Section 1: Basic Algebra

Introduction

This section will cover how to:

Each topic is introduced with a theory section including examples and then some practice questions. At the end of the page there is an exercise where you can test your understanding of all the topics covered in this page.

Remember that you are NOT allowed to use calculators in this topic.

 

Collecting Like Terms

Definitions
terma collection of one or more numbers/letters multiplied or divided together
e.g. 3, x, 5y, 4x, 15xy
expressiona collection of one or more terms added or subtracted together
e.g. 3 + x, 5y 4x, x 3x + 5
coefficientthe number part of a particular term
e.g. in the expression 5x x + 3, the coefficient of x is 5 and the coefficient of x is -1
like termsterms which have exactly the same letters in
e.g. 2x and 3x are like terms, but 5y and 3y are not like terms
Skills
To collect like terms, look for the terms which have exactly the same letters in, and add them together by adding their coefficients. You can not add together terms unless they have exactly the same letters in them. Number terms with no letters in can be added as normal. Be careful with + and – signs; each sign goes with the term it is in front of.
Examples
Make sure you have understood how these work before moving on:
(a)3a + 12b – 5a + 7b – 2a=– 4a + 19b
(b)x² + 5x – 5 + 4x² – 3x + 5=5x² + 2x     (these two terms can not be added as one is x and the other is x²)
(c)2p + 3pq – 5pq² + 6ppq=8p + 2pq – 5pq²
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Simplify these expressions by collecting like terms:

(a) 3p + 6q – 5p + 3q

– 2p + 9q
The answer could also be written as: 9q – 2p

(b) x3 – 4x2 + 7x – 3 + 7x2 – 9x + 1

x3 + 3x2 – 2x – 2
The terms can be in any order as long as they have the right signs in front of them.

(c) ab2 + 3ab - 3a + 2ab2 - 5ba + 4a

Remember that if a term appears to have no coefficient then the coefficient is 1; we never write a coefficient of 1. Also note that the terms 3ab and – 5ba are like terms because they have exactly the same letters in.
The answer is: 3ab2 – 2ab + a

 

Laws of Indices

Rules to learn
positive powers an means multiply together n lots of a
e.g. a5 = a × a × a × a × a
multiplying powersIf the bases are the same then you add the powers: am × an = a(m + n)
e.g. x5 × x3 = x8
dividing powers If the bases are the same then you subtract the powers: am ÷ an = a(mn)
e.g. y2 ÷ y6 = y– 4
This also works for powers in fractions:
e.g.p7 = p3
p4
powers of powers The powers are multiplied: (am)n = a(mn)
e.g. (z2)5 = z10
special powers a1 = a and a0 = 1
e.g. b4 ÷ b3 = b1 = b
e.g. d 3 × d – 3 = d 0 = 1
multiplying termsTo multiply terms, multiply the coefficients to get the new coefficient then multiply each letter in turn:
e.g. 3x2 × 5x3 = 15x5
e.g. 4xy 2 × x3 × 3x2y = 12x6y3
Other Notes
When you see a value multiplied by a power, remember that the power has higher precedence than the multiplication and as such you calculate the power first. This means that ax2 is equivalent to a × (x2) and not (a × x)2
Examples
Make sure you have understood how these work before moving on:
(a)35= 3 × 3 × 3 × 3 × 3 = 243
(b)p5 × p–2= p(5 + –2) = p3
(c)t7 ÷ t5= t(7 – 5) = t2
(d)(v2)–3= v(2 × –3) = v– 6
(e)
b3 × b5
b7
=b8 ÷ b7 = b1 = b
(f)5xy2 × 2y × 3x2y–3= (5 × 2 × 3) × (x1 × x2) × (y2 × y1 × y–3) = 30x3
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Simplify as far as possible:

(a) m– 3 × m 8

m– 3 × m 8 = m 5

(b) r 9 ÷ r 2

r 9 ÷ r 2 = r 7

(c) (s 3) 4

(s 3) 4 = s 12

(d) 2g 4 × 3g 3

2g 4 × 3g 3 = 6g 7

(e) 2xy 3 × x 6 × 7y

2xy 3 × x 6 × 7y = 14x 7y 4

 

