Download Factorise different types of algebraic expressions

Factorise different types of algebraic
expressions
Factorisation
3
Introduction
3
Factors
3
Common factors
4
Algebraic factors and products
6
More on common factors
7
Factorising an expression
8
Factorising quadratics
16
Finding the factors
19
Factorising a special quadratic: the difference of two squares
28
Algebraic fractions
Reducing a fraction
39
More fractions
41
How to add and subtract algebraic fractions
52
Changing to a common denominator
62
Multiplying algebraic fractions
68
Dividing algebraic fractions
72
Introduction to equations
83
Solving equations
84
Introduction
84
What is an equation?
85
Finding the solution of quadratic equations
107
Multiplying by nought
108
Factorising quadratic expressions
113
More on solving equations
123
Simultaneous equations
123
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Algebraic formulas
141
Introduction
141
Algebraic formulas in real life
142
Some examples
142
Evaluating formulas
144
Transformation of formulas
152
Tips for changing the subject of formulas
2
154
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Factorisation
Introduction
So far in this module you have learned how to simplify and expand
algebraic expressions.
In this third section of the module you will learn to factorise different types
of algebraic expressions.
You will be introduced to some important ideas about factors, such as
common factors, highest common factor (HCF) and factorising quadratics.
Factors
You remember from previous modules that the factors of a number are all
the whole numbers which can be multiplied to equal our original number.
Example
eg
12 = 1  12
12 = 2  6
12 = 3  4
So the factors of 12 are 1, 2, 3, 4, 6, 12.
Try these
Write out the factors of
(a) 15
did you get 1, 3, 5, 15
Good!
(b) 18
hopefully you got 1, 2, 3, 6, 9, 18
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Common factors
Sometimes we need to know which factors are common to two or more
numbers.
For small numbers, common factors are often easy to see.
Example
What is the common factor of 6 and 10?
Note: The number 1 is a common factor of every number.
Writing out the factors of each of these numbers:
The factors of 6 are 1, 2, 3, 6
The factors of 10 are 1, 2, 5, 10
We can easily see that only 1 and 2 are in both lists.
1 and 2 are common factors of 6 and 10
Example
What are the common factors of 24 and 36?
Write out the factors of each number.
Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
The common factors in both lists are 1, 2, 3, 4, 6, 12.
Highest common factor
The highest or largest factor that is common to a series of numbers is called
the highest common factor (HCF).
In the example above, the largest factor that is common to both 24 and 36 is
12. So the number 12 is the highest common factor of 24 and 36.
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Example
What are the common factors of 45 and 60?
What would be the highest common factor?
The factors of 45 are 1, 3, 5, 9, 15, 45.
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.
The common factors are 1, 3, 5, 15.
The highest common factor (HCF) is 15.
Activity 1
1
Find the common factors of each of these pairs of numbers. Then work out the highest
common factor (HCF) of each pair.
(a) 6 and 12
(b) 15 and 21
(c) 14 and 42
(d) 13 and 52
(e) 70 and 28
(f) 16 and 64
2
Can you find the highest common factor of these sets of three numbers?
(a) 16, 20 and 32
(b) 18, 36 and 72
(c) 48, 64 and 100
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Answers to Activity 1
Since 1 is always a factor, it has not been included.
1
(a)
(b)
(c)
(d)
(e)
(f)
The common factors of 6 and 12 are 2, 3 and 6. The HCF is 6.
The common factor of 15 and 21 is 3. The HCF is 3.
The common factors of 14 and 42 are 2, 7 and 14. The HCF is 14.
The common factor of 13 and 52 is 13. The HCF is 13.
The common factors of 70 and 28 are 2, 7 and 14. The HCF is 14.
The common factors of 16 and 64 are 2, 4, 8 and 16. The HCF
is 16.
2
(a) The HCF of 16, 20 and 32 is 4.
(b) The HCF of 18, 36 and 72 is 18.
(c) The HCF of 48, 64 and 100 is 4.
Algebraic factors and products
Just as in arithmetic, one of the processes that you often use in algebra is
multiplication. Algebraic multiplication is a quick way of adding together,
say, the same pronumeral a number of times.
For instance, a + a + a + a + a + a + a can be thought of as 7 lots of a and
written simply as 7a. Here 7 and a are factors and 7a is the product.
Similarly this algebraic expression:
(y – 8)  (y – 8)  (y – 8)  (y – 8)  (y – 8)  (y – 8)  (y – 8)  (y – 8)  (y – 8)
can be thought of as 9 lots of (y – 8) and written simply as 9(y – 8) or 9y – 72.
Here 9 and (y – 8) are factors and 9y – 72 is the product.
In the previous section we investigated finding the product of pairs of
factors like these:

x and 6x

4a and (2a –9)

(z – 3) and (z + 6)
Do you remember how to do this? Try this activity to check.
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Activity 2
1
Find the product of each of these three pairs of factors:
(a) x and 6x
(b) 4a and (2a – 9)
(c) (z – 3) and (z + 6)
2
Show that these statements are correct:
(a) the product of the two factors (c + 4) and (c + 2) is c2 + 6c + 8
(b) the product of the two factors (a – 6) and (a + 7) is a2 + a – 42
(c) the product of (x – 4) and (x – 4) is the perfect square x2 – 8x + 16
Answers to Activity 2
1
(a) The product of x and 6x is 6x2
(b) The product of 4a and (2a – 9) is 8a2 – 36a
(c) The product of (z – 3) and (z + 6) is z2 + 3z – 18
More on common factors
We can write the factors of an algebra expression such as 7x by writing all
the factors of 7, and also the x.
Factors of 7x are 1, 7, x
Factors of 10a are 1, 2, 5, 10, a
We can write common factors just as we did before.
For example; find the highest common factor (HCF) of 12x and 15.
Factors of 12x are 1, 2, 3, 4, 6, 12, x
Factors of 15 are 1, 3, 5, 15
HCF is 3
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Factorising an expression
We know how to multiply a binomial factor
3(6 –2x) = 18 –6x
eg
We are now going to see how to reverse this process.
Going from 18 –6x (two terms subtracted)
to 3(6 –2x) (two factors multiplied)
is called factorisation.
For example, factorise 21 – 6x.
We look at both terms, 21 and –6x and find the HCF.
The HCF of 21 and –6x is 3
This gives the number to take outside the bracket
3(? – ?) = 21 –6x
We ask, what must I multiply 3 by to get 21? Well 3  7 = 21
3(7 –?) = 21 –6x
What must I multiply 3 by to get –6x? Well 3  –2x gives –6x
So
3(7 –2x) = 21 –6x
Another example: factorise
15x + 20
Look for the HCF of 15x and 20 or simply ask which is the biggest number
which will divide into both 15 and 20.
ie
5
5(? + ?) = 15x + 20
5 times 3x gives 15x
5(3x + ?) = 15x + 20
5 times 4 gives 20
so
8
5(3x + 4) = 15x + 20
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Example
Factorise
18x –24
The biggest number which divides into both 18 and 24 is 6.
6  3x = 18x
6  –4 = –24
so
18x –24 = 6(3x –4)
Try these!
Factorise
(a) 14x + 21
(b) 4 –18m
(c) 6 –30y
Check the answers
(a) 7(2x + 3)
(b) 2(2 –9m)
(c) 6(1 –5y)
Did you get them all correct? Remember, in the last one, the biggest number
which divides into both 6 and 30 is 6 itself!
We have been taking out a common factor of both numbers. Sometimes this
common factor may be a pronumeral.
eg
factorise
5x + 3xy
We can write out the factors of both 5x, and 3xy and the only common factor
is x.
Or we can simply notice that x is common to both terms.
5x + 3xy = x(? + ?)
What must we multiply by x to get 5x? Just 5
What must we multiply by x to get 3xy? Just 3y
5x + 3xy = x(5 + 3y)
You’ll notice the x has just come out the front of the bracket.
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You should always check your answer by multiplying your answers.
Example
ax + 5a
The common pronumeral is a
a(x + 5)
Try these!
(a) 6x –bx
(b) 10m + mn
(c) 3a –4ab
did you get
(a) x(6 –b)
(b) m(10 + n)
(c) a(3 –4b)?
Did you check by multiplication?
Good.
Example
3a –4a2
The common pronumeral is a
3a –4a2 = a(3 –?)
What do we have to multiply a by to get 4a2. Well a  4a gives 4a2
so
3a –4a2 = a(3 –4a)
Try these.
Factorise
(a) 3x –8x2
(b) 4am + 7a2
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and check by multiplication
Did you get
(a) x(3 –8x)
(b) a(4m + 7a)
Good!
Sometimes we may have a power of a pronumeral as the HCF.
eg factorise
2x2 –x3
The common factor will be the lowest of the two powers; ie x2
2x2 –x3 = x2(2 –?)
What do we multiply x2 by to get x3:
Well x2 × x gives x3
so
2x2 –x3 = x2(2 –x)
Example
4m2 + 3am3 = m2(4 + 3am)
Example
Factorise 7m3 – 2m2
7m3 = 7  m  m  m
We have:
2m2 =2  m  m
We can therefore take m2 out as a common factor.
Therefore, 7m3 – 2m2 = m2(7m – 2).
Check:
m2(7m – 2) = 7m3 – 2m2
Correct!
Example
Factorise ab2 + a2b
Here we have:
ab2 = a  b  b
a2b = a  a  b
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Here a  b or ab is common!
Therefore, ab2 + a2b = ab(b + a)
Check:
ab(a + b) = ab2 + a2b
Correct!
Example
Factorise 3p + mp + np
Here p is the common factor and, therefore, we can write
3p + mp + np = p(3 + m + n)
Check:
p(3 + m + n) = 3p + mp + np
Correct!
Note: Often you will be able to factorise an expression just by looking at it.
Example
16x –24ax
Find the largest number which will divide into both 16 and 24, and also then
look for a common pronumeral.
16x –24ax = 8x(2 –3a)
Try these and check each result
factorise
(a) 25x + 30mx
(b) 8x2 –4x
(c) 6x3 + 9x2
Did you get
(a) 5x(5 + 6m)
(b) 4x(2x –1)
(c) 3x2(2x + 3)
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Activity 3
1
Factorise each of these algebraic expressions. Remember to check your answers.
(a) 6a + 12
(b) 4s + 8
(c) 10p + 50
(d) 7f + 14
(e) 5t + 10
(f) 9 + 18k
(g) 24m + 18
(h) 16 + 32d
(i) 15m + 25
2
Try these! Check each answer as you go.
(a) 4a + at
(b) bn + 6n
(c) 12c + dc
(d) 2e + 4ae
(e) 7z + 7zy
(f) 3w + wz
And now try these:
(g) ab + a2
(h) cd2 + d
(i) p2 + pr
(j) wy2 + 2y
(k) s3t + 2s2
(l) mn + m2n2
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3
Negatives! Just take care. Factorise these. Check each one!
(a) 3s –15
(b) 4 – 16t
(c) 6p – 2
(d) 2gh – 6g
(e) 5p2– 10p
(f) qz – 5q3
(g) 4n2 – 8n3
(h) 16p4– 8p2
(i) 4ab2 – 4b
More difficult examples
4
Factorise these expressions:
(a) ap + aq + ar
(b) ab + ca + ad
(c) 7m + 7n – 14p
(d) 9r – 3t – 12s
(e) 4a + 2ab + 6ac
(f) 15st + 3s – 6sr
(g) 5ay – 10yb – 15yc
(h) 2pq2 – 4p2q –8pq
(i) 3klm + 9klm2 – 6kl2m
Answers to Activity 3
1
(a) 6(a + 2)
(b) 4(s + 2)
(c) 10(p + 5)
(d) 7(f + 2)
(e) 5(t + 2)
(f) 9(1 + 2k)
(g) 6(4m + 3)
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(h) 16(1 + 2d)
2
(i) 5(3m + 5)
(a) a(4 + t)
(b) n(b + 6)
(c) c(12 + d)
(d) 2e(1 + 2a)
(e) 7z(1 + y)
(f) w(3 + z)
(g) a(b + a)
(h) d(cd + 1)
(i) p(p + r)
(j) y(wy + 2)
(k) s2(st + 2)
3
(l) mn(1 + mn)
(a) 3(s – 5)
(b) 4(1 – 4t)
(c) 2(3p – 1)
(d) 2g(h – 3)
(e) 5p(p – 2)
(f) q(z – 5q2)
(g) 4n2(1 – 2n)
(h) 8p2(2p2 – 1)
4
(i) 4b(ab – 1)
(a) a(p + q + r)
(b) a(b + c + d)
(c) 7(m + n – 2p)
(d) 3(3r – t – 4s)
(e) 2a(2 + b + 3c)
(f) 3s(5t + 1 – 2r)
(g) 5y(a – 2b – 3c)
(h) 2pq(q – 2p – 4)
(i) 3klm(1 + 3m – 2l)
I’m sure that you handled at least the first three questions very well.
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Factorising quadratics
In Section 2 we investigated finding the product of two factors like (x + a)
and (x + b). When we multiplied these factors together, we obtained the
following product:
(x + a)(x + b)
= x2 + bx + ax + ab
= x2 + (a + b)x + ab
This says that the number in front of x in the answer is the sum of the
numbers in the brackets and the number at the end of the answer is the
multiplication (product) of the two numbers in the brackets.
eg (a + 3)(a + 4) = a2 + 4a +3a + 12
= a2 + 7a + 12
Note: 7 = 3 + 4
and
12 = 3 × 4
Let’s look at some examples just to illustrate this idea. Study each example
carefully.
Example
(a)
(b) (x + 6)(x – 4)
= x2 – 4x + 6x – 24
= x2 + (–4 + 6)x – 24
= x2 + 2x – 24
(c) (x – 8)(x + 12) = x2 + 12x – 8x – 96
= x2 + (12 – 8)x – 96
= x2 + 4x – 96
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(d) (x – 10)(x + 7) = x2 + 7x – 10x – 70
= x2 + (7 – 10)x – 70
= x2 – 3x – 70
(e) (x – 6)(x – 8)
= x2 – 8x – 6x + 48
= x2 + (–8 – 6)x + 48
= x2 – 14x + 48
Note: Expressions like x2 + 8x + 15, x2 + 2x –24, x2 + 4x – 96, x2 – 3x – 70
and x2 – 14x + 48 (which we found in this example) are called
quadratic expressions.
Notice that in each of these quadratics:
 the coefficient of x is the sum of the two numbers given in the factors

