seconds
6 The speed of sound in water is about 5 times the speed of sound in air. A person standing on the deck of the ship sends a sonar pulse through the air to a nearby cliff face. If the pulse takes 3 seconds to travel to the cliff face and return, calculate the distance to the cliff face. Write a rule to represent this situation.
History of mathematicsT H E R H I N D PA P Y R U S ( c . 1 8 5 0 B C )
The ancient Egyptians differed from the ancient Greeks in that Egyptians thought about mathematics in a practical rather than an abstract way. They didn’t like fractions which had numerators other than one (except the fraction two-thirds for reasons still unknown). They found that fractions with numerators of one, unit fractions, were easy to multiply, since the numerator would
always be one: for example × = .
The Egyptians developed ingenious methods to avoid using any fraction other than those with a numerator of one. Solutions to many Egyptian problems concerned with beer and bread were recorded on papyri. The most famous of these is the Rhind papyrus, which contains 84 problems and their solutions including the calculation of the ancient Egyptian value for pi (π) of 3.1605. A part of the papyrus is shown in the photograph above.
The Rhind papyrus was named after the Scottish Egyptologist, A. Henry Rhind, who bought the 6 m scroll in 1858. A scribe named Ahmes is believed to have copied it in around 1650 BC from a document originally written about 200 years before that. This papyrus shows a method for multiplying numbers using only addition and subtraction. Also known as the aha papyrus: aha meaning unknown quantity to be determined, an early pronumeral, it is now in the British Museum in London.
Questions1. Which numerator did the Egyptians use
in their calculations with fractions?2. Which fraction was an exception to this
rule?3. What practical problems did most of the
solutions deal with?
ResearchHow was Egyptian multiplication done with only addition and subtraction?
12--- 1
3--- 1
6---
During this time . . .
The Sumerians built the first cities, invented writing and made wheels from date palm trunks.
Papyrus reeds were used to make boats, baskets and paper.
Dividing pronumeralsWhen dividing pronumerals, rewrite the expression as a fraction and simplify bycancelling.
Remember that when the same pronumeral appears on both the top and bottom linesof the fraction, it may be cancelled. Follow the worked examples given below.
a Simplify .
b Simplify 15n ÷ 3n.
THINK WRITE
a Write down the expression. a
Simplify the fraction by cancelling 16 with 4 (divide both by 4).
=
No need to write the denominator since we are dividing by 1.
= 4f
b Write down the expression and then rewrite it as a fraction.
b 15n ÷ 3n
=
Simplify the fraction by cancelling 15 with 3 and n with n.
=
No need to write the denominator since we are dividing by 1.
= 5
16 f4
----------
116 f
4---------
24 f1
------
3
1
15n3n
---------
251---
3
8WORKEDExample
Simplify −12xy ÷ 27y.
THINK WRITE
Write down the expression and then rewrite it as a fraction.
−12xy ÷ 27y
=
Simplify the fraction by cancelling 12 with 27 (divide both by 3) and y with y.
1 If Betty is now x years old, how old was Betty 6 years ago?
2 Find the area of a rectangle with length of 225 cm and width of 1.3 m.
3 Evaluate if p = 4, q = 2 and r = 7.
4If m = −6 and n = −3 are substituted into the expression , it would have
a value of:
A −2 B −3 C −4 D −5 E −6
5 Simplify 11x − 8y − 9x + 4y − 3.
6 Simplify 10z2 − 5y − 3z2 + 4y + 4.
7 True or false? −6p × −4q × r × 2t = 48pqrt
8 Simplify .
9 Find the missing term from the list −2, −4, 12pq, −48pq to replace ∇ in
.
10 Simplify .
Expanding bracketsWe have seen that the expression 3(a + 5) means 3 × (a + 5) or (a + 5) + (a + 5) +(a + 5). Simplifying this expression further gives us the expression 3a + 15:
(a + 5) + (a + 5) + (a + 5) = a + a + a + 5 + 5 + 5= 3a + 15
Look at the pattern below:With brackets Expanded form1. 3 × (2 + 1) 3 × 2 + 3 × 1
Removing brackets from an expression is called expanding the expression. The rulethat we have used to expand the expressions above is called the Distributive Law.
1 Use the Distributive Law to expand the following expressions.a 3(d + 4) b 2(a + 5) c 4(x + 2)d 5(r + 7) e 6(g + 6) f 2(t + 3)g 7(d + 8) h 9(2x + 6) i 12(4 + c)j 7(6 + 3x) k 45(2g + 3) l 1.5(t + 6)m 11(t − 2) n 3(2t − 6) o t(t + 3)p x(x + 4) q g(g + 7) r 2g(g + 5)s 3f(g + 3) t 6m(n − 2m)
2 Expand the following.a 3(3x − 2) b 3x(x − 6y) c 5y(3x − 9y)d 50(2y − 5) e −3(c + 3) f −5(3x + 4)g −5x(x + 6) h −2y(6 + y) i −6(t − 3)j −4f(5 − 2f) k 9x(3y − 2) l −3h(2b − 6h)m 4a(5b + 3c) n −3a(2g − 7a) o 5a(3b + 6c)p −2w(9w − 5z) q 12m(4m + 10) r −3k(−2k + 5)
Use the Distributive Law to expand the following expressions.a 3(a + 2) b x(x − 5)
THINK WRITE
a Write down the expression and replace the hidden multiplication sign.
a 3(a + 2)= 3 × (a + 2)
Use the Distributive Law to expand the brackets.