Expanding Brackets and Factorising Expressions

Expanding Brackets
To multiply a term outside brackets by an expression inside the brackets, simply multiply the term outside by each term inside and add together the results.
e.g. 2x2(3x – 2)
   = [2x2 × 3x] + [2x2 × –2]
   = 6x3 – 4x2
e.g. –2pq(5p – 2q + q2)
   = [–2pq × 5p] + [–2pq × –2q] + [–2pq × q2]
   = –10p2q + 4pq2 – 2pq3
Other Notes
Expanding brackets is also called multiplying out brackets.
Factorising Expressions
Factorising an expression is the opposite of expanding brackets. Firstly you find the largest common factor of all the terms in the expression and place it in front of the brackets, then you work out what you need to multiply it by to get each term in the expression. You can check your answer by multiplying it out again.
e.g. 6x2 – 9x3
   = 3x2( ........ )
   = 3x2(2 – 3x)
e.g. 10ab2 + 4a2b – 2a2b2
   = 2ab( .................. )
   = 2ab(5b + 2aab)
Other Notes
You must make sure you take out the largest possible common factor so that the expression is fully factorised. You will know if it is not fully factorised because the expression inside the brackets will still have common factors.
e.g. 8y3 – 12y
   = 2y(4y2 – 6)       [2 is still a common factor inside the brackets]
   = 4y(2y2 – 3)       [now the expression is fully factorised]
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Multiply out the brackets:

(a) –2p(5p – 3q)

–10p2 + 6pq

(b) 5x(x2 + 3x – 2)

5x3 + 15x2 – 10x
Fully factorise:

(a) 6x4 – 18x3

6x3(x – 3)

(b) 10u2v2 – 5uv2 + 30u2v2

5uv2(2u – 1 + 6u)

 

Negative and Fractional Powers

Rules to learn
negative powers Negative powers can be found by calculating the positive power and then finding the reciprocal of your answer (the reciprocal of a number is 1 divided by that number):
an =
1
a n
e.g. 5–3 =
1
53
=
1
125
As an is just the reciprocal of a n, when you calculate the negative power of a fraction you can simply flip the fraction over and calculate the positive power:
e.g.
2 –2
3
=
3 2
2
=
9
4
e.g.
1 –3
4
=
4 3
1
= 43 = 64
This even works for the whole number example above:
e.g. 5–3 =
5 –3
1
=
1 3
5
=
1
125
fractional powers Fractional powers with a "1" on top represent roots:

a
1
 n
= na

e.g. 25
1
2
= √25 = 5

e.g. 32
1
5
= 532 = 2

Other fractional powers represent two operations; the top number is a power and the bottom number is a root. They can be done in either order:


a
 m
 n
=
n am
= ( na) m

e.g. 27
2
3
=
3 272
=
3 729
= 9     (doing the power first is harder)

e.g. 27
2
3
= ( 327) 2 = (3)2 = 9            (doing the root first is easier)

In general you are better off doing the root first then the power, especially as you will not have a calculator for these questions.

all rational powers Using the above rules you can calculate any power which can be written as a fraction, whether positive or negative. There will only ever be three components:
  • if the power is negative, you will need to do the reciprocal (flip it over)
  • if the power is a fraction with n on the bottom, you will need to calculate the nth root
  • if the power is a fraction with m on the top, you will need to calculate the mth power

These can be done in any order so pick the order which is easiest. Have a look at these examples and see the easiest order to get each answer:


e.g. 16
1
2
=
1
4
    (start with 16, square root to get 4, then find the reciprocal)

e.g. 32
3
5
=
1
8
    (do the fifth root, then raise to the power of 3, then find the reciprocal)

e.g.
1
25
3
2
= 125  (find the reciprocal, then square root, then raise to the power of 3)
Practice Questions
Work out the answer to each question then click on the button marked
to see if you are correct.
Work out the value of:

(a)
3–5

Start by working out the power as if it were positive, then find the reciprocal by flipping the answer over:
3–5 =
1
243

(b)

27
1
3

The power of a third is the same as a cube root:

27
1
3
= 327 = 3

(c)

16
3
4

Find the fourth root of 16 first, then cube the answer:

16
3
4
= ( 416) 3 = (2)3 = 8

(d)

125
1
3

Do the cube root first, then find the reciprocal:

125
1
3
=
1
5

(e)
1
27
2
3

Find the reciprocal first, then cube root it, then square it:
1
27
2
3
= 9

 


Exercise

Work out the answers to the questions below and fill in the boxes. Click on the
button to find out whether you have answered correctly. If you have then the answer will be ticked
and you should move on to the next question. If a cross
appears then your answer is wrong. Click on
to clear the incorrect answer and have another go, or you can click on
to get some advice on how to work out the answer and then have another go. If you still can't work out the answer after this then you can click on
to see the solution.