the constant term (on the end) is given by the product of these two
numbers.
You need to be very careful with signs.
This rule gives us a quick way to multiply out brackets.
Example
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Activity 4
1
From the numbers given in the two factors, check for yourself that the coefficient of x
and the constant term are correct. See if you can do the calculations mentally.
(a) (x + 4)(x + 7) = x2 + 11x + 28
(b) (x + 9)(x + 12) = x2 + 21x + 108
(c) (x + 6)(x – 5) = x2 + 1x – 30
= x2 + x – 30
(d) (x – 8)(x + 7) = x2 – 1x – 56
= x2 – x – 56
(e) (x – 11)(x + 4) = x2 – 7x – 44
(f) (x – 7)(x – 7) = x2 – 14x + 49
2
Use mental calculations to expand these expressions:
(a) (x + 3)(x + 1)
(b) (x + 4)(x + 10)
(c) (x – 10)(x + 1)
(d) (x – 6)(x – 5)
(e) (x – 12)(x + 2)
(f) (x + 7)(x + 8)
(g) (x + 6)(x + 9)
(h) (x – 7)(x – 3)
(i) (x + 6)(x + 3)
(j)
(x – 10)(x + 3)
(k) (x + 5)(x + 5)
(l)
18
(x – 6)(x – 6)
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Answers to Activity 4
2
(a) x2 + 4x + 3
(b) x2 + 14x + 40
(c) x2 – 9x – 10
(d) x2 – 11x + 30
(e) x2 – 10x – 24
(f) x2 + 15x + 56
(g) x2 + 15x + 54
(h) x2 – 10x + 21
(i) x2 + 9x + 18
(j) x2 – 7x – 30
(k) x2 + 10x + 25
(l) x2 – 12x + 36
Finding the factors
Example
In finding the product of two factors like (x + 2) and (x – 6), you have just
learned how to quickly calculate the coefficient of x and the constant term
using the numbers given in the two factors.
In the work that follows, we want to investigate reversing that process.
That is, given the product of two factors, how do we find the two factors
that were multiplied together to give that product?
This is called factorising a quadratic. What you have just learned will really
help you to do just that!
Since I’m sure that you have been able to follow the work so far, you should
have no trouble in finding the required factors.
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Let’s investigate this method of factorising quadratics with three examples.
1 Factorise x2 + 4x + 3
The given quadratic is x2 + 4x + 3
Here, the important numbers are 4 and 3
Think: What two numbers add together to give 4 and multiply together to
give 3?
Starting point: Find two numbers that multiply to give 3.
Answer: 3 and 1. Do they add to give 4? Yes!
The required factors are (x + 3) and (x + 1)
That is x2 + 4x + 3 = (x + 3)(x + 1)
2 Factorise a2 + 6a + 8
The given quadratic is a2 + 6a + 8
The important numbers are 6 and 8
Think: What two numbers add together to give 6 and multiply together to
give 8?
Starting point: Find two numbers that multiply to give 8.
Well
81 = 8
24 = 8
Which pair add together to give 6?
Answer: 4 and 2
The required factors are (a + 4) and (a + 2)
That is a2 + 6a + 8 = (a + 4)(a + 2)
Try one!
m2 + 8m + 12
Which two numbers multiply to give 12?
Which pair add to give 8?
Then
(m2 + 8m + 12) = (
)(
)
Did you get (m + 2)(m + 6) – in any order
20
Good!
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Example
x2 – 7x + 12
If there is a minus sign, be careful to include minus numbers in your factors
if 12.
Starting point: What two numbers multiply to give 12
3  4 = 12
12  1 = 12
2  6 = 12
But also!
–3  –4 = 12
–12  –1 = 12
–2  –2 = 12
Which pair add to give –7.
Answer:
–3 and –4
So
(x2 –7x + 12) = (x –3)(x –4)
Try a couple!
(a) Factorise m2 –6m + 8
What two numbers multiply to give 8 _________________
Which pair add to give –6
(m2 –6m + 8) = (m
)(m
)
Answer: (m –2)(m –4)
(b) Factorise x2 –8x +15
What two numbers multiply to give 15 ________________
x2 –8x + 15 = (x
)(x
)
Answer: (x –3)(x –5)
Note: (x –3)(x –5) is exactly the same as (x –5)(x –3)
Harder examples
1
Factorise x2 –x –6
Starting point: What two numbers multiply to give –6
–6  1
–3  2
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But also 6  –1
3  –2
Which pair add to give –1
Answer: –3 and 2
So
2
x2 –x –6 = (x –3)(x + 2)
Factorise p2 + 2p – 8
What two numbers multiply to give –8
–4  2
–1  8
4  –2
1  –8
=
=
=
=
–8
–8
–8
–8
Which pair add to give 2?
Answer: 4 and –2
So
3
(p2 + 2p –8) = (p + 4)(p –2)
Factorise x2 – 3x – 10
What two numbers multiply to give –10 and add to give –3?
Answer: –5 and 2
(x2 –3x –10) = (x –5)(x +2)
4
Factorise p2 – p – 30
Think: What two numbers add together to give –1 and multiply together to
give –30?
Starting point: Find two numbers that multiply to give –30 and add
together to give –1
Answer: –6 and 5
The required factors are (p – 6) and (p + 5)
That is p2 – p – 30 = (p – 6)(p + 5)
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Activity 5
1
Complete these factorisations by filling in the spaces:
(a) Factorise t2 – 9t + 14
Think: What two numbers add together to give –9 and multiply together to
give 14?
Answer: –7 and………
The required factors are (t………) and (t……….)
That is t2 – 9t + 14 = (…………)(…………)
2
(b) Factorise m + 7m – 18
Think: What two numbers add together to give 7 and multiply together to give –
18?
Answer: 9 and………
The required factors are (m………) and (m……….)
That is m2 + 7m – 18 = (…………)(…………)
Answers to Activity 5
(a) Answer: –7 and –2
Required factors are (t – 7) and (t – 2)
That is t2 – 9t + 14 = (t – 7)(t – 2)
(b) Answer: 9 and –2
Required factors are (m + 9) and (m – 2)
That is (m + 9)(m – 2) = m2 + 7m – 18
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Activity 6
1
Quick Number Quiz
Find two numbers such that:
(a) their sum is 8 and their product is 15
(b) their sum is 16 and their product is 15
(c) their sum is 4 and their product is –45
(d) their sum is 8 and their product is 12
(e) their sum is –13 and their product is 40
(f) their sum is 15 and their product is 54
(g) their sum is –11 and their product is 24
(h) their sum is –18 and their product is 72
(i) their sum is –3 and their product is –28
(j) their sum is –7 and their product is –18
And four more for the thinkers
(k) their sum is 19 and their product is 60
(l) their sum is 24 and their product is 108
(m) their sum is –15 and their product is 36
(n) their sum is 13 and their product is –48
Check your answers! Correct any errors.
2
Use the answers from the ‘Quick Number Quiz’ to factorise these quadratics:
(a) x2 + 8x + 15
(b) x2 + 16x + 15
(c) x2 + 4x – 45
(d) x2 + 8x + 12
(e) x2 – 13x + 40
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(f) x2 + 15x + 54
(g) x2 – 11x + 24
(h) x2 – 18x + 72
(i) x2 – 3x – 28
(j) x2 – 7x – 18
Harder examples
(k) x2 + 19x + 60
(l) x2 + 24x + 108
(m) x2 – 15x + 36
(n) x2 + 13x – 48
3
Factorise each of these quadratics:
(a) x2 + 6x + 9
(b) x2 + 13x + 12
(c) x2 + 10x + 21
(d) x2 – 10x + 25
(e) x2 – 17x + 30
(f) x2 – 7x – 30
(g) x2 – 5x – 14
(h) x2 + 3x – 70
(i) x2 + 15x + 14
(j) x2 + 2x – 3
(k) x2 – x – 2
(l) x2 + 10x + 9
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4
Now factorise each quadratic in this mixed bag!
(a) a2 + 3a + 2
(b) z2 + 9z + 20
(c) p2 + 3p – 28
(d) d2 + 2d + 1
(e) b2 – 5b + 4
(f) t2 – 12t + 36
(g) w2 – 18w + 80
(h) m2 – 16m + 64
(i) y2 – 11y + 18
(j) s2 – 7s – 18
(k) q2 – 11q + 24
(l) n2 + 18n + 81
Answers to Activity 6
1
(a) 5 and 3
(b) 15 and 1
(c) 9 and –5
(d) 2 and 6
(e) –5 and –8
(f) 6 and 9
(g) –3 and –8
(h) –6 and –12
(i) 4 and –7
(j) –9 and 2
(k) 4 and 15
(l) 6 and 18
(m) –3 and – 12
(n) –3 and 16
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2
(a) (x + 3)(x + 5)
(b) (x + 15)(x + 1)
(c) (x + 9)(x – 5)
(d) (x + 6)(x + 2)
(e) (x – 8)(x – 5)
(f) (x + 6)(x + 9)
(g) (x – 3)(x – 8)
(h) (x – 6)(x – 12)
(i) (x – 7)(x + 4)
(j) (x – 9)(x + 2)
(k) (x + 4)(x + 15)
(l) (x + 18)(x + 6)
(m) (x – 3)(x – 12)
(n) (x – 3)(x + 16)
3
(a) (x + 3)(x + 3)
(b) (x + 1)(x + 12)
(c) (x + 7)(x + 3)
(d) (x – 5)(x – 5)
(e) (x – 2)(x – 15)
(f) (x + 3)(x – 10)
(g) (x + 2)(x – 7)
(h) (x – 7)(x + 10)
(i) (x + 14)(x + 1)
(j) (x + 3)(x – 1)
(k) (x – 2)(x + 1)
(l) (x + 9)(x + 1)
4
(a) (a + 1)(a + 2)
(b) (z + 4)(z + 5)
(c) (p + 7)(p – 4)
(d) (d + 1)(d + 1)
(e) (b – 4)(b – 1)
(f) (t – 6)(t – 6)
(g) (w – 8)(w – 10)
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(h) (m – 8)(m – 8)
(i) (y – 2)(y – 9)
(j) (s – 9)(s + 2)
(k) (q – 3)(q – 8)
(l) (n + 9)(n + 9)
Factorising a special quadratic: the
difference of two squares
In Section 2 we looked at the multiplication of two factors like (x + a) and
(x – a) which produced a special quadratic, x2 – a2. You will remember that
this is called the difference of two squares.
Check for yourself that the product is correct!
Some examples
1
(x –4)(x + 4) = x2 –42
= x2 –16
2
(2a –9)(2a + 9) = (2a)2 –(9)2
= 4a2 –81
3
(4m + 3n)(4m –3n) = (4m)2 – (3n)2
= 16m2 –9n2
We now want to investigate the reverse of that process. That is, given a
quadratic x2 – a2, we want to be able to factorise it. To help us, we know that
the two factors of x2 – a2 are (x + a) and (x – a). Let’s consider an example.
Example
Factorise x2 – 9
Can you see that x2 – 9 could be written as x2 – 32?
Now we can easily write down the factors. They are (x + 3) and (x – 3)
So x2 – 9 = (x + 3)(x – 3)
Another way for you to get the same answer would be to write x2 – 9 as x2 +
0x – 9 and then ask yourself this question:
Think: What two numbers add to give 0 and multiply to give –9?
Answer: 3 and – 3 because 3 + (–3) = 0 and 3  (–3) = –9
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Therefore, the factors are (x + 3) and (x – 3)
Either way we have that x2 – 9 = (x + 3)(x – 3)
Even from just this one example I’m sure that you will be able to
successfully complete the following activity.
Activity 7
1
Factorise these quadratics:
(a) x2 – 4
(b) y2 – 36
(c) a2 – 49
(d) p2 – 25
(e) r2 – 100
(f) m2 – 81
(g) n2 – 16
(h) q2 – 64
(i) c2 – 144
Answers to Activity 7
1
(a) (x + 2)(x – 2)
(b) (y + 6)(y – 6)
(c) (a + 7)(a – 7)
(d) (p + 5)(p – 5)
(e) (r + 10)(r – 10)
(f) (m + 9)(m – 9)
(g) (n + 4)(n – 4)
(h) (q + 8)(q – 8)
(i) (c + 12)(c – 12)
I’m sure that you found those expressions fairly easy to factorise.
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There are some more for you to try in Activity 8. They are fairly
straightforward too. You just need to take your time and think. Study these
examples first.
Example
Factorise 4a2 – 9
To make this factorisation easier, let’s write 4a2 – 9 as (2a)2 – 32. Now we
can factorise the expression like this:
4a2 – 9
= (2a)2 – 32
= (2a + 3)(2a – 3)
Example
Factorise 16b2 – 49
16b2 – 49 = (4b)2 – 72
= (4b + 7)(4b – 7)
Example
Complete the factorisation of 36x2 – 25
36x2 – 25 = (6x)2 – 52
= (6x +…….)(6x –…….)
Did you write this? (6x + 5)(6x – 5)
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Activity 8
1
Factorise these expressions:
(a) 4a2 – 25
(b) 9s2 – 64
(c) 25c2 – 64
(d) 36z2 – 1
(e) 16y2 – 49
(f) 81t2 – 4
(g) 64m2 – 9
(h) 49b2 – 25
(i) 100x2 – 1
(j) 16n2 – 121
(k) 64w2 – 9
(l) 144u2 – 121
Answers to Activity 8
1
(a) (2a + 5)(2a – 5)
(b) (3s + 8)(3s – 8)
(c) (5c + 8)(5c – 8)
(d) (6z + 1)(6z – 1)
(e) (4y + 7)(4y – 7)
(f) (9t + 2)(9t – 2)
(g) (8m + 3)(8m – 3)
(h) (7b + 5)(7b – 5)
(i) (10x + 1)(10x – 1)
(j) (4n + 11)(4n – 11)
(k) (8w + 3)(8w – 3)
(l) (12u + 11)(12u – 11)
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Well done in getting successfully to this point in the materials. I’m sure that
your confidence in handling algebraic expressions is growing.
In the next part of Section 3, we’ll be looking at simplifying algebraic fractions
as well as adding and subtracting them. The work that you have just completed
will help you do just that.
Take time now to go over what you have done in this section. When you feel you are
ready, go on to the following Check your progress questions.
Check your progress 1
1
Determine whether or not:
(a) 2 and 3 are common factors of 6 and 18
(b) 6 and 7 are common factors of 21 and 42
(c) 6 and 8 are common factors of 24 and 48
(d) 3 and b are common factors of 3b and 6ab
(e) 2a and 3b are common factors of 6ab2 and 6a2b
(f) x and y are common factors of x2wy and xz2y
2
Find the highest common factor of:
(a) 8 and 40
(b) 18 and 27
(c) 36 and 90
(d) 4a and 6b
(e) 5pq and 6p2r
(f) 6ab and 4ac
3
Factorise each expression:
(a) 2a + 4
(b) 6c – 9
(c) 14cd + 2c
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(d) 6y2 + 5y
(e) 5c + 10d
(f) x2z + z2x
(g) 7d3 – d2
(h) m2 + nm2
(i) t2s2 + s2
4
(a) Find two numbers whose sum is 11 and whose product is 24
(b) Find two numbers whose sum is 4 and whose product is –45
(c) Find two numbers whose sum is –10 and whose product is 25
(d) Find two numbers whose sum is 0 and whose product is –16
Now use that information to factorise these quadratics:
(e) m2 + 4m – 45
(f) m2 – 16
(g) m2 + 11m + 24
(h) m2 – 10m + 25
5
Factorise
(a) x2 + 7x + 12
(b) m2 + 8m + 12
(c) a2 –7a + 10
(d) a2 –3a –10
(e) k2 + k –12
(f) y2 –25
(g) m2 –100
(h) 36 –x2
(i) 4m2 –a2
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Answers to Check your progress 1
1
(a) 2 and 3 are common factors of 6 and 18
(b) 6 and 7 are not common factors of 21 and 42
(c) 6 and 8 are common factors of 24 and 48
(d) 3 and b are common factors of 3b and 6ab
(e) 2a and 3b are common factors of 6ab2 and 6a2b
(f) x and y are common factors of x2wy and xz2y
2
(a) 8
(b) 9
(c) 18
(d) 2
(e) p
(f) 2a
3
(a) 2(a + 2)
(b) 3(2c – 3)
(c) 2c(7d + 1)
(d) y(6y + 5)
(e) 5(c + 2d)
(f) xz(x + z)
(g) d2(7d – 1)
(h) m2(1 + n)
(i) s2(t2 + 1)
4
(a) 8 and 3
(b) 9 and –5
(c) –5 and –5
(d) –4 and 4
(e) (m + 9)(m – 5)
(f) (m – 4)(m + 4)
(g) (m + 8)(m + 3)
(h) (m – 5)(m – 5)
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(a) (x + 3)(x + 4)
(b) (m +6)(m + 2)
(c) (a –5)(a –2)
(d) (a –5)(a + 2)
(e) (k + 4)(k –3)
(f) (y –5)(y + 5)
(g) (m –10)(m + 10)
(h) (6 –x)(6 + x)
(i) (2m –a)(2m + a)
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Algebraic fractions
In algebra, expressions such as these:
x
x – 6 2x + 4 x 2 – 6 x 2 – 6x + 8
,
,
,
,
2
3
x – 5
x2
x 2 – 16
are called algebraic fractions.
Consider the fraction
x 2 + 4x + 4
3x + 8
The numerator of this algebraic fraction is x2 + 4x + 4 and the denominator
is 3x + 8.
As with fractions in arithmetic, if the numerator and the denominator of an