= 3 × a + 3 × 2
Simplify by multiplying. = 3a + 6
b Repeat the steps in part a. b x(x − 5)= x × (x − 5)= x × x + x × −5= x2 − 5x
1
2
3
10WORKEDExample
remember1. Brackets are grouping symbols2. Removing brackets from an expression is called expanding the expression. 3. When expanding brackets, put the × sign before the bracket.4. The rule that is used to expand brackets is called the Distributive Law.
During his time . . .Space travel — men walk on the moon.The Cold War ends.Ecological awareness grows.Miniaturisation of computers.
John Coates, a world-renowned Australian mathematician, was born in 1945. He attended Taree High School and studied for his Bachelor of Science at the Australian National University (ANU). After further studies in Paris, he completed a PhD at the University of Cambridge in England, where he later lectured.
He taught mathematics at Harvard and Stanford, both very prestigious universities in the United States. Later he held positions as a professor at the ANU and two institutes in France. In 1986 he returned to Cambridge as Sadleirian Professor and was appointed Head of Department.
He still works at Cambridge in arithmetical algebraic geometry and his research interests include elliptic curves, the Iwasawa theory, Fermat’s Last Theorem and explicit reciprocity laws! As well as this, his work includes the algebraic approximation of functions.
Coates is not just a brilliant mathematician and outstanding researcher, he is also praised for being a great teacher who has inspired many students to pursue careers in mathematical research. He is also known for his valuable contributions as an editor of one of the best known journals in research mathematics, Inventiones Mathematicae.
During his international career he has also received numerous awards, including election as a fellow of the Royal Society of London in 1985 and the Senior Whitehead Prize from the London Mathematical Society in 1997.
Questions1. What country did John Coates grow
up in?2. Reciprocity is about expressions
involve reciprocals. What are reciprocals?
3. What three career areas does John Coates work in?
Algebraic terms can also be broken down into factors. For example, the factors of 3xare 3 and x. The expression, 6m, can be broken down into factors as shown below:
6m = 6 × m= 3 × 2 × m
Here are some other examples:8x = 8 × x
= 4 × 2 × x = 2 × 2 × 2 × x
3ab = 3 × a × b6a2b = 6 × a × a × b
= 3 × 2 × a × a × b
To find the highest common factor, HCF, of algebraic terms follow these steps.1. Find the highest common factor of the number parts.2. Find the highest common factor of the pronumeral parts.3. Multiply these together.
Find the highest common factor (HCF) of 6x and 10.
THINK WRITE
Find the highest common factor of the number parts.Break 6 down into factors.Break 10 down into factors.The highest common factor is 2.
6 = 3 × 210 = 5 × 2
HCF = 2Find the highest common factor of the pronumeral parts.There isn’t one, because only the first term has a pronumeral part! The HCF of 6x and 10 is 2.
1
2
12WORKEDExample
Find the highest common factor (HCF) of 14fg and 21gh.
THINK WRITE
Find the highest common factor of the number parts.Break 14 down into factors.Break 21 down into factors.The highest common factor is 7.
14 = 7 × 221 = 7 × 3HCF = 7
Find the highest common factor of the pronumeral parts.Break fg down into factors.Break gh down into factors.Both contain a factor of g.
fg = f × ggh = g × h
HCF = gMultiply these together. The HCF of 14fg and 21gh is 7g.
C h a p t e r 4 A l g e b r a 145To factorise an expression we place the highest common factor of the terms outside the brackets, and the remaining factors for each term inside the brackets.
Factorise the expression 2x + 6.
THINK WRITE
Break down each term into factors. 2x + 6= 2 × x + 2 × 3
Write the common factor outside the brackets and the other factors inside the brackets.
= 2 × (x + 3)
Remove the multiplication sign. = 2(x + 3)
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14WORKEDExample
Factorise 12gh − 8g.
THINK WRITE
Break down each term into its factors. 12gh − 8g= 4 × 3 × g × h − 4 × 2 × g
Write the highest common factor outside the brackets.Write the other factors inside the brackets.
= 4 × g × (3 × h − 2)
Remove the multiplication signs. = 4g(3h − 2)
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15WORKEDExample
remember1. Factorising is the opposite process to expanding.2. Factorising a number or expression involves breaking it down into smaller
factors.3. To find the highest common factor, HCF, of algebraic terms, follow these steps.
(a) Find the highest common factor of the number parts.(b) Find the highest common factor of the pronumeral parts.(c) Multiply these together.
4. To factorise an expression we place the highest common factor of the terms outside the brackets, and the remaining factors for each term inside the brackets.