 

Question 1

Simplify the following expressions by collecting like terms:

(a)

3a + 12b – 5a + 7b =
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Add together the a terms, then add together the b terms. Remember that the sign goes with the term it is in front of. 3a – 5a = –2a and 12b + 7b = 19b

(b)

x2 + 3x – 5x – 15 =
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Remember like terms must have exactly the same letters in. Only two of these terms are like terms. +3x and –5x are like terms, and they add together to make –2x.

(c)

2p – 5q – 4r – 2q + 3p + 5r + 7q =
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If a term has a coefficient of 1 then we don't write the coefficient. If a term has a coefficient of 0 then we don't write the term at all. 2p + 3p = 5p
–5q – 2q + 7q = 0q = 0
–4r + 5r = 1r = r

(d)

2ab + 3a – 5b =
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Check carefully to see if there are any like terms. There are no like terms so the expression is already simplified.

Question 2

Simplify as far as possible:

(a)

p2 × p4 =
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If you multiply numbers in index form which have the same base, you add the powers. p2 × p4 = p(2 + 4) = p6

(b)

5y × 2y3 =
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Start by multiplying the coefficients, then multiply together the powers of y. Remember that y on its own is actually y1. 5 × 2 = 10 and y × y3 = y4

(c)

3b2 × (b3)5 × 5b3 =
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Start by simplifying the middle term, then multiply the three parts together. (b3)5 = b15
3b2 × b15 × 5b3 = 15b20

(d)

(2x2)3 ÷ 4x3 =
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Remember that (2x2)3 means 2x2 × 2x2 × 2x2. (2x2)3 = 2x2 × 2x2 × 2x2 = 8x6
8x6 ÷ 4x3 = 2x3

Question 3

Expand the brackets in the questions below and simplify your answers where necessary:

(a)

5(2b – 7) =
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Multiply 5 by each of the terms inside the brackets and add the results together. 5 × 2b = 10b and 5 × –7 = –35 so the answer is 10b – 35

(b)

3y2(5 + 2y – 7y2) =
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Multiply each term in the brackets by 3y2 3y2 × 5 = 15y2
3y2 × 2y = 6y3
3y2 × –7y2 = –21y4
The answer is 15y2 + 6y3 – 21y4

(c)

2(x + 5) – 3x(5 – 2x) =
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Multiply the terms in the first bracket by 2 and the terms in the second bracket by –3x, then add all the results together. 2(x + 5) = 2x + 10
–3x(5 – 2x) = –15x + 6x2
Adding these gives 6x2 – 13x + 10

(d)

2p(7 – 3p) – 6p2(3p – 1) =
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Multiply out the brackets and simplify. 14p – 6p2 – 18p3 + 6p2 = 14p – 18p3
(the p2 terms cancel each other out)

Question 4

Factorise these expressions:

(a)

5y – 7y2 =
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Take y outside the brackets. The answer is y(5 – 7y). This can be checked by multiplying out the brackets again.

(b)

9y + 12y3 =
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Take 3y outside the brackets. The answer is 3y(3 + 4y2)

(c)

8xy + 8x2 + 4x2y =
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Take 4x outside the brackets. The answer is 4x(2y + 2x + xy)

(d)

6xy2 + 10x2y =
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Take 2xy outside the brackets. The answer is 2xy(3y + 5x)

Question 5

Simplify as far as possible:

(a)

( y– 2 )5 =
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(am)n = a(mn) ( y– 2 )5 = y(–2 × 5) = y–10

(b)

x– 4 ÷ x– 5 =
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am ÷ an = a(mn) and a1 = a x– 4 ÷ x– 5 = x(–4 – –5) = x1 = x

(c)

y × y– 1 =
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am × an = a(m + n) and a0 = 1 y1 × y– 1 = y0 = 1

(d)

( x3 )2 × x–8 =
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Work out the answer in stages. ( x3 )2 × x–8 = x6 × x–8 = x–2

Question 6

Evaluate the following, giving your answers as whole numbers or single fractions:

(a)

4–3 =
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an =
1
a n
4 –3 =
1
43
=
1
64

(b)

2 –3
5
=
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Flip the fraction over (because of the
negative power), then raise the resulting
fraction to the power 3.
2 –3
5
=
5 3
2
=
125
8

(c)


8
1
3
=
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The third power is the same as a cube root.

8
1
3
= 38 = 2

(d)


49
3
2
=
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Square root first, then cube the answer you get.

49
3
2
= (49) 3 = (7)3 = 343


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