algebraic fraction are multiplied by or divided by the same number or by the
same algebraic expression, then the original algebraic fraction and the final
algebraic fraction are said to be equivalent.
You remember how to create equivalent fractions in arithmetic.
We simply multiply or divide both top and bottom by the same number.
Algebraic fractions work the same way.
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Example
Create a fraction equivalent to
Just ask the question, what did we have to multiply 5m by to get 15m?
Obviously 3!
Then 3 is the number we must multiply both top and bottom by
Try the next ones yourself.
Check your answers.
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Did you get
Good!
Of course, since pronumerals just represent numbers, we could multiply
both top and bottom by the same pronumeral to get equivalent fractions.
Example
Try these
Check your answers
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Another example
What did we multiply 3x by to get 12x2?
Some for you!
Check your answers
All correct? Good!
Reducing a fraction
Instead of multiplying the numerator and denominator of a fraction to create
an equivalent fraction, we often want to reduce a fraction to its simplest
form. As in arithmetic we do that by dividing the numerator and
denominator of the given fraction by the highest common factor (HCF).
Sometimes with simple algebraic fractions you can see what the highest
common factor is. To be quite sure, however, it is often best to write the
numerator and the denominator in terms of their prime factors. In doing that
you will be able to see quite easily what the highest common factor is.
Once you find the HCF, the fraction can be reduced to its simplest form by
cancelling out the HCF from both the numerator and the denominator.
Remember that prime factors for algebraic fractions will often consist of
both numerals and pronumerals.
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Example
Find the HCF of each of these algebraic fractions. Then reduce the fraction
to its simplest form.
(a)
6y 3
3  2  y  y  y
=
9y
3  3  y
=
2  y  y
3
=
2y 2
3
You can see that 3 and y are common to both the numerator and the
denominator.
Since they are common, they can be cancelled out.
6y 3
2y 2
Then
and
are equivalent fractions.
9y
3
(b)
y4
y  y  y  y
=
2
2y
2  y  y
=
y  y
2
y2
=
2
Then
y4
2y
2
and
y2
2
are equivalent fractions.
Always divide by the biggest number that goes into top and bottom. Here,
3 divides into 12 and 18, but 6 is bigger!
Try these
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Check your answers
How did you go?
More fractions
Sometimes a fraction may have a binomial numerator or denominator.
We can still form equivalent fractions by multiplying top and bottom by
the same number
Example
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(d) If we multiply both the numerator and denominator of
we get
Then
(a – 6)  5 5a – 30
=
(a + 7)  5 5a + 35
a – 6
by 5,
a+ 7
a – 6
5a – 30
and
are said to be equivalent fractions.
a+ 7
5a + 35
(e) If we multiply both the numerator and denominator of
x (x + 3) x 2 + 3x
we get
=
2 (x + 3)
2x + 6
x
by (x + 3),
2
x
x 2 + 3x
Then and
are equivalent fractions.
2
2x + 6
Of course, you can reduce a fraction to an equivalent fraction, by dividing
top and bottom.
Example
You must factorise top and bottom before cancelling!
You can only cancel the common factors!
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4x 2 + 5x x(4x + 5)
(d)
=
x(x – 2)
x 2 – 2x
(here we’ve taken x out as a common factor)
=
4x + 5
x – 2
(on cancelling out the common factor x from both the numerator and the denominator)
So
4x + 5
4x 2 + 5x
and
are equivalent fractions.
2
x – 2
x – 2x
(x + 1)(x + 2)
x 2 + 3x + 2
(e)
=
2
(x + 1)(x + 3)
x + 4x + 3
=
x + 2
x + 3
(on cancelling out the common factor (x + 1) from both the numerator and the denominator)
x + 2
x 2 + 3x + 2
Then 2
and
are equivalent fractions.
x + 4x + 3
x + 3
(f)
25abc
25  a  b  c
=
500abt
20  25  a  b  t
c
=
20t
Then
25abc
c
and
are equivalent fractions.
500abt
20t
Sometimes we can cancel by a binomial term.
Example
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I’m sure that you are feeling reasonably confident about creating equivalent
algebraic fractions for yourself.
To get the correct answer just think and take your time. By now you have all
of the skills necessary to obtain the right answer every time.
We’ll start off slowly with questions that you will be able to manage. Let’s
go!
Activity 9
1
44
Create an equivalent algebraic fraction by multiplying the numerator and denominator
of:
(a)
2x
by x
3
(b)
4x
by x2
5
(c)
(2x – 1)
by 2x
7x
(d)
4x 2
by x2
7
(e)
(3x + 5)
by 4x
5x
(f)
x2 – 4
by 2x
x + 6
(g)
5 – x
by 4x
2 + x
(h)
9x
by x + 1
x + 1
(i)
x2 + 6
by x2
2
x
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2
Decide what is the highest common factor of each fraction. Then reduce each fraction
to its lowest term or simplest form.
3  2  a a
6a2
(a)
=
3 a
3a
(b)
2 2  3bcc
12bc 2
=
3 3 b c
9bc
5c cd
5c 2 d
(c)
2 =
5 2cd d
10cd
(d)
33 2 a
18a
3 =
5 3aaa
15a
(e)
2b + 6 2(b + 3)
=
(b + 3)
b + 3
(f)
5x + 20 5(x + 4)
=
(x + 4)
(x + 4)
(g)
4z – 8 4(z – 2)
=
3z – 6 3(z – 2)
x(x + 2)
x 2 + 2x
(h) 2
=
x(x – 9)
x – 9x
(i)
(j)
(k)
(l)
(m)
3
p 2 – 4p
2p – p
2
=
p(p – 4)
p(2 – p)
(m – 6)(m + 6)
m 2 – 36
=
2
m(m + 6)
m + 6m
y 2 + 5y + 6
2
y + 2y
t 2 – 7t + 12
2
t – 16
=
(y + 2)(y + 3)
y(y + 2)
=
(t – 3)(t – 4)
(t + 4)(t – 4)
n 2 + 5n – 24
2
n + n – 56
=
(n + 8)(n – 3)
(n + 8)(n – 7)
Reduce each fraction to its lowest term or simplest form:
(a)
ax
ay
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(b)
xy
3y
(c)
4x
2
x
(d)
12pq
9pr
(e)
q2
qt
(f)
d
cd
(g)
–10
5c
–m 5
(h)
m3
(i)
–4rs
2
2rs
(j)
–x 5
4x
(k)
–14mn
2
–7mn
(l)
4
46
16x 2 y 2
32xy
Reduce each of the following algebraic fractions to its simplest form. Be careful with
the factorisation.
(a)
2a – 2
8a – 8
(b)
6a + 6
2a + 2
(c)
8
4x – 12
(d)
3x – 6
5(x – 2)
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(e)
3k – 3
k – 1
(f)
x + 2
2x + 4
(g)
3s + 15
6s + 30
(h)
7n – 21
3n – 9
(i)
24c – 12
10c – 5
(j)
x2 – 9
x 2 – 3x
(k)
k + 2
(k + 2)(3k + 8)
t 2 – 25
(l) 2
t – 8t + 15
x 2 + 12x
(m) 2
x + 7x
5
(n)
3x 2 + 3
6x 2 + 6
(o)
c 2 – 36
(c + 6)(c + 8)
Simplify each of the following. Some numerators and denominators will need to be
factorised. Be careful!
(a)
(x – 2)(x – 4)
x – 4
(b)
(y + 3)(y + 4)
y(y + 3)
b 2 – 6b + 5
(c)
b – 1
c 2 – 49
(d) 2
c – 14c + 49
(e)
x 2 + x – 20
x 2 – x – 12
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(f)
x2
x 2 – 36
+ 12x + 36
(g)
x 2 + 2x – 24
x 2 – 16
(h)
r 2 – 64
– 16r + 64
r2
m 2  10m  24
(i)
m 2 + 8m + 12
Answers to Activity 9
1
2x 2
(a)
3x
4x 3
(b)
5x 2
(c)
4x 2 – 2x
14x 2
4x 4
(d)
7x 2
(e)
12x 2 + 20x
20x 2
2x 3 – 8x
(f)
2x 2 + 12x
(g)
20x – 4x 2
8x + 4x 2
9x 2  9x
(h) 2
x + 2x + 1
(i)
48
x 4 + 6x 2
x4
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2
(a)
HCF = 3a
Simplest form is 2a
(b)
HCF = 3bc
Simplest form is
(c)
HCF = 5cd
Simplest form is
(d)
6
2
5a
(e)
HCF = (b + 3)
Simplest form is 2
(f)
HCF = (x + 4)
Simplest form is 5
(g)
HCF = (z – 2)
Simplest form is
(h)
(i)
HCF = p
Simplest form is
(j)
HCF = (m + 6)
Simplest form is
(k)
x + 2
x – 9
p – 4
2 – p
m – 6
m
HCF = (y + 2)
Simplest form is
(l)
4
3
HCF = x
Simplest form is
HCF = (t – 4)
Simplest form is
(m) HCF = (n + 8)
Simplest form is
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c
2d
HCF = 3a
Simplest form is
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4c
3
y + 3
y
t – 3
t + 4
n – 3
n – 7
49
3
(a)
x
y
(b)
x
3
(c)
(d)
(e)
3r
q
t
(f)
1
c
(g)
–2
c
(h)
–m 2
= –m2
1
(i)
–2
s
(j)
–x 4
4
(k)
(l)
4
4
x
4q
(a)
2
n
xy
2
1
4
(b) 3
(c)
2
(x – 3)
(d)
3
5
(e) 3
50
(f)
1
2
(g)
1
2
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5
(h)
7
3
(i)
12
5
(j)
(x + 3)
x
(k)
1
(3k + 8)
(l)
(t + 5)
(t – 3)
(m)
(x + 12)
(x + 7)
(n)
1
2
(o)
(c – 6)
(c + 8)
(a) (x – 2)
(b)
(y + 4)
y
(c) (b – 5)
(d)
(c + 7)
(c – 7)
(e)
(x + 5)
(x + 3)
(f)
(x – 6)
(x + 6)
(g)
(x + 6)
(x + 4)
(h)
(r + 8)
(r – 8)
(i)
(m + 4)
(m + 2)
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How to add and subtract algebraic
fractions
Now, let’s explore adding and subtracting simple algebraic fractions.
Finding the solution to these questions is just as straightforward as adding
or subtracting ordinary fractions.
As a start let’s work through some typical examples and then there are some
similar questions for you to do.
Study each of the following examples.
Example
x
x
2x
+
=
3
3
3
Here we are simply adding thirds. We have x ‘lots of a third’ plus another x
‘lots of a third’. Answer: 2x ‘lots of a third’.
Remember, if the denominators are the same then to obtain the answer
simply add the numerators.
Example
Because they have the same denominator, we need only subtract the top
lines.
3p
12
–
2p
12
=
p
12
Here we are subtracting twelfths. We have 3p twelfths minus 2p twelfths.
Answer: p twelfths.
Example
7m
6m
4m
9m
+
–
=
10
10
10
10
Here all of the fractions are tenths.
Question: What is 7m + 6m – 4m tenths?
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Answer: 9m tenths
Any problem with these three examples?
If you need to, work through them again before going on.
Adding and subtracting when the denominator is an
algebraic expression
Now let’s examine some questions where the denominator is an algebraic
expression.
Example
3
5
8
+
=
y
y
y
Here all three denominators are y, which represents some number. Finding
the answer is as simple as adding the numerators, 3 and 5.
Example
5
7
1
11
+
–
=
3m
3m
3m
3m
Here all four denominators are 3m, which represents some number. Again,
finding the answer is as simple as adding and subtracting the numerators.
Check the answer to this question to see if you agree.
Example
6
8
5
9
+
–
=
(x + 2)
(x + 2)
(x + 2)
(x + 2)
If you have been able to follow the last six examples quite easily, then you
are ready to try the next activity.
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Activity 10
1
Complete the following fraction patterns:
(a)
(b)
54
a+a
a
a
+
2
2
a
a
+
3
3
a
a
+
4
4
a
a
+
5
5
a
a
+
6
6
a
a
+
7
7
= 2a
2a
=
2
2a
=
3
2a
=
4
=
=
=
2c + c
2c
c
+
2
2
2c
c
+
3
3
2c
c
+
4
4
2c
c
+
5
5
2c
c
+
6
6
2c
c
+
7
7
5
6
7
= 3c
3c
=
2
3c
=
3
3c
=
4
=
=
=
5
6
7
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(c)
6d + 3d
6d
3d
+
2
2
6d
3d
+
3
3
6d
3d
+
4
4
6d
3d
+
5
5
6d
3d
+
6
6
6d
3d
+
7
7
= 9d
9d
=
2
9d
=
3
9d
=
4
=
5
=
6
=
7
4a – 3a = a
(d)
4a
3
4a
5
4a
7
4a
9
4a
11
3a
3
3a
5
3a
7
3a
9
3a
11
–
–
–
–
–
a
3
a
=
5
=
=
=
=
7
9
11
7e – 4e = 3e
(e)
7e
3
7e
5
7e
7
7e
9
7e
11
–
–
–
–
–
4e
3
4e
5
4e
7
4e
9
4e
11
3e
3
3e
=
5
=
=
=
=
7
9
11
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(f)
10k – 8k = 2k
10k
3
10k
5
10k
7
10k
9
10k
11
–
–
–
–
–
8k
3
8k
5
8k
7
8k
9
8k
11
2k
3
2k
=
5
=
=
7
=
=
9
11
These patterns are really just reinforcing the idea that, if the fractions are similar, you really
only have to add or subtract the numerator.
2
56
Add these algebraic fractions. Where possible, reduce each answer to its simplest
form.
(a)
f
4f
+
=
5
5
5
(b)
3m
5m
+
=
10
10
10
(c)
2n
2n
+
=
6
6
6
(d)
2w 6w
+
=
8
8
8
(e)
7v
v
+ =
3
3
3
(f)
9z
3z
+
=
12 12
12
(g)
4k
4k
+
=
16
16
(h)
8g
2g
+
=
2
2
(i)
12 y 3 y
+
=
20 20
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3
4
Add these algebraic fractions. Where possible reduce each answer to its simplest form.
(a)
5
3
+
=
c
c
(b)
6
3
+
=
m
m
(c)
8
7
+
=
e
e
(d)
9
5
+
=
z
z
(e)
10
3
+
=
w
w
(f)
7
2
+
=
3x
3x
(g)
3
4
+
=
(b + 2)
(b + 2)
(b + 2)
(h)
12
4
+
=
(3c – 5)
(3c – 5)
(3c – 5)
(i)
10
5
+
=
(y + 12)
(y + 12)
(j)
1
14
+
=
(13c – 6)
(13c – 6)
c
m
e
w
Subtract these algebraic fractions. Where possible, reduce each answer to its simplest
form.
(a)
2f
f
–
=
4
4
(b)
7m
m
–
=
10
10
10
(c)
8x
6x
–
=
5
5
(d)
12w
5w
–
=
6
6
4
5
6
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57
5
58
(e)
4v
v
–
=
12
12
12
(f)
8z
3z
–
=
2
2
(g)
9t
4t
–
=
20
20
(h)
8s
5s
–
=
3
3
(i)
24y
3y
–
=
4
4
2
Subtract these algebraic fractions. Where possible, reduce each fraction to its simplest
form.
(a)
5
3
–
=
b
b
b
(b)
9
7
–
=
n
n
n
(c)
8
5
–
=
h
h
h
(d)
9
8
–
=
y
y
(e)
12
3
–
=
t
t
(f)
6
4
–
=
5k
5k
(g)
8
4
–
=
(d – 8) (d – 8) (d – 8)
(h)
9
3
–
=
(6c + 6) (6c + 6) (6c + 6)
(i)
11
6
–
=
(z – 7)
(z – 7)
(j)
18
12
–
=
(3c – 9) (3c – 9)
t
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6
7
Simplify each of these:
(a)
4
5
7
+
–
=
t
t
t
(b)
8
6
12
+
–
=
s
s
s
(c)
12
8
10
+
–
=
(z + 5) (z + 5) (z + 5)
(d)
4
20
16
+
–
=
(k – 8) (k – 8) (k – 8)
t
Simplify each of these:
(a)
9
2
3
–
–
=
t
t
t
(b)
18
6
3
–
–
=
s
s
s
(c)
24
4
12
–
–
=
(c – 9) (c – 9) (c – 9)
(d)
14
2
10
–
–
=
(p + 6) (p + 6) (p + 6)
t
Answers to Activity 10
1
(a)
2a 2a 2a
;
;
5
6
7
(b)
3c 3c 3c
;
;
5
6
7
(c)
9d 9d 9d
;
;
5
6
7
(d)
a a
a
;
;
7 9 11
(e)
3e 3e 3e
;
;
7
9
11
(f)
2k 2k 2k
;
;
7
9
11
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59
2
(a)
5f
= f
5
(b)
8m
4m
=
10
5
(c)
4n
2n
=
6
3
(d)
8w
= w
8
(e)
8v
3
(f)
12z
= z
12
8k
k
=
16
2
10g
(h)
= 5g
2
15y
3y
(i)
=
20
4
(g)
3
60
(a)
8
c
(b)
9
m
(c)
15
e
(d)
14
z
(e)
13
w
(f)
9
3
=
3x
x
(g)
7
(b + 2)
(h)
16
(3c – 5)
(i)
15
(y + 12)
(j)
15
(13c – 6)
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4
(a)
f
4
(b)
6m
3m
=
10
5
(c)
2x
5
(d)
7w
6
(e)
3v
v
=
12
4
(f)
5z
2
(g)
5t
t
=
20
4
(h)
(i)
5
3s
= s
3
21y
4
(a)
2
b
(b)
2
n
(c)
3
h
(d)
1
y
(e)
9
t
(f)
2
5k
(g)
4
(d – 8)
(h)
6
1
=
(6c + 6)
(c + 1)
(i)
5
(z – 7)
(j)
6
2
=
(3c – 9)
(c – 3)
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61
6
7
(a)
2
t
(b)
2
s
(c)
10
(z + 5)
(d)
8
(k – 8)
(a)
4
t
(b)
9
s
(c)
8
(c – 9)
(d)
2
(p + 6)
Did you notice that sometimes it was possible to cancel the answer you got?
It’s a good idea to always write the answer in its simplest form.
Changing to a common denominator
Let’s now look at some simple examples where we have to change to a
common denominator. Here we’ll only use denominators that belong to
particular ‘fraction families’ that we have explored earlier.
Example
k
k
2k
k
+
=
+
2
4
4
4
3k