1a The highest common factor (HCF) of 12 and 16 is:
b The highest common factor (HCF) of 10 and 18 is:
c The highest common factor (HCF) of 4 and 16 is:
d The highest common factor (HCF) of 2x and 8xy is:
e The highest common factor (HCF) of 4f and 12fg is:
2 Find the highest common factor (HCF) of the following.a 4 and 6 b 6 and 9 c 12 and 18 d 13 and 26e 14 and 21 f 2x and 4 g 3x and 9 h 12a and 16
3 Find the highest common factor (HCF) of the following.
a 2gh and 6g b 3mn and 6mp c 11a and 22bd 4ma and 6m e 12ab and 14ac f 24 fg and 36ghg 20dg and 18ghq h 11gl and 33lp i 16mnp and 20mnj 28bc and 12c k 4c and 12cd l x and 3xz
4 Factorise the following expressions.a 3x + 6 b 2y + 4 c 5g + 10d 8x + 12 e 6f + 9 f 12c + 20g 2d + 8 h 2x − 4 i 12g − 18j 11h + 121 k 4s − 16 l 8x − 20m 12g − 24 n 14 − 4b o 16a + 64p 48 − 12q q 16 + 8f r 12 − 12d
5 Factorise the following.a 3gh + 12 b 2xy + 6yc 12pq + 4p d 14g − 7ghe 16jk − 2k f 12eg + 2gg 12k + 16 h 7mn + 6mi 14ab + 7b j 5a − 15abck 8r + 14rt l 24mab + 12abm 4b − 6ab n 12fg − 16gho ab − 2bc p 14x − 21xyq 11jk + 3k r 3p + 27pqs 12ac − 4c + 3dc t 4g + 8gh − 16u 28s + 14st v 15uv + 27vw
The fThe factorised form of the eactorised form of the exprxpressions andessions andthe letter beside each givthe letter beside each gives the puzzle codees the puzzle code..
D = 6x – 3 =
E = 8 – 2x =
G = 15x + 10 =
H = –2x + 1 =
I = 12x + 20 =
L = 2xy – 3y =
N = 18 – 42x =
O = –2x – 2 =
P = 2xy – 8y =
S = –12x – 21 =
T = xy – 2x =
V = 56x – 35 =
E = 6xy + 3x =
L =
N =
E =
O =
P =
E =
S =
T =
O =
–20x – 25 =
8xy – 72y =
–x + 2 =
6x – 14 =
–xy + y =
3x – 12 =
–8x – 24 =
3xy – 2y =
6xy – 10y =
–49 – 28x =
H =
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1 Using x and y to represent numbers, write expressions for:a the sum of x and yb the difference between y and xc five times y subtracted from three times xd the product of 5 and xe twice the product of x and yf the sum of 6x and 7yg y multiplied by itself.
2 If tickets to the school play cost $15 for adults and $9 for children, write an expression for the cost of:a x adult ticketsb y child ticketsc k adult tickets and
m child tickets.
3 Find the value of the following expressions, if a = 2 and b = 6.a 2a b 6a
c 5b d
e a + 8 f b − 2g a + b h b − a
i j 3a + 7
k 2a + 3b l
4 The formula C = 2.2k + 4 can be used to calculate the cost in dollars, C, of travelling by taxi for a distance of k kilometres. Find the cost of travelling 4.5 km by taxi.
5 Substitute r = 3 and s = 5 into the following expressions and evaluate.a 2(r + s) b 2(s − r)c 5(r + s) d 8(s − r)e s(r + 4) f s(2r − 3)g 2r(r + 1) h rs(7 + s)
6 Find the value of the following expressions, if a = 2 and b = −5.a a + b b b + a
c ab d
e 2ab f 5 − ag 12 − ab h a2 − 2i 3(a + 2) j b(a − 4)k 12 − a(b − 3) l 5a + 6b
7 Simplify the following by collecting like terms.a 4d + 3d b 3c − 5cc 3d + 5a − 4a d 6g − 4ge 4x + 11 − 2x f 2g + 5 − g − 6g 2xy + 7xy h 12t2 + 3t + 3t2 − t
8 Simplify the following.a 3 × 7g b 6 × 3yc 7d × 6 d −3z × 8
9 Simplify the following.
a b
c 6rt ÷ −2t d −3gh ÷ −6g
e f −36xy ÷ −12y
g h
10 Use the Distributive Law to expand the following expressions.a 2(x + 3) b 5(2x − 1)c −2(f + 7) d 3m(b − m)e −3y(7 − y) f 9b(c − 2)
11 Expand the following and then simplify by collecting like terms.a 3(4v + 5) − 15 b 6t + 5(2t − 7)c 23 + 5(3p − 4) + 2p d 2(x + 5) + 5(x + 1)e 2g(g − 6) + 3g(g − 7) f 3(3t − 4) − 6(2t − 9)
12 Factorise the following expressions.a 3g + 12 b xy + 5yc 5n − 20 d 12mn + 4pne 12g − 6gh f 12xy − 36yz