4
Here our denominators tell us that we are dealing with halves and quarters.
This suggests that we should change everything to quarters before adding.
Example
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The simple rule is to multiply top and bottom of each fraction by the
denominator of the other fraction.
So
(x + 6)
(x + 6)
5(x + 6)
2(x + 6)
+
=
+
2
5
10
10
7(x + 6)

10
Here our denominators tell us that we are dealing with halves and fifths.
This suggests that we should change everything to tenths before adding.
Example
(4z – 10)
(4z – 10)
4(4z – 10)
3(4z – 10)
–
=
–
3
4
12
12
(4z – 10)

12
Here our denominators tell us that we are dealing with thirds and quarters.
This suggests that we should change everything to twelfths before
subtracting.
Complete these two exercises by filling in the blank spaces.
1
Show that the answer to the following question is
4a
10
a
a
a
+
–
=
+
–
2
10
5
10
10
10
=
10
Could you reduce this answer to its simplest form?
Yes. It would reduce to
2a
5
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2
Show that the answer to the following question is
3(c + 2)
4
+
(c + 2)
2(c + 2)
–
6
3
=
=
12
+
3(c + 2)
12
12
–
12
12
Could you reduce this answer to its simplest form?
Yes. It would reduce to
c + 2
4
Activity 11
1
Complete these:
(a)
p
p
+
=
2
4
4
=
(b)
4
4
m
m
+
=
3
6
=
64
+
6
+
6
6
(c)
2f
f
+
=
4
8
(d)
3n
n
+ =
5
2
(e)
(x + 3)
(x + 3)
+
=
2
3
(f)
(b – 5)
2(b – 5)
+
=
2
3
(g)
(2 p + 8)
(2 p + 8)
+
=
12
3
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2
Complete these:
(a)
m
m
–
=
–
5
10
10
10
=
(b)
z
z
–
=
2
3
=
10
–
6
6
6
(c)
u u
– =
3 4
(d)
5n n
– =
4
8
(e)
(t + 1)
(t + 1)
–
=
2
10
=
3
(f)
(3 f – 2) (3 f – 2)
–
=
3
12
(g)
(5q + 1) (5q + 1)
–
=
5
20
10
–
10
10
Complete and simplify these algebraic fractions:
(a)
c
c
c
+
–
=
+
–
2
3
4
12
12
12
=
(b)
12
m
m
m
+
–
=
+
–
2
10
20
20
20
20
=
(c)
a
a
a
+
–
=
2
8
4
(d)
3u
u
u
+
– =
2
10 5
20
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65
(e)
3(d + 4) 2(d + 4) (d + 4)
–
–
=
4
3
12
(f)
5(a – 9) (a – 9) 2(a – 9)
–
–
=
3
2
6
Answers to Activity 11
1
(a) Numerators 2p; p; 3p
(b) Numerators 2m; m; 3m
(c) Numerators 4f; f; 5f
(d) Numerators 6n; 5n; 11n
(e) Numerators 3(x + 3); 2(x + 3); 5(x + 3)
(f) Numerators 3(b – 5); 4(b – 5); 7(b – 5)
(g) Numerators (2p + 8); 4(2p + 8); 5(2p + 8)
2
(a) Numerators 2m; m; m
(b) Numerators 3z; 2z; z
(c) Numerators 4u; 3u; u
(d) Numerators 10n; n; 9n
(e) Numerators 5(t + 1); (t + 1); 4(t + 1)
(f) Numerators 4(3f – 2); (3f – 2); 3(3f – 2)
(g) Numerators 4(5q + 1); (5q + 1); 3(5q + 1)
3
(a) Numerators 6c; 4c; 3c; 7c
(b) Numerators 10m; 2m; m; 11m
(c) Numerators 4a; a; 2a; 3a
(d) Numerators 15u; u; 2u; 14u
(e) Numerators 9(d + 4); 8(d + 4); (d + 4); 0
(f)
66
10(a – 9) 3(a – 9) 2(a – 9) 5(a – 9)
;
;
;
6
6
6
6
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Some more examples
Your turn
Did you get:
Good!
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Multiplying algebraic fractions
Do you remember how to multiply fractions?
We multiply algebraic fractions the same way.
Example
Now some for you to try.
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Check your answers
How did you go? Hope you got them all correct.
We can make cancelling a little simpler by cancelling before we multiply
You may cancel a common factor in any top line with the same common
factor in any bottom line.
Example
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69
Try these:
Check your answers:
You multiply binomial fractions exactly the same way.
Example
Hint: always factorise the binomial if you can.
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Now you try some
Check your answers
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71
Dividing algebraic fractions
Do you remember how to divide arithmetic fractions?
We do algebraic fractions the same way.
Example
You can try the rest
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Check your answers:
Activity 12
1
Add or subtract the following fractions and cancel if possible.
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73
2
Multiply these fractions (be careful to cancel where you can!)
3
Divide the following, cancelling where possible.
A bit of a challenge!
4
74
Complete and simplify (cancel down).
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Answers to Activity 12
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75
Check your progress 2
1
2
3
76
Complete these:
(a)
3
=
4
(b)
1
2
4
=
=
=
=
5
10
50
100
8
=
9
=
=
12
20
24
Complete these:
(a)
a
4a
10a
=
=
=
2
16
(b)
b
=
4
(c)
c
5c
6c
=
=
=
d
10d
(d)
3k
9k
=
=
=
5m
20m
30m
8
=
12
=
5b
Write both the numerator and the denominator of each of these fractions in terms of
prime factors and then reduce each fraction to its simplest form.
(a)
9
12
(b)
16
20
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4
5
(c)
50
60
(d)
24
36
(e)
4x
2
(f)
12e
3e
(g)
6r
3s
(h)
8rst
2
6r s
Reduce these fractions to their simplest form:
(a)
5x + 10
5
(b)
9c + 3
6
(c)
16
4v – 8
(d)
a2 + ab
a
Add or subtract each of these fractions:
(a)
6
5
+
12
12
(b)
2
3
+
5
10
(c)
7
2
–
8
8
(d)
4
1
–
3
2
(e)
x
3x
+
10
10
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77
(f)
(g)
6
5y
12
2p
3
–
+
y
12
p
4
(h)
k
2k
k
+
–
2
3
4
(i)
7
5
+
y
y
(j)
9
4
–
2m
2m
(k)
(x + 5)
(x + 5)
+
8
8
(l)
(2w + 5)
(2w + 5)
–
5
10
Factorise the numerator and the denominator of each of these fractions and then reduce
each fraction to its simplest form.
(a)
px
pz
(b)
6ab
3bc
(c)
6pq
9qr
(d)
3a + 12
6a + 24
(e)
a2 + 2ab
a
(f)
a + a2 b
ab
(g)
4p + 4q
6q + 6p
x2 + x
(h) 2
x – x
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x 2 + 7x + 12
(i)
(x + 3)
r2 – r – 6
(j)
(r – 3)
x 2 + 15x + 36
(k) 2
x + 14x + 24
(l)
7
x 2 – 25
x 2 + 2x – 35
Carry out the operation required in each question.
Answers to Check your progress 2
1
(a) Numerators 6, 15, 18
(b) Numerators 10, 20; Denominator 20
2
(a) Numerator 8a; Denominators 8, 20
(b) Numerators 2b, 3b; Denominator 20
(c) Numerator 10c; Denominators 5d, 6d
(d) Numerators 12k, 18k; Denominator 15m
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3
(a)
3
4
(b)
4
5
(c)
5
6
(d)
2
3
(e) 2x
(f) 4
4
(g)
2r
s
(h)
4t
3r
(a) x + 2
(b)
3c + 1
2
(c)
4
v – 2
(d) a + b
5
(a)
11
12
(b)
7
10
(c)
5
8
(d)
5
6
4x
2x
=
10
5
4y
y
(f)
=
12
3
11p
(g)
12
(e)
(h)
80
11k
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6
(i)
12
y
(j)
5
2m
(k)
2(x + 5)
(x + 5)
=
8
4
(l)
2w + 5
10
(a)
x
z
(b)
(c)
(d)
2a
c
2p
3r
1
2
(e) a + 2b
(f)
1 + ab
b
(g)
4 2

6 3
(h)
x + 1
x – 1
(i) x + 4
(j) r + 2
(k)
x + 3
x + 2
(l)
x + 5
x + 7
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Introduction to equations
So far in this module, you have learned how to simplify algebraic
expressions and how to factorise algebraic expressions. I am sure that you
are feeling confident about the work thus far and will handle this section of
work with the same confidence.
In this section, you will learn how to solve linear equations, simple
quadratic equations and linear simultaneous equations.
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Solving equations
Introduction
First in this section, you will be investigating how to solve simple algebraic
equations such as these:
a 
 b
c
2a =6
2 x –3=7
3  x  4   15
or equations involving simple algebraic fractions such as:
a 
b
5x
 20
2
1 2x
 11
3
Later in this section, you will be investigating how to solve quadratic
equations similar to these:
a 
b
x 2  5 x +6=0
x 2 – 7 x +12=0
With quadratic equations, the question that you will be trying to answer is
what value or values of x will make, say, x 2  5x  6 equal to nought?
Once you have become skilled in finding such solutions, we will look at two
other ideas related to finding the solution of other types of algebraic
equations which I think you’ll find interesting and challenging.
Note: ‘Challenging’ does not mean difficult. It simply means that it will
require just a little more thinking on your part.
Enough about that at this stage. Let’s start by asking a simple question.
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What is an equation?
Answer: An equation is a mathematical statement like 2 x + 4 = 18 in which
two algebraic expressions are said to be equal.
Finding the value or values that make the equation true
To find the solution of an equation, the task is to find a value or values of
the pronumeral that will make the mathematical statement or equation true.
For instance:
1 Is the statement 3x + 4 = 2x + 5 true if x = 1?
Check
Left Hand Side (LHS)
 3x + 4
= 3  1+ 4 when x = 1
=7
Right Hand Side
RHS   2x + 5
= 2  1+ 5 (when x = 1)
=7
 LHS  RHS when x  1
So the statement is true if x = 1
The roots or the solutions of an equation
The number or numbers that satisfy or make an equation true are called the
roots or the solutions of the equation.
For instance (as we have just found out):
x = 1 is the root or solution of the equation 3x + 4 = 2x + 5
Some simple rules
Rule 1:
You can add the same number to both sides of an equation and the
equation remains true.
Example
add 5
2  8 = 16
2  8 + 5 = 16 + 5
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You’ll notice it is still true that
21 = 21
We can use this to solve very simple equations.
Example
x – 4 = 10
Now let’s add 4 to both sides
x – 4 + 4 = 10 + 4
x = 14
The idea is to get x alone on one side of the equation.
So how do we ‘undo’ the – 4?
+ 4 undoes the – 4.
Example
a – 10 = 1
Now we want to ‘undo’ – 10
+ 10 will undo it.
But you must add it to both sides.
Example
Rule 2:
You can subtract the same number from both sides of an equation and
the equation remains true.
Example
It is still true that 13 = 13
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Example
m + 10 = 14
How can we ‘undo’ the + 10.
– 10 will ‘undo’ the + 10.
Example
a + 5 = 11
Subtract 5 from both sides.
Rule 3:
You may multiply both sides by the same number, and the equation
remains true.
It is still true
48 = 48
We can use this when we need to ‘undo’ a division.
Example
we need to ‘undo’ the division by 3 to get x alone on one side.
Multiplication undoes division
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Example
multiply both sides by 10
Rule 4:
You may divide both sides by the same number, and the equation
remains true.
is still true
Example
5x = 30
we need to ‘undo’ the multiplication by 5 to get x alone.
So divide both sides by 5
Example
6x = 24
Divide both sides by 6
Some for you:
Find the value of the pronumeral needed to make each equation true.
(a) x – 6 = 5
(b) a + 10 = 24
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(d) 4a = 44
Check your answers
(a) x = 11
(b) a = 14
(c) a = 36
(d) a = 11
How did you go!
All correct?
They get a lot more interesting, don’t worry.
Some helpful hints
Remember, to help you find the solution of an equation, you may need to:

add the same number to each side

subtract the same number from each side

multiply each side by the same number

divide each side by the same number.
Let’s combine the operations!
Example
3x + 4 = 16
We aim to get x alone on one side. First we need to ‘undo’ the +4.
Subtract 4 from both sides.
Now, we need to ‘undo’ the multiplication by 3, by dividing both side by 3.
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Another example
‘undo’ the – 3 by adding 3 to both sides.
‘undo’ the divide by multiplying both sides by 5.
Example
First, ‘undo’ the + 10 by subtracting 10 from both sides
Now, undo the divide by multiplying both sides by 3
Now divide both sides by 2
Note every time we get a solution, we should test it in the original equation
to check our answer.
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Some for you
Solve each one
(a) 5m + 7 = 22
(b) 3x – 11 = 4
Check your answers
(a) m = 3
(b) x = 5
(c) k = 10
(d) x = 16
(e) x = 20
(f) y = 6
How did you go. All correct?
The answers to equations are not always positive whole numbers.
Example
(a) 5x + 7 = 20
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(b) 2.4x – 1.8 = 4.2
add 1.8 to both sides
Now divide both sides by 2.4.
(c) 2 –k = 8
Subtract 2 from both sides
There is a trick to know here. If you end up with – k and you want k, just
multiply both sides by – 1.
Equations containing brackets
Example
Find the solution to 5(x + 4) = 10
Solution
Write the equation
Removing the brackets
Subtracting 20 from each side
Dividing each side by 5
5(x + 4)
5x  20
5x  20 – 20
5x
x
 10
 10
 10 – 20
= –10
 –2
The root or solution is x = –2
Check the solution:
LHS = 5(x + 4)
= 5(–2 + 4)
(when x = –2)
= 10
RHS = 10
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LHS = RHS when x = –2
So x = –2 is the solution.
Example
Subtract 36 from both sides
divide both sides by – 12
Check
Note: In getting this solution, I have put in all of the main steps for you.
Of course, you could do a lot of the work in your head. If you can get the
solution mentally, then by all means do so. If you feel happier using pen
and paper, then that’s fine. The aim is to get the correct solution!
Some to try: (check your answers in the equation)
(b) 2.5x + 4 = 12.25
(c) 13 – x = 5
(d) 2(x – 4) = 12
(e) 5(13 – 2a) = 25
Answers
(a) 4.5
(b)
(c)
(d)
(e)
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3.3
8
10
4
93
All correct? Good!
Equations with fractions
Example
Find the solution of
1  2x
 11
3
Solution
Write the equation
Multiplying each side by 3
Subtracting 1 from each side
Dividing each side by 2
1 2x
 11
3
3(1+2x)
 3 11
3
1  2 x  33
2 x  32
x  16
The solution is x = 16
Check the solution:
1+ 2x
3
1 2  16

3
33

3
 11
LHS =
when x  16
RHS  11
LHS = RHS when x = 16
So x = 16 is the solution.
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Example
x 3x

 22
2 5
Find the solution of
Solution
x 3x

 22
2 5
Write the equation
5x 6x

 22
10 10
Making similar fractions
11x
 22
10
Multiplying each side by 10 11x  220
Dividing each side by 11 x = 20
The solution is
x = 20
Check the solution:
LHS =

x

2
20
3x
5
60

2
5
= 10 + 12
(when x = 20)
= 22
RHS = 22
LHS = RHS when x = 20
So x = 20 is the solution.
What happens when there is a pronumeral on both sides of the equation.
Examples
(a) 5m – 2 = 3m + 8
Select the smallest term in m and subtract it from both sides. Here we’d
subtract 3m from both sides.
5m – 3m – 2 = 3m – 3m + 8
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(b) 6x + 4 = 28 – 2x
add 2x to both sides
(c) 5(x – 3) = 2(x + 3)
multiply out
5x – 15 = 2x + 6
subtract smallest number of x’s
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Example
Subtract 4x from both sides.
Add 21 to both sides.
Divide both sides by 31.
Check the solution:
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LHS = RHS so x = 1 is the solution.
Simple problems can be solved if they are written out as an equation.
Example
Think of a number, add 8, double the answer and the result is 6 times the
number. Find the number.
First pick a letter to represent the number. Let x = the number.
Now read the question a bit at a time and replace words with mathematical
symbols.
number plus 8
x+8
double the answer
2(x + 8)
result is 6 times the number
RHS = 6x
Now we have our equation: 2(x + 8) = 6x
Solve the equation:
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The solution is x = 4. Now check the solution.
Example
I am 19 years older than my son. Together our ages add to 31. How old is
my son?
Let my son’s age = a.
My age is 19 more = a + 19
Add ages to get 31:
My son’s age is 6 years.
Activity 12
1
Check that the following statements are true:
(a) c = 7 is a root or solution of the equation 3c + 4 = 25
(b) s = 2 is a root or solution of the equation 5s – 6 = 4
(c) a = –3 is a root or solution of the equation a2  9
(d) b = 1 is a root or solution of the equation 5b – 2 = 3b
(e) m = 13 is a root or solution of the equation
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m +2
5
3
99
2
Work through each of these solutions and make sure that they are correct. Decide what
was done at each step. Finally check the solution by substituting the answer into the
original equation.
(a) 2a+5  21
2a  16
a  8
(b)
62s – 3 = –42
12s – 18 = –42
12s = –24
s = –2
(c) 6a – 9 = 2a + 7
4a – 9 = 7
4a = 16
a= 4
(d)
3
2y y
–
3 2
4y 3y
–
6
6
y
6
y

5

5

5

30
Solve each of the following equations. Check each solution.
(a) 6c = 36
(b) 8b  72
(c) 3p = 39
(d) n + 8 = 28
(e) w – 6  42
(f) 8  t  –4
(g) 5 – y = 17
(h) 1 – k  –9
(i) 1.2  x  5.6
(j) –5  m = 6.7
(k) 3.5  b = –6.5
(l) –1.9  g  –0.9
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4
(m)
e
9
3
(n)
p
 1.5
4
(o)
u
 0.4
2
Solve each of the following equations. Check each solution.
(a) 2w + 4 = 6
(b) 3x – 12 = 9
(c) 5  10r  20
(d) 15  4c  7
(e) 26  12k  2
(f) 8 – 2p  4
(g) 2 – 3p = 8
(h) 2 – 6q  –10
(i) 16 – 3s  22
(j) 1.9  7a  22.9
(k) –30  8m  –10
(l) 6 – 6m  45
5
Solve each of the following equations. Check each solution.
(a) 3b + 5 = 2b + 7
(b) 4a+ 10 = a+ 1
(c) z + 6 = 5z – 6
(d) 6d – 14 = 4d – 8
(e) 12p – 20 = 3p + 7
(f) –p + 9 = –5p – 7
(g) 3x + 2  2x – 11
(h) 72m – 1  4m  13
(i) 3 – 3p = 32p – 8
(j) 4t + 5  2t – 4 
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(k) 9x – 5  6x
(l) –2w – 12  2w + 6
(m) –12  3r – 1 – 2r + 7   0
(n) 2m + 5  32m – 1  23
6
Solve each of the following equations. Check each solution.
(a)
m
5
4
(b)
k
 12
2
(c)
p
 10
10
(d)
x +3
2
3
(e)
z–1
5
5
(f)
p7
4
8
(g)
2a – 8
2
5
(h)
3t + 5
2
10
(i)
6k  4
 –2
4
Remember, for each of these create similar fractions first and then solve.
102
(j)
s s
 5
2 3
(k)
2p p
– 3
4 4
(l)
2n 2n

6
5 10
(m)
x –1 x +3

2
4
(n)
s –2 s 5

5
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(o)
2y – 2 y + 1

2
3
More difficult examples
7
For each of these statements, write an equation and then find the solution. Don’t forget
to check your solution.
(a) Think of a number n and subtract 8 from it. The result is 14. What is the value
of n?
(b) Think of a number n and then double it. The result is 34. What is the value of n?
(c) Think of a number n, divide it by 3 and then add 6. The result is 18. What is the
value of n?
(d) Think of a number n and add 12 to it. The result is the same as multiplying the
number n that you thought of by 4. What is the number?
(e) Think of a number n and halve it. Add to that a quarter of n. The result is 6. What
is the number?
(f) The sum of two consecutive numbers n and (n + 1) is 35. What is the value of n?
(g) The sum of three consecutive numbers is 63. What are the three numbers?
Answers to Activity 12
3
(a) c  6
(b) b = 9
(c) p  13
(d) n = 20
(e) w = 48
(f) t = –12
(g) y = –12
(h) k  10
(i) x = 4.4
(j) m  11.7
(k) b  –10
(l) g  1.0
(m) e  27
(n) p  6
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(o) u = 0.8
4
(a) w  1
(b) x = 7
(c) r  1.5
(d) c = 2
(e) k  2
(f) p  2
(g) p  –2
(h) q  2
(i) s  –2
(j) a = 3
(k) m  2.5
(l) m = –6.5
5
(a)
3b  5
3b  5 – 5
3b
3b – 2b
b





2b  7
2b  7 – 5
2b  2
2b – 2b  2
2
(subtract 5 from each side)
(subtract 2b from each side)
(b) a = –3
(c) z = 3
(d) d = 3
(e) p  3
(f) p = –4
(g)
3x  2
3x  6
3x  6 – 6
3x
3x – 2x
x



=


2x – 11
2x – 11
2x – 11 – 6
2x – 17
2x – 17 – 2x
–17
(expanding brackets)
(subtract 6 from each side)
(subtract 2x from each side)
(since 2x – 2x  0)
(h) m = 2
(i) p = 3
(j) t = –14
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(k) x  15
(l) w  –6
(m )
– 12 + 3(r – 1) – (2r + 7)
– 12 + 3r – 3 – 2r – 7
r – 22
r – 22 + 22
r
=
=
=
=
=
0
0
0
22
22
(expand brackets)
(collect like terms)
(add 22 to each side)
(n) m = 2
6
(a) m  20
(b) k  24
(c) p  100
(d) x = 3
(e) z  26
(f) p  25
(g) a  9
(h) t = 5
(i) k  –2
(j) s  6
(k) p  12
(l) n  10
(m )
x –1

2
2x – 1

4
2x – 1 
2x – 2 
2x – 2 + 2 =
2x 
2x – x 
x 
x3
4
x3
4
x  3
x3
x +3+2
x5
x 5 –x
5
(change to equivalent fractions)
(multiply each side by 4)
(expand brackets)
(add 2 to each side)
(subtract x from each side)
(n) s  9
(o) y  2
7
(a)
n – 8  14
n  22
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(b)
(c)
(d)
(e)
(f)
2n  34
n  17
n
 6  18
3
n  36
n  12  4n
n4
n n
 6
2 4
n =8
n  (n  1)  35
n  17
(g) n  n  1  n  2  63
n = 20
The three numbers are 20, 21, 22.
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Finding the solution of quadratic
equations
I hope that you enjoyed the work related to solving simple algebraic
equations like the ones we have looked at so far in this section. You should
now feel confident to move on.
What we now want to investigate is how to find the solution of quadratic
equations similar to these:

x2 + 15x + 56 = 0

x2 + 4x – 32 = 0

x2 – 2x – 24 = 0

x2 – 16x + 63 = 0
What we are interested in here is finding values of x that will make the
algebraic expression on the left-hand side equal to nought when we put that
value of x into the equation.
Usually there are two such values but, for some quadratic equations, there
may be only one value.
Question:
How do we find those values of x?
Answer:
We could guess!
For easy equations, you would probably get the right answer. For more
difficult equations, you could waste a lot of time and get the wrong answer
or no answer at all. Obviously you want to avoid doing that.
Question:
Is there an easy method that will always give you the correct
answers (provided you are careful and think about what you
are doing)?
Answer:
Yes, there is. To find out how to do it, read on.
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Multiplying by nought
As a first step towards finding solutions to quadratic equations like the ones
above, let’s look at the simple idea of multiplying by nought.
Activity 13
1
Multiplying numbers by nought
Write down the answer to each of these:
(a) 0 15 
(b) 0  507 
(c) 0 (–21) 
(d) 0  10
–3

(e) 0  42 =
Did you get nought for every answer? Good!
2
Multiplying pronumerals by nought
Write down the answers to the following. (Remember that when multiplying algebraic
expressions, we often leave out the multiplication sign.)
(a) 0  6q  
(b) 0  7t –14 
(c) 0  9b 2  
(d) 0  4 s 2 – 9  
(e) 0  5a –1  
(f) 0  f 3   0  9 f  
Was every one of your answers nought? Good!
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More multiplying by nought
Write down the answer to each of these:
(a) 3  0 
(b) 35  0 
(c) –986  0 
(d) 105  0 
and these:
(e)
 p 0 
(f)
 k  4 0 
(g)
x 0 
(h)
x
(i)
8a  0   6a  0 
2
2
– 65  0 
2
Again, was every answer nought?
An important idea
Did you also notice in Activity 13 that it didn’t matter on which side of the
multiplication sign the nought was placed. The answer was always nought.
The idea that, if you multiply by nought, then your answer is always
nought is a very important one. From what we have observed, we could
write this important statement:
If you have the equation a × b = 0, then either a = 0 or b = 0
Let’s use this idea in the next activity.
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Activity 14
1
Complete these statements. You’ll have to think about some of them just a little, but
I’m sure that you won’t have any trouble. Remember to check your answers as you go.
(a) In the equation 3a = 0, the only solution is that a = ____
(b) In the equation –34s2  0 , the only solution is that s = ____
(c) In the equation 2p + 4   0 , the only possibility is that p + 4 = 0 .
That is, p = ____
(d) In the equation 9m – 7  0 , the only possibility is that m – 7 = 0 .
That is, m = ____
(e) In the equation n  24   8  0 , the only possibility is that n  24  ____ .
That is, n = ____
(f) In the equation k – 61  56  0 , the only possibility is that k – 61  ____ .
That is, k = ____
2
If we have the equation w w + 5  0 , then there are two possible answers. What do
you think they are?
_____________________________________________________________________
Check below to see if you were correct.
Would you agree that the possible answers are that w could be equal to nought or
(w + 5) could equal nought?
Answers: w  0 or w + 5 = 0
That is, w = 0 or w = –5
Now try these questions. Can you go straight to the answers?
(a) In the equation d – 14  d = 0 , the two possible answers are that d – 14 = ____
or d = ____
That is, d = ____ or d = ____
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(b) In the equation t t + 5  0 , the only possible answers are that t = ____ or t = ____
(c) In the equation c – 12c = 0 , the only possible answers are that c = ____
or c = ____
3
Read each of these statements. See if you agree with each one and then complete it.
(a) In the equation m + 6m + 2 = 0, the only possible answers are m + 6 = 0 or
m+2=0
That is, m = ____ or m = ____
(b) In the equation a + 4 a– 8  0 , the only possible answers are a + 4 = 0 or
a–8=0
That is, a = ____ or a = ____
(c) In the equation n – 10 n  12  0 , the only possible answers are n – 10 = 0 or
n + 12 = 0
That is, n = ___ or n = ____
(d) In the equation (z – 3)(z – 7) = 0 , the only possible answers are z – 3 = 0 or
z–7=0
That is, z = ____ or z = ____
4
Write a simple answer to each of these:
(a) If x = 2 , what is the value of x – 2x + 7? _____
(b) If y = 4, what is the value of y + 6y – 4 ? _____
(c) If k = –2, what is the value of k + 2k + 16? _____
(d) If k = –13, what is the value of k – 2k  13? _____
5
Write answers to each of these:
(a) If w = 3, what is the value of w + 8w – 3? _____
What would be the value of the expression if w = –8? _____
(b) If a = –5, what is the value of a  5 a – 1? _____
What would be the value of the expression if a = 1? _____
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(c) What values of m would make the expression m + 7m + 5 equal
to nought? _____ _____
(d) What values of s would make the expression s  10s  15 equal
to nought? _____ _____
Answers to Activity 14
1
(a) a = 0
(b) s = 0
(c) p = –4
(d) m = 7
(e) n = –24
(f) k = 61
2
(a) d = 14 or d = 0
(b) t = 0 or t = –5
(c) c = 0 or c = 12
3
(a) m = –6 or m = –2
(b) a = –4 or a = 8
(c) n = 10 or n = –12
(d) z = 3 or z = 7
4 (a)
(b)
If x  2, then x – 2x  7   2 – 22  7 
 0(9)
 0
If y  4, then y  6y – 4   (4  6)(4 – 4)
 (10)0
 0
(c)
(d)
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If k  –2, then k  2 k  7   (–2  2)(–2  7)
 0(5)
 0
If k  –13, then (k  2)(k  13)  (–13  2)(–13  13)
 (–11)0
 0
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(a) If w = 3, then w + 8w – 3  0
It also equals zero when w = –8
(b) If a = –5, then a + 5a– 1  0
It also equals zero when a = 1
(c)
m + 7m + 5  0 when m = –7 or m = –5
(d)
s  10s + 15  0 when s = –10 or s = –15
Factorising quadratic expressions
As we said earlier, multiplying by nought is the first mathematical idea that
you need to help you find the solutions of quadratic equations.
The other mathematical idea that you will need is one that we have already
looked at; that is, factorising quadratic expressions. Before going on, let’s
brush up your skills in factorising quadratics.
Activity 15
Factorise the following quadratics:
(a) x 2  9x + 14
Remember: To factorise x 2  9x + 14 , you have to find two numbers that add together to
give 9 and multiply together to give 14.
The two numbers are _____ and _____
The factors are (x _____) and (x _____)
So x 2  9x  14  (______)(_____)
(b) x 2 – x  72
Remember: To factorise x 2 – x –72 , you have to find two numbers that add together to
give –1 and multiply together to give –72.
The two numbers are ____ and ____
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The factors are (x ____) and (x ____)
So x 2 – x – 72  (_____)(_____)
(c) x 2  x – 56
Remember: To factorise x 2  x – 56 , you have to find two numbers that add together to
give 1 and multiply together to give –56.
The two numbers are ____ and ____
The factors are (x ____) and (x ____)
So x 2  x – 56  (_____)(_____)
(d) x 2 – 12x + 35
Remember: To factorise x 2 – 12x + 35 , you have to find two numbers that add together to
give –12 and multiply together to give 35.
The two numbers are ____ and ____
The factors are (x ____) and (x ____)
So x 2 – 12x + 35 = (_____)(_____)
Answers to Activity 15
(a)
7 and 2
(x  7) and (x  2)
(x  7)(x  2)
(b)
–9 and 8
(x – 9) and (x  8)
(x – 9)(x + 8)
(c)
–7 and 8
(x – 7) and (x  8)
(x – 7)(x  8)
(d)
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–7 and – 5
(x – 7) and (x – 5)
(x – 7)(x – 5)
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Now let’s put these ideas together to find the roots of quadratic equations.
Finding the roots of quadratic equations
As a start we’ll work through the first two examples and then you can
complete the solution of the next two examples. These questions use the
quadratics that you have just factorised.
Example
Find the roots or solutions of x 2  9x + 14 = 0
Solution
If x 2 +9x  14  0,
then factorising gives:
(x  2)( x  7)  0
This means that either x + 2 = 0 or x + 7 = 0
Do you agree?
If
x20
then
x  –2
If
then
x 70
x  –7
Now we check the solutions by substituting each of these values in turn in
the expression x2 + 9x + 14
when x  –2, x 2  9x  14  (–2)2  9(–2)  14
 4 – 18  14
0
when x  –7, x 2  9x  14  (–7)2  9(–7)  14
 49 – 63  14
0
So the solutions x  –2 or x = –7 are correct .
Any problems so far?
If you’re a little unsure, go back and work through the example again using
pencil and paper. If you are feeling confident, then go on.
If you did the above calculations using a calculator and did not get the right
answer, it means that you didn’t put brackets around the negative numbers.
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Example
Find the roots or solutions of x 2 – x – 72 = 0
Solution
If x 2  x  72  0,
then factorising gives
:
(x  8)(x  9)  0
This means that either x + 8 = 0 or x – 9 = 0
Do you agree?
If
x80
then
x  –8
If
x – 9 0
then
x9
Now we check the solutions by substituting each of these values in turn in
the expression x 2 – x – 72
when x  –8, x 2 – x – 72  (–8) 2 – (–8) – 72
 64  8 – 72
0
when x  9, x 2 – x – 72  9 2 – 9 – 72
 81 – 9 – 72
0
So the solutions x = –8 or x = 9 are correct.
Example
Find the roots or solutions of x 2  x – 56 = 0
Solution
If x 2 + x  56  0,
then factorising gives
:
(x  8)(x  7)  0
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Now complete the solution for yourself.
This means that ______ = 0 or ______ = 0
The two solutions are x = ______ or x = ______
You should have x = –8 or x = 7
Now check your solutions:
Substitute each of these values of x in turn in the expression x 2  x – 56
when x  –8, x 2  x – 56  (–8)2  (–8) – 56
 64 – 8 – 56
0
when x  7, x 2  x  56 
Do you agree that the solutions x = –8 and x = 7 arecorrect?
Example
Find the roots or solutions of x 2 – 12x + 35 = 0
Solution
If x 2  12x  35  0,
then factorising gives
:
(x  5)(x  7)  0
This means that _____ = 0 or _____ = 0
The two solutions are x = _____ or x = _____
Now check your solutions:
when x  5, x 2  12 x  35 
when x  7, x2  12 x  35 
Do you agree that the solutions x = 5 and x = 7 are correct?
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Example
Find the roots or solution of
2x2 – 4x = 0
Do you remember common factors?
2x(x – 2) = 0
This means that 2x = 0
Therefore x = 0
or x – 2 = 0
or x = 2
Check the solution
Both solutions are correct.
Example
x2 – 16 = 0
Do you remember the difference of two squares
x2 –42 = 0
(x –4)(x + 4) = 0
Check the solution
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Activity 16
1
Check that:
(a) x = –2 and x = –4 are solutions of the equation x 2  6x + 8 = 0
(b) x  –9 and x = –3 are solutions of the equation x 2  12x + 27 = 0
(c) x  8 and x = –5 are solutions of the equation x 2 – 3x – 40 = 0
(d) x = 0 and x = 4 are solutions of the equation x 2 – 4x = 0
(e) x = 6 and x = 7 are solutions of the equation x 2 – 13x + 42 = 0
Note: The next equation will have only one solution since it is a perfect square.
(f) x = 10 is the solution of the equation x 2 – 20x + 100 = 0
2
Solve the following equations that have already been expressed in factor form:
(a)
x – 6x – 9   0
(b)
x – 4x – 12   0
(c)
x – 7x + 3  0
(d)
x + 9x – 1  0
(e)
x – 11x + 2   0
(f)
x – 8x – 4  0
(g)
x – 6x – 6   0
(h) x x – 3  0
3
Solve the following equations:
(a) x 2 – 5x + 6 = 0
(b) x  5x + 6 = 0
2
(c) x 2  10x + 9 = 0
2
(d) x – 4x – 21 = 0
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(e) x 2  19x +18 = 0
(f) x 2  2x – 15 = 0
(g) x 2  8x + 16 = 0
(h) x 2 – 8x + 16 = 0
(i) x 2  11x + 28 = 0
(j) x 2 – 16x + 63 = 0
(k) x 2 – 13x + 22 = 0
(l) x 2  18x + 72 = 0
(m) x 2  19x – 20  0
(n) x 2 – 6x = 0
More difficult examples
Look at this example first.
Example
If the roots of a quadratic equation are –2 and –1, find the equation.
Solution
The factors are (x  2) and (x  1)
So the equation is
or
4
(x  2)(x  1)  0
x 2  3x  2  0
(a) If the roots of a quadratic equation are 7 and 5, find the equation.
(b) If the roots of a quadratic equation are –3 and 4, find the equation.
(c) If the roots of a quadratic equation are 7 and –2, find the equation.
Answers to Activity 16
2
(a) x = 6 or x = 9
(b) x = 4 or x = 12
(c) x = 7 or x = –3
(d) x = –9 or x = 1
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(e) x = 11 or x = –2
(f) x = 8 or x = 4
(g) x = 6 or x = 6
(h) x  0 or x = 3
3
(a) (x – 3)(x – 2)  0
x  3 or x  2
(b)
(x  3)(x  2)  0
x  –3 or x  –2
(c) (x  1)(x  9)  0
x = –1 or x  –9
(d)
(x  3)(x – 7)  0
x  –3 or x  7
(e) (x  1)(x  18)  0
x  –1 or x  –18
(f)
(x – 3)(x  5)  0
x  3 or x  –5
(g) (x  4)(x  4)  0
x  –4
(h)
(x – 4)(x – 4)  0
x 4
(i) (x  4)(x  7)  0
x  –4 or x  –7
(j)
(x – 7)(x – 9)  0
x  7 or x  9
(k) (x – 2)(x – 11)  0
x  2 or x  11
(l)
(m)
(n)
(x  6)(x  12)  0
x  –6 or x  –12
(x – 1)(x  20)  0
x  1 or x  –20
x(x – 6)  0
x  0 or x  6
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4
(a) Factors are x – 7, x – 5
Equation is x 2 – 12x + 35 = 0
(b) Factors are x + 3, x – 4
Equation is x 2 – x – 12 = 0
(c) Factors are x – 7, x + 2
Equation is x 2 – 5x – 14  0
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More on solving equations
Let’s start by considering several questions and then discussing their
solutions. Question 2 and Question 3 are a little more challenging and will
need just a little more thinking about. You should be able to follow all the
solutions quite easily.
Question 1:
The sum of two numbers is 16 The difference between the
two numbers is 2 Find the two numbers.
Question 2:
The cost of one pencil and one pen is $0.95 The cost of
one pencil and two pens is $1.41 Find the cost of each.
Question 3:
Find the perimeter and the area of this rectangle.
Simultaneous equations
If you look at each of the questions carefully, you will see that each contains
two pieces of information. To find each answer to each question, we need to
consider both pieces of information.
Go back and read each of the questions carefully and decide for each
question what are the two pieces of information that have been given.
Now check to see if you were correct.
Question 1:
You were told the sum of the two numbers and the
difference between the two numbers.
Question 2:
You were told the cost of one pencil and one pen and
you were also told the cost of one pencil and two pens.
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Question 3:
You were given two expressions for the length of the
rectangle and you were given two expressions for the
width of the rectangle.
Now for each piece of information (in each question) a mathematical
equation can be written. Since the two pieces of information are given in the
same question, the two equations are called simultaneous equations.
Solution to question 1
I’m sure that you can tell me what the answer is going to be; however, let’s
see how the answer is worked out mathematically. Knowing the answer will
help check our solution.
Let’s pretend that we don’t know what the two numbers are. Then we can
use pronumerals to represent those two numbers.
Let x be one number and y the other number.
Then from the information given, we can write two mathematical equations.
(These equations are usually numbered.) The equations are:
x + y = 16 (1)
x – y = 2 (2)
Remember here that x in the two equations stands for the same number.
Similarly y in the two equations stands for the same number.
These equations (together) can be solved in one of two ways. The first way
of solving these equations is called the substitution method.
First we need to find x in terms of y. This means we solve the equation to
get x by itself.
x–y=2
x–y+y=2+y
x= 2 + y
Substitution method
From equation (2), we have x = 2 + y
Now we substitute or put this expression for x in equation (1). This gives us:
2 + y + y = 16
2y = 14
y=7
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Since y = 7, then from equation (1) we have:
x  7  16
x 9
Therefore, the solution is x  9, y = 7.
Check that these values for x and y satisfy equation (2).
Therefore, the two numbers in Question 1 are 9 and 7.
The second way of solving these equations is called the elimination method.
Elimination method
With this method, we attempt to eliminate either x or y from the equations.
We usually do this by adding the two equations together or by subtracting one
equation from the other. Which of these two we use depends on the signs (+
or –) that are attached to x and/or y. The equations are:
x + y = 16 (1)
x – y = 2 (2)
In relation to these two equations, you can see that there is a positive y in
equation (1) and a negative y in equation (2). Therefore, if we add the two
equations together, the y’s will be eliminated.
Adding (1) and (2) we get
x + x  y – y  16  2
2x  18
x 9
Since x = 9, then from equation (1) we have:
9  y  16
y 7
Therefore, the solution is x  9, y = 7.
Check that these values for x and y satisfy equation (2).
Now let’s go back and check that we have answered Question 1. The sum of
9 and 7 is 16 and the difference between 9 and 7 is 2. Correct!
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Therefore, the two numbers are 9 and 7.
Note: Take another look at the equations. Subtracting them would have
eliminated x. Try it.
Solution to question 2
Let x cents be the cost of the pencil and let y cents be the cost of the pen.
From the statements in the question, we have the following two
mathematical equations:
x + y = 95
x + 2y = 141
(1)
(2)
Substitution method
From equation (1), we have x = 95 – y (here x is in terms of y).
Substituting this value for x in equation (2), we have:
95 – y  2y
95  y
y
y
 141
 141
 141 – 95
 46
Since y = 46, then from equation (1) we have:
x  46  95
x  49
Therefore, the solution is x  49, y = 46
So the pencil cost 49 cents and the pen cost 46 cents.
Check that these values for x and y satisfy equation (2). Note: We could
have obtained a value for x from equation (2) and substituted this in
equation (1). Doing this would have given you the same answer. Try this for
yourself.
Elimination method
The equations are:
x + y = 95
x + 2y = 141
(1)
(2)
Observe that we have a single x in both equation (1) and equation (2) so let’s
eliminate the x’s by subtracting equation (1) from equation (2).
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Subtracting (1) from (2), we have:
x + 2y – x + y   141– 95
x  2y – x – y  46
y  46
Since y = 46, then from equation (1) we have:
x  46  95
x  49
Therefore, the solution is x  49, y = 46.
Check that these values for x and y satisfy equation (2).
Now let’s go back and check that we have answered Question 2. The cost of
one pencil and one pen is 49c + 46c which equals 95c and the cost of one
pencil and two pens is 49c + 92c which equals 141c = $1.41 Correct!
Therefore, one pencil costs 49c and one pen costs 46c.
Solution to question 3
Using the figure below, form two equations involving x and y. Find the
values of x and y and then find the area and the perimeter of the rectangle.
Note: The figure is a rectangle, so the opposite sides are equal.
Using this, we can write down the following two equations:
x + y = 17
2x + y = 29
(1)
(2)
Substitution method
From equation (1), we have y = 17 – x
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Substituting or putting this expression for y in equation (2), we have:
2x + 17 – x   29
2x  17 – x  29
x + 17 = 29
x = 12
Since x = 12, then from equation (1) we have:
12  y  17
y=5
Therefore, the solution is x  12, y = 5.
Check that these values for x and y satisfy equation (2).
Elimination method
The equations are:
x + y = 17
2x + y = 29
(1)
(2)
Subtracting equation (1) from equation (2), we have:
2 x +y  29
  
x  y  17
x  12
Since x = 12, then from equation (1) we have:
12  y  17
y 5
Therefore, the solution is x  12, y  5.
Check that these values for x and y satisfy equation (2).
Now let’s go back and check that we have answered question 3.
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length
Therefore:
perimeter
area
= 2x  y
 2  12  5
 29 cm
=
=
width
 xy
 12  5
 17 cm
2  29 + 2  17
92 cm
 29  17
 493 cm 2
Note: After you have solved the equations, go back and read the question to
find out what they wanted you to find—in this case the perimeter and area.
Some more examples using the elimination method:
Since the y’s have different signs, we add the equation.
Put this in either equation, but here we’ll choose equation 2.
Therefore the solution is
x = 3;
y = 1
Test these yourself. Do they satisfy both equations?
Neither adding or subtracting will eliminate the y’s this time because the
number of y’s is different in both equations.
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We can fix this by multiplying both sides of (1) by 2. Call the new
equation (3).
put this into (1)
Solutions are x = 2;
y = 4
Now check these solutions in the equations.
Activity 17
1
130
Use the substitution method to find the value of the second pronumeral in each of
these pairs of simultaneous equations:
(a)
p  q  13
q 4
(b)
a b  25
b  13
(c)
m – n  60
m  92
(d)
r  s  14
s  –8
(e)
z – y  –33
z  –6
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(f)
2
3
–c  d  –12
d  10
Use the substitution method to find the value of the two pronumerals in each of these
pairs of simultaneous equations:
(a)
x  y  13
y x 5
(b)
a b  26
a  b  20
(c)
p – q  16
p  2q  8
(d)
r  2s  11
s r 1
(e)
2t – u  3
t 9 –u
(f)
w – 5v  4
w  2v  16
Use one of the two equations (you decide which one you will use) to express y in terms
of x. Then use the substitution method to find the values of the pronumerals.
(a)
x –y 6
x  y  14
(b)
x  y  21
x – y  15
(c)
x – 2y  6
x  y  15
(d) 2x  y  40
– x  y  –2
(e)
x – y  28
2x  y  38
(f)
3x  2y  69
x  y  26
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4
5
Use the elimination method to solve these pairs of simultaneous equations by first
adding the equations together.
(a)
x  y  20
x – y  10
(b)
a– b  6
a b  22
(c)
p  q  66
–p  q  –2
(d)
z  2y  26
z – 2y  –2
(e)
2f – g  28
f  g  20
(f)
–s  t  8
s  3t  72
Use the elimination method to solve these pairs of simultaneous equations by first
subtracting one equation from the other.
(a)
2m  n  10
m n6
(b)
y z 1
2y  z  7
(c)
c  2d  32
cd4
(d)
2y – 3z  13
2y – 5z  7
(e)
r  4s  9
r – 2s  3
(f)
2k  j  0
2k  2 j  –15
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More difficult examples
6
Find the solution to these questions:
(a) The sum of two numbers is 27 and the difference between the numbers is 9. Find
the two numbers.
(b) Shaun is 8 years older than Maria. If the total of their ages is 64 years, find the age
of each person.
(Hint: Let the age of Shaun be x years and of Maria be z years.)
(c) For the figure below, form two equations and find the values of m and n.
(d)
(i)
What type of triangle is shown?
(ii)
What is the size of each base angle?
(iii) Find the value of x and y.
(e) A resort offers two holiday packages: 10 nights plus airfare for $1520 or 14 nights
plus airfare for $2000. Find the cost per night and the airfare.
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Answers to Activity 17
1
(a) p  9, q = 4
(b) a = 12, b = 13
(c) m = 92, n = 32
(d) r = 22, s = –8
(e) y  27, z = –6
(f) c = 22, d = 10
2
(a) x = 4, y = 9
(b) a  23, b = 3
(c) p = 24, q = 8
(d) r = 3, s = 4
(e) t = 4, u = 5
(f) v  4, w = 24
3
(a) x = 10, y = 4
(b) x = 18, y = 3
(c) x  12, y = 3
(d) x = 14, y = 12
(e) x = 22, y = –6
(f) x  17, y = 9
4
(a) x = 15, y = 5
(b) a = 14, b = 8
(c) p  34, q = 32
(d) y  7, z = 12
(e) f  16, g  4
(f) s  12, t = 20
5
(a) m  4, n = 2
(b) y  6, z = –5
(c) c = –24, d = 28
(d) y  11, z = 3
(e) r = 5, s = 1
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(f)
6
j  –15, k = 7
1
2
(a) Numbers are 18 and 9.
(b) Maria is 28 years old and Shaun is 36 years old.
(c) m = 14, n = 7
(d) (i)
isosceles triangle
(ii)
each angle is 45
(e) Airfare = $320; nightly rate = $120
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Check your progress 3
1
Solve each of these simple equations:
(a) 3n = 27
(b) b – 7 = 0
(c)
t–
7 15

2 2
(d) k + 9 = 17
(e) w + 6.8 = 10
(f)
p
4
8
(g)
2m
 12
3
(h) 18  k + 14
(i) 13  w – 3
(j) 16  10y
2
Solve these equations:
(a) 3k + 34 = 40
(b) 8  18  5t
(c) 2x + 4 = x – 5
(d) 3q – 16 = q – 8
(e) t + 9 = 19 – 4t
(f) 5z – 2  15
(g)
m
 6  11
2
(h) 3x = x
(i)
136
n n
 6
2 4
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(j)
3t  2 t  11

5
5
(k)
(2r  6)
3
8
(l)
3
z–
z
6
4
Solve the following equations. First expand and collect like terms.
(a) 2b + 2   3b + 2  20
(b) 42p – 4 p  2  0
(c) 53k + 6  33k – 13
(d) 8x – 32x – 9  3
4
Solve the following quadratic equations:
(a) y 2  6y + 9 = 0
(b) p 2 – 5p – 50 = 0
(c) n 2  6n – 27 = 0
(d) c 2  4c – 5 = 0
(e) h 2 – 49  0
(f) b2  8b + 12 = 0
(g) m 2 – 11m  30  0
(h) r 2 – 2r – 63  0
5
Find the solution to these questions:
(a) Forty three adults attend a meeting. It is known that five more women than men
attended the meeting.
(i) If m men attended the meeting, how many women attended the meeting?
(ii) Write an equation that you can solve to determine how many women and men
attended the meeting. Then solve your equation to find the number of women
and men who attended the meeting.
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137
(b) The length of a rectangle is four times its width.
Write down an expression for the perimeter of the rectangle and an expression for
the area of the rectangle.
If the perimeter is 60 cm, what is the area?
6
7
Solve the following simultaneous equations using the elimination method:
(a)
y  z  13
y –z 3
(b)
b  c  12
2b – c  6
(c)
2c  3d  16
2c – 4d  9
Solve the following simultaneous equations using the substitution method:
(a)
w  6v
w  30 – 4v
(b)
3m  n  13
m n9
(c)
x – 2y  5
x  2y  1
Answers to Check your progress 3
1
(a) 9
(b) 7
(c) 11
(d) 8
(e) 3.2
(f) 32
(g) 18
(h) 4
(i) 16
(j)
138
16 8

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2
(a) 2
(b) –2
(c) –9
(d) 4
(e) 2
(f) 5
(g) 10
(h) 0
(i) 8
(j)
9
2
(k) 9
(l) 8
3
(a) b  2
(b) p  2
(c) k =
–23
2
(d) x  –12
4
(a) y  –3
(b) p  10 or p  –5
(c) n  –9 or n  3
(d) c  –5 or c  1
(e) h = 7 or h = –7
(f) b = –6 or b = –2
(g) m = 6 or m = 5
(h) r = 9 or r = –7
5
(a) (i) If there are m men, then there are (m + 5) women, since there
are 5 more women than men.
(ii) Since the total number of adults is 43, then:
m + m + 5  43
2m + 5 = 43
2m = 38
m = 19
There are 24 women and 19 men.
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139
(b) Let the width of the rectangle be W.
Then the length of the rectangle is 4W.
Therefore:
Perimeter P = 4W + W + 4W + W
= 10W
Area A = 4W×W
= 4W2
If the perimeter is 60, then
10W = 60
W=6
Therefore: A = 4 × 62
= 4 × 36
= 144
6
(a) y = 8 and z = 5
(b) b = 6 and c = 6
(c) d = 1 and c 
7
13
2
(a) v = 3 and w = 18
(b) m = 2 and n = 7
(c) x = 3 and y = –1
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Algebraic formulas
Introduction
So far you have investigated a variety of algebraic expressions and
equations. In doing so you have developed skill in manipulating terms in
expressions and finding solutions of equations.
In this final topic of Algebraic Processes and Pythagoras, you will learn
how to evaluate the subject of a formula, change the subject of a formula
and evaluate an unknown variable in a simple formula.
When you have completed this topic, you will be ready to do the last
assessment.
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Algebraic formulas in real life
In real life there are many algebraic formulas that people use daily in
industry, science and mathematics. The pronumerals in these formulas
represent real-life physical quantities.
Formulas simply show the relationships between the pronumerals involved.
Some formulas are very simple while others are quite complex. Every year
new formulas are being developed from the needs of industry and science.
Most formulas representing the relationship among physical quantities are
given or remembered in a standard form. For example, the formula for the
area A of a rectangle of length L and breadth B is given as A  L  B .
In relation to this formula, we are usually given the length and breadth of a
rectangle and asked to find its area. If the area A is what has to be found,
then A is called the subject of the formula.
Note: The subject of any formula is usually written by itself on the left-hand
side of the equals sign while the remaining numerals and pronumerals are
written on the right-hand side of the equals sign.
Some examples
How many formulas do you know and use every year?
Here are a few simple examples you probably recognise or have used at
some stage:
142
1
A  LB
(Formula for the area of a rectangle)
2
P  2L  B
(Formula for the perimeter of a rectangle)
3
C  2r
(Formula for the length of the circumference of a
circle)
4
V  Ah
(Formula for the volume of a prism. A is the area
of the base of the prism and h is the height of the
prism)
5
D  ST
(Formula for the distance travelled at a certain
speed S over a period of time T)
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6
E  mc2
7
I
Prn
100
(Formula for the energy contained in a given mass
m; c is the speed of light which is approximately
300 000 km/s. This is Einstein’s famous formula)
(Formula for calculating interest. P is the
principal or amount invested; r is the rate of
interest and n is the time)
Activity 18
For each of the formulas above (1 to 7), write down the subject of the formula.
Answers to Activity 18
1
The subject of the formula A  L  B is A
2
The subject of the formula P  2L  B  is P
3
The subject of the formula C  2r is C
4
The subject of the formula V  Ah is V
5
The subject of the formula D  ST is D
6
The subject of the formula E  mc2 is E
7
The subject of the formula I 
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Prn
is I
100
143
Evaluating formulas
To find the value of the subject of a formula, you merely substitute in the
value of each of the other pronumerals, which you must know.
Example
Given A  L  B , find the value of A when L = 4.9 and B = 5.8
Here
A = 4.9  5.8
= 28.42
Example
Given this formula:
S
n
2a  n – 1d
2
find S when n  16, a  3, d  5
Here
S =
16
2  3  16 – 1  5
2
 8 6  15  5
 648
Example
Write down a simple formula that will enable you to find the cost (C) of p
peaches at 25 cents each. Find C when p = 6.
Here
C  p  25
When p = 6,
C  6  25 cents
 150 cents
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Activity 19
1
Study each of these formulas.
Write down the subject of each formula and then find its value, for the given value of
the pronumerals.
Feel free to use a calculator when you want to.
(a) If P = 2L + 2B, find the value of P when L = 3.6 and B = 9.7.
(b) It is known that I 
Prn
100
Find I when P = 1000, r = 5.5 and n = 9.
(c) Given that v = u + at, evaluate v when u = 12.4, a = 6.8 and t = 5.
(d) If s 
u  v 
2
 t , find the value of s given u = 34, v = 67 and t = 60.
(e) To convert temperature from degrees Centigrade to degrees Fahrenheit, we use the
9C
formula F 
 32 .
5
Find the temperature in degrees Fahrenheit when the temperature is 27°C.
(f) The energy of a particle of mass m which is related to its speed v (called kinetic
1
energy) is given by the formula E  mv 2 . Find the kinetic energy of a particle
2
given that m = 8 and v = 7.5.
(g) The distance s travelled by a body over a given time t is given by the formula
1
s  ut  at 2 where u is the initial velocity and a is its acceleration. Find the
2
distance travelled given that u = 30, a = 4.5 and t = 20.
(h) If v2 = u2 + 2as, find v given that u = 12, a = 6 and s = 27.
4 3
r gives the volume V of a sphere of radius r. Find the volume of a sphere
3
of radius 5 cm. Use the  key on your calculator.
(i) V 
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145
2
Write down a formula that relates the following pronumerals. Then find the value of
the subject of the formula given the value of the other pronumeral(s).
(a) The number of minutes (M) in H hours: find M when H = 43.
(b) The distance D travelled by a car in T hours travelling at K km/h: find the distance
travelled if T = 5.5 and K = 80.
(c) The perimeter P of an isosceles triangle with two sides having length m and one
side of length n: find the perimeter of the triangle where m = 9.8 and n = 6.7.
(d) The profit P made on an item that is bought for B and sold for S: find the profit
when B = $1237 and S = $1703.
3
Evaluate each of these. Use your calculator if you want to. Check your answers.
1
h(mn)
3
(a)
V=
(b)
T  2
Find V when h = 24, m = 64 and n = 10.7.
Find T when
= 22.5 and g = 10.
g
Use the  key on your calculator.
4
nn  1n  2
6
(c)
T
(d)
W  dr 2h
(e)

r 
A  P 1
 100
(f)
V=
4 3
r
3
Find T when n = 25.
Find W when d = 4.5, r = 0.75 and h = 8.9.
n
Find A when P = 10 000, r = 20 and n = 2.
Find V when r = 3.5.
For each of these figures, construct a formula using the given pronumerals that would
enable you to determine the area of each figure, once the value of each pronumeral is
known.
Find the area of each figure when x = 6.7 and y = 4.82.
(a)
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(b)
(c)
(d)
More difficult examples
5
(a) The perimeter of this triangle is given as D cm.
Complete this formula:
D=
which expresses D in terms of
p and b.
Find D if p is 34 cm and b is
13 cm.
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147
(b) Two angles of a triangle are given as x° and y°. If the third angle is given as z°,
complete this formula:
z=
which expresses z in terms of x
and y.
Find z if x = 56° and y = 67°.
(c) Consider this box. The total length of all of its edges is T cm.
If the box has a length of m cm, a
width of m cm and height of p cm,
complete this formula:
T=
which expresses T in terms of m
and p.
Find T if m = 8 and p = 4.
(d) If the volume of the box in question 5(c) is V cm3, complete this formula:
V=
which expresses V in terms of m and p.
Find the volume of the box when m = 8 and p = 4.
More difficult examples
6
(a) We are told that the sum N of the cubes of the first n whole numbers 13, 23, 33, 43,
53, 63…. is given by this formula:
N=
1 2
n (n + 1)2
4
Check that this formula is correct for:
(i)
13 + 23 (here n = 2)
(ii)
13 + 23 + 33 (here n = 3)
(iii) 13 + 23 + 33 + 43 (here n = 4)
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(b) Find the sum of:
7
(i)
the cubes of the first ten whole numbers
(ii)
the cubes of the first 20 whole numbers.
(a) The sum N of the first n whole numbers 1, 2, 3, 4, 5, 6,…. is given by this
formula:
N=
1
n(n + 1)
2
Check that this formula is correct by finding:
(i)
the sum of the first two whole numbers
(ii)
the sum of the first three whole numbers
(iii) the sum of the first four whole numbers
(iv) the sum of the first five whole numbers
by adding the numbers together and by using the formula.
(b) Find the sum of:
(i)
the first ten whole numbers
(ii)
the first 20 whole numbers
(iii) the first 50 whole numbers
(iv) the first 100 whole numbers.
8
Consider the set of numbers 2, 9, 16, 23, 30, …
Write down the next four numbers in the set.
Then write down a formula for the nth number in the set.
Check your formula by putting n equal to each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9
in turn.
Answers to Activity 19
1
(a) 26.6
(b) 495
(c) 46.4
(d) 3030
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149
(e) 80.6
(f) 225
(g) 1500
(h) 21.63 (rounded off to 2 decimal places)
(i) 523.60 cm3 (rounded off to 2 decimal places)
2
3
(a)
M = 60H
When H = 43, then M = 2580
(b)
D = TK
When T = 5.5 and K = 80, then D = 440
(c)
P = 2m + n
When m = 9.8 and n = 6.7, then P = 26.3
(d)
P=S–B
When S = $1703 and B = $1237, then P = $466
(a) 5478.4
(b) 9.42
(c) 2925
(d) 70.77 (correct to two decimal places)
(e) 14 400
(f) 179.59 (correct to two decimal places)
4
(a) Area of A = (y + 3)(x + 2)
= 68.03
(b) Area of B = (x + 6)(x + 6)
= 161.29
(x – 4)(y – 3)
2
= 2.457
(c) Area of C =
(d) Area of D = (2x – 2)(x + 2) + (x – 1)(x + 1)
= 143.07
5
(a) D = 2p + b
= 81
(b) z = 180 – x – y
= 57
(c) T = 8m + 4p
= 80
(d) V = m2p
= 256
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6
(b) (i) 3025
(ii) 44 100
7
(b) (i) 55
(ii) 210
(iii) 1275
(iv) 5050
8
The next four numbers in the set are: 37, 44, 51, 58.
Then the nth number is given by the formula N = 7n – 5.
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Transformation of formulas
As was pointed out earlier, most formulas are usually given or remembered
in a standard form, for example A = L ×B. Here A is called the subject of the
formula.
For a particular problem, however, it may be more convenient to express the
formulas differently. For example, suppose that you are given the area A of a
rectangle and also its length L and are asked to find its breadth B. In this
case B becomes the subject of the formula and you therefore need to write B
in terms of the other pronumerals.
To find B we start off by writing our well-known formula:
A  L  B (This is often written as A = LB.)
Now, dividing each side by L, we have:
B
A
L
This is a formula whose subject is B. Then, given a value for A and a value
for L we can calculate a value for B.
This process is called changing the subject of the formula.
Exercise
(a) Given that the length of a rectangle is 15 cm and the width is 12 cm,
show that the area is 180 cm2.
Hint: Use A = LB.
(b) Given that the area of a rectangle is 84 cm2and the breadth of the
rectangle is 7 cm, find the length of the rectangle.
Hint: This time change the subject and use L 
152
A
.
B
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Did you write this?
(a)
 LB
 15  12
 180
A
Area = 180 cm2
(b)
A
B
84

7
 12
L 
Length = 12 cm
Well done!
Exercise
The volume V of a rectangular block whose length is L, width is B and
height is H is given by this formula:
V = LBH
Given that the volume of the block is 360 cm3, the length is 9.4 cm and the
breadth is 4.7 cm, show that the height of the block is 8.15 cm (correct to
two decimal places).
Hint: Here make H the subject by dividing each side by LB.
Did you write this?
V
LB
360

9.4  4.7
H
(Remember to put brackets around the bottom line when using the calculator.)
 8.15
Height is 8.15 cm.
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Good!
153
Tips for changing the subject of
formulas
To change the subject of formulas that you will come across, you may
need to:

add or subtract the same quantity from each side

multiply or divide each side by the same quantity

square each side of the formula

take the square root or the cube root of each side.
Study these examples.
Example
Consider the formula v = u + at. Make:
(i) u the subject; and
(ii) t the subject.
If v  u  at,
(i)
then v – at  u
(subtracting
at from each side)
and u  v – at
(ii)
If v  u  at ,
then v – u  at
v–u
t
a
v–u
and t 
a
(subtracting u from each side)
(dividing each side by a)
Example
Given the formula F 
qm
, make:
r
(i) m the subject, and
(ii) r the subject.
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If F 
(i)
qm
r
,
then Fr  qm
and m 
Fr
q
If F 
qm
(ii)
r
(multiplying each side by
(dividing each side by
q)
,
then Fr  qm
and r 
r)
(multiplying each side by
qm
F
(dividing each side by
r)
F)
Example
Make h the subject of the formula A  2rr  h .
If A  2 r (r  h),
then A  2 r 2  2 rh
A  2 r  2 rh
2
A  2 r 2
h
2 r
A  2 r 2
and h 
2 r
(multiplying each term inside the brackets by 2 r )
(subtracting 2 r 2 from each side)
(dividing each side by 2 r )
Example
Make r the subject of the formula for volume of a cylinder:
If V   r 2 h,
V
then
 r2
h
V
r
h
and r =
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(dividing each side by  and h)
(taking the square root of each side)
V
h
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V = r2h.
155
Example
the subject of the pendulum formula: T  2
Make
If T  2
T

2
then
T2
 2 
2

2

T 2g
 2 
and

g
g
.
,
(dividing each side by 2 )
g
(squaring each side)
g
(multiplying each side by g )
T 2g
 2 
2
Activity 20
1
Make x the subject of each formula. After each answer, check that you were right.
Rework any question where you got an incorrect answer.
(a) z = y + x
(b) b = cd + x
(c) z = abc – x
(d) q = px
(e) s = a2x + t
(f) s = 2yx – rt
(g)
x
=p
r
(h) a 
156
t
3x
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(i) n 
mx
v
(j) c = 4(x + 5)
(k) t = 2(7 – x)
2
(l) y = x + 4
2
Change the subject of each of the following formulas to the indicated pronumeral (in
brackets). Check each of your answers as you go!
(a) c  ax  b (x)
(b) F = ma (a)
(c) P  2  2b (b)
(d) v = 3(u – 2) (u)
(e) 8m = 12 – 4n (n)
(f) p =
k
– 3 (k)
4
(g) F = 1.8C + 32 (C)
(h) b = 60t – 8 (t)
(i) I =
Prn
(P)
100
2
(j) ax = y (y)
2
(k) P = RI (I)
In the following questions, substitute the given values then solve the equation.
3
(a) The area of a triangle is given by A =
1
bh.
2
Find b if A = 9 and h = 12.
(b) If v = u + at, find t when v = 18, u = 10 and a = 4.
(c) If s =
1
7
(u + v)t, find v if s = 84, u = 9 and t = .
2
2
(d) If A 
h
b  b , find b2 when b1 = 50, A = 300 and h = 4.
2 1 2
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(e) Given that F = 1.8C + 32, find C when F = 82.
(f) Einstein’s formula states that E 
1 2
mc . Find c when E = 12.5 and m = 2.5.
2
How are you going so far?
If you have any problems, read over the notes again and rework any questions that caused
you trouble. If you are feeling confident, then try Question 4.
More difficult examples
4
(a) Given that A = (R2 – r2)h, find R when A = 21, r = 1.3 and h = 6.
(b) Given that P =
(c) If f =
(d) If
5
2R – V
, find V when P = 0.6 and R = 10.
5
vu
, find v when f = 40 and u = 50.
v+u
x x x
  = 1, find b when x = 25, a = 40 and c = 30.
a b c
If D = A – a, then find the value of:
(i) A + a
(ii)
A+a
given that D = 60 and a = 20.
6
If A = LB, then find the value of:
(i) L
(ii) 2(L + B)
given that A = 25 when B = 10.
7
If Q =
X
, then find the value of:
Y
(i) XY
(ii)
XY
given that Y = 34 when Q = 12.
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8
Given that R =
P
, find the value of:
Q
(i) Q
(ii) (P + Q) 2
if R = 2.4 when P = 12.
9
(a) Write down a formula that relates:
(i) the perimeter P of a square, and
(ii) the area A of a square
to the length L of the side of the square.
(b) If the perimeter of a square is 20.8, find the area of the square.
10 (a) Write down a formula that relates the volume V of a cube to the length L of the
side of the cube.
(b) Find the length of the side of the cube whose volume is 216 cubic centimetres.
(c) Consider a cube with sides of length L and volume V. If you create a cube whose
sides are double this length, what do you think the volume of the new cube will
be? Take a guess!
(d) Now see if you were right in part (c) by calculating the volume of a cube whose
sides are double the length of the sides of the cube that you investigated in part (b)
above. What is the volume of this cube?
11 (a) A box without a lid is cm long, cm wide and h cm high. Show that the surface
area A of the outside of the box is given by this formula:
A = ( + 4h)
(b) Use this formula to find the surface area of the outside of an open box that is 6 cm
long, 6 cm wide and 8 cm high.
(c) If an open box has a surface area of 44k2 (square units) and the open box is 4k cm
long and 4k cm wide, what is the height of the box?
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159
Even more difficult examples
12 (a) Consider the numbers 3, 7, 11, 15, 19, 23, 27.
The nth number in the sequence is given by the formula N = 4n – 1. For n = 1 (ie
the 1st number in the sequence) this statement is true because when n = 1, we
have N = 4(1) – 1
=3
and when n = 2, N = 4(2) – 1
=7
Verify that the formula is true for the 3rd, 4th, 5th, 6th and 7th numbers.
What is the 50th number in the sequence?
What is the 100th number?
(b) The nth number in this sequence is known to be 131. What position is this number
in the sequence? That is, what is the value of n?
13 (a) Consider the sequence of odd numbers 1, 3, 5, 7, 9,…….
The sum of the first n odd numbers is given by the formula N = n2.
Verify that this formula is true for the sum of the first two odd numbers. Then
verify that it is true for:
(i) the sum of the first three odd numbers
(ii) the sum of the first four odd numbers
(iii) the sum of the first five odd numbers.
What is the sum of:
(i) the first 25 odd numbers
(ii) the first 50 odd numbers
(iii) the first 100 odd numbers?
(b) The sum of the first n odd numbers is 729. How many odd numbers were added to
give that total?
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Answers to Activity 20
1
(a) x = z – y
(b) x = b – cd
(c) x = abc – z
(d) x =
q
p
(e) x =
s –t
a2
(f) x =
s  rt
2y
(g) x = rp
(h) x =
t
3a
(i) x =
nv
m
(j) x =
c – 20
4
(k) x =
14 – t
2
(l) x =
2
y–4
c –b
a
F
(b) a =
m
(a) x =
(c) b =
P –2
2
(d) u =
v 6
3
(e) n =
12  8m 4(3–2m)

 3  2m
4
4
(f) k = 4p + 12
(g) C =
(h) t =
F – 32
1.8
b8
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161
3
(i) P =
100I
rn
(j) y =
ax
(k) I =
P
R
(a) b = 1.5
(b) t = 2
(c) v = 39
(d) b2 = 100
(e) C = 27.78 (correct to two decimal places)
(f) c = 3.162 (correct to three decimal places)
4
(a) 2.278 (correct to three decimal places)
(b) 18.20
(c) v = 200
(d) b = –54.55 (correct to two decimal places)
5
(i) 100
(ii) 10
6
(i) 2.5
(ii) 25
7
(i) 13 872
(ii) 117.78
8
(i) 5
(ii) 289
9
(a) (i) P = 4L
(ii) A = L2
(b) A = 27.04
10 (a) v = L3
(b) 6
(c) eight times the volume of the original cube
(d) 1728 cm3
11 (b) 228 cm2
(c) h =
162
7k
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12 (a) 50th number is 199 and the 100th number is 399.
(b) 131
n
= 4n – 1
= 33
13 (a) (i) 625
(ii) 2500
(iii) 10 000
(b) n2 = 729
n 
If you got down to and tried questions 4–11, well done and congratulations
on giving them a go. If you tried questions 12 and 13, then double
congratulations!
I hope that you found this section interesting and just a little challenging in
parts. Remember, that the important thing is to try questions by at least
thinking about them. Getting the right answer is a bonus.
Congratulations! You have now finished the module resource for Algebraic
Processes and Pythagoras. We hope you enjoyed it.
When you feel you are ready, you can move on to the Check your progress
questions that follow. They will help you to review the whole section before
you do your final Module assessment task.
Keep up the good work! Now that you’ve come this far, we know you’ll do
very well.
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Check your progress 4
1
Find the value of w if x = 6 and y = 8:
(a) w = 4x + 2
(b) w = 7x + y
(c) w 
x y

2 2
(d) w = 5y 2
(e) w = 2x 2 y
(f) w = xy – yx
(g) w = x(y – x)
2
(h) w = x – y
(i) w  xy 
2
2
yx
2
If m = 3, n = 6, p = 8 and q = 0, find the value of r:
(a) r 
n–m
nm
p2
(b) r  2
m
3
(c) r 
48
mn
(d) r 
m–q
pn
(e) r 
mpq
20
(f) r 
1 1

m n
(a) If b = 5c + 8 and c = 6, find b.
2
(b) If y = 2x – 9 and x = 5, find y.
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(c) If y = (x + 6)(x – 7), find y when:
(i) x = 3
(ii) x = –6
(iii) x = 7
(d) If cd = 48 and d = 8, find the value of:
(i) c
(ii) c 2
(e) If b  24 and
 3 , find the value of:
(i) b
(ii) 2  b
(f) If S = n(n + 1), find S when n = 16.
4
(g) If T 
nn  1n  2
, find T when n = 5.
6
(h) If A 
hx  y 
, find A when h = 24, x = 6 and y = 10.
2
(a) If x°, y° and z° are the angles of a triangle, then x + y + z = 180 Make y the
subject of the formula.
(b) Given the formula r  2n – 4, make n the subject of the formula.
(c) The perimeter of a rectangle with length and breadth b is given
P  2  b. Make b the subject of the formula.
(d) Make t the subject of the formula s 
by the formula
uv
t .
2
(e) Make r the subject of the formula for the area of a circle, A  r2 .
(f) Change the subject of the formula V  bh to h.
gt 2
(g) Change the subject of the formula s 
to t.
2
5
(a) The area of a triangle is given by A 
1
bh. .
2
Find b if A = 27 and h = 9.
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165
(b) If v = u + at, find t when v = 18, u = 10 and a = 4.
(c) If s 
1
7
u  v t , find v if s = 84, u = 9 and t = .
2
2
(d) If s  ut 
1 2
at , find t if a = 10, u = 0 and s = 125.
2
(e) To convert k kilometres to m miles we use the formula:
How many kilometres are equivalent to 125 miles?
(f) A computer repair company charges $70 for the first hour and $30 for each hour
after that. This can be expressed by:
C = 70 + 30 (h – 1)
If a repair costs $190, how many hours did it take?
(g) A taxi charges $2.45 flag fall plus $1.55 per kilometre.
(i)
Write a formula for the cost C of a trip of k kilometres.
(ii)
If a trip cost $11.75, how long was the trip?
Answers to Check your progress 4
1
(a) 26
(b) 50
(c) 7
(d) 320
(e) 576
(f) 0
(g) 12
(h) –28
(i) 72
2
166
3 1

9 3
64
(b)
9
(a)
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48 8

18 3
3
(d)
14
(c)
(e) 0
(f)
3
3 1

6 2
(a) 38
(b) 41
(c) (i) –36
(ii) 0
(iii) 0
(d) (i) 6
(ii) 36
(e) (i) 8
(ii) 22
(f) 272
(g) 35
(h) 192
4
(a) y = 180 – x – z
(b) n 
r 4
2
(c) b 
P2
2
(d) t 
2s
uv
(e) r 
A

(f) h 
V
b
(g) t 
5
2s
g
(a) b = 6
(b) t = 2
(c) v = 39
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(d) t = 5
(e) 200 km
(f) 5 hours
(g) (i) C = 2.45 + 1.55k
(ii) 6 km
You have now completed Algebraic Processes and Pythagoras.
You should now complete Assignment 2 and send it in for assessment.
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