Expressions and Formulae – Algebraic Expressions
EXPRESSIONS AND FORMULAE
ALGEBRAIC EXPRESSIONS
EVALUATING ALGEBRAIC EXPRESSIONS
If we are given numerical values for the variable(s) in an expression,
we will be able to evaluate the expression.
SIMPLIFYING ALGEBRAIC EXPRESSIONS
You should be able to simplify algebraic expressions which have more than one
variable, for example 5a + 4b + 2a.
EXAMPLE
In the above expression there are two variables, a and b.
The expression has three terms.
Each term includes its sign (+ or –), for example 5a means the same as + 5a.
The terms can be added in any order, for example 5a + 4b + 2a = 5a + 2a + 4b.
Terms such as 5a and 2a are called like terms and so 5a + 2a can be
simplified leading to 7a.
• Terms such as 7a and 4b are not like terms so 7a + 4b cannot be simplified.
Hence if you were asked to simplify 5a + 4b + 2a, the solution would be 7a + 4b.
It will help when you are simplifying expressions if you remember that a term like
a means the same as 1a, while a term like –a means the same as –1a. A term such
as 0a is the same as 0.
2a + 3p
EXAMPLE
Simplify:
SOLUTION
(a) 4x + 5x – x
(b)10p + 2q – 7p
(c) a + 6b + 5a – 4b
(d)6x + 5y + 3x – y
(a) 4x + 5x – x = 4x + 5x – 1x = 8x
(b)10p + 2q – 7p = 10p – 7p + 2q = 3p + 2q
(c) a + 6b + 5a – 4b = 1a + 5a + 6b – 4b = 6a + 2b
(d)6x + 5y + 3x – y = 6x + 3x + 5y – 1y = 9x + 4y
JUST A W
EE NOTE
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With pract
.
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step can b
Some examples may require you to multiply out brackets before simplifying.
Expand the brackets and simplify:
SOLUTION
(a) 4(x + 5) + 3x
(b)7(x + 3y) + 2(x – 6y)
(a) 4(x + 5) + 3x = 4x + 20 + 3x = 4x + 3x + 20 = 7x + 20
(b)7(x + 3y) + 2(x – 6y) = 7x + 21y + 2x – 12y = 7x + 2x + 21y – 12y = 9x + 9y
1_ExpressionsFormulae.indd 8-9
If c = 6 and d = 4, evaluate:
SOLUTION
(a) 5c (b) d2 (c) dc .
(a) 5c = 5 × 6 = 30 (b) d2 = 4 × 4 = 16 (c) c
d
=
6
4
= 6 ÷ 4 = 1∙5.
Order of Operations
DON’T FORGET
Some expressions involve several operations – addition, subtraction, multiplication
and division, as well as brackets, powers of numbers and roots. With these more
complicated expressions the operations must be carried out in the correct order.
You should do things in brackets first before dealing with any orders, that is powers
or roots, and then you should multiply or divide before you add or subtract.
The correct order of
operations is essential when
carrying out calculations
and using formulae. A useful
memory aid is the mnemonic
BODMAS:
B
Brackets first
O
Orders (Powers, Roots)
EXAMPLE
DM Division and
Multiplication
If p = 5 and q = 3, evaluate
(without a calculator):
AS
SOLUTION
(a) 6 + 7p
(b)3p2
(c) 4(p – q)
(d)p2 + q2
(a) 6 + 7p = 6 + 7 × 5 = 6 + 35 = 41 (multiplication before addition)
(b)3p2 = 3 × 52 = 3 × 25 = 75 (powers before multiplication)
(c) 4(p – q) = 4 × (5 – 3) = 4 × 2 = 8 (brackets before multiplication)
(d)p2 + q2 = 52 + 32 = 25 + 9 = 34 (powers before addition)
Addition and Subtraction
Check all the examples carefully to ensure you follow the correct order of
operations. Make sure you follow the powers in parts (b) and (d), for example 52
means ‘5 to the power 2’ or ‘5 squared’ and equals 5 × 5 = 25.
COURSE IDEA
Many calculators are programmed to carry out
calculations in the correct order automatically. Try the
above calculations on your own calculator, working
from left to right and see if you get the correct solution.
You may have to use some different keys on your
calculator such as brackets and the squaring key (x2).
Find out more about these keys.
THINGS TO DO AND THINK ABOUT
1. Simplify:
(a) 7y – y + 5y (b) 5p + 4q + 3p – 2q
EXAMPLE
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SAMPLE PAGES – N4 MATHEMATICS
•
•
•
•
•
2. Expand the brackets and simplify:
(a) 4(3x + 5) + 2x (b) 4(x + 3) + 3(x – 1)
3. If a = 4 and b = 7, evaluate:
3(a + b)
(a) 5a + 2 (b) 6a – 2b (c) a2 + b2 (d) 11
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Expressions and Formulae – Gradient
EXPRESSIONS AND FORMULAE
GRADIENT
COORDINATES
The gradient, or slope, of a straight line is a measurement which tells us how
steep a straight line is.
•
•
•
•
•
When we consider gradients on a coordinate grid, for example finding the gradient of
the straight line joining two given points, some special rules apply.
The gradient of a straight line can be stated in the following formula:
Vertical height
Horizontal distance
Vertical height
DON’T FORGET
Horizontal distance
To simplify a fraction,
divide the numerator and
denominator by their HCF. In
this example, you can divide
40 and 200 by 40 leading
to the solution.
GRADIENT EXAMPLES
EXAMPLE
A sketch of a ramp is shown below.
40 cm
2m
Find the gradient of the ramp. Express your answer
as a fraction in its simplest form.
SOLUTION
The measurements must have the
same units, so convert metres to
centimetres, changing 2 m to 200 cm.
Gradient =
Vertical height
Horizontal distance
=
40
200
=
1
5
COURSE IDEA
Ramps are used in many real-life situations,
for example, to allow wheelchair access to buildings,
to load large items onto trucks and vans and even
in skateboard parks. Due to health and safety
considerations there are strict regulations concerning
the gradient of such ramps. Investigate this.
EXAMPLE
Safety regulations state that the maximum gradient on a railway network should be 0·027.
Measurements are taken on a particular slope on the network.
15 m
SAMPLE PAGES – N4 MATHEMATICS
Gradient =
Parallel lines have the same gradient.
Lines which slope upwards from left to right, are said to have a positive gradient.
Lines which slope downwards from left to right, are said to have a negative gradient.
Lines which are horizontal, are said to have a zero gradient.
Lines which are vertical, are said to have an undefined gradient.
These results for gradients are summarised below:
Positive
Negative
Zero
Undefined
When you are asked to find the gradient of the straight line joining two given points,
you should plot the points, join them, form a right-angled triangle (if possible) and
calculate the gradient using the formula given at the start of this section. You must
also remember the rules given above.
EXAMPLE
Find the gradient of the straight line joining the points A (–4, 5) and B (3, 2).
JUST A W
ical height
B, the vert
For line A
t goes
h
the heig
is –3 since
orizontal
h
e
th
d
an
d o w n by 3
istance
d
e
7 as th
distance is
te that
o
N
7.
y
b
ard
goes forw
written
e
b
ly
–3 would normal
7
3
as – .
SOLUTION
y
6
A
5
7
4
3
2
B
1
–5
–4
–3
–2
–1
0
–1
1
2
3
4
5 x
Give a reason for your answer.
SOLUTION
Vertical height
15
Gradient = Horizontal distance = 600
= 15 ÷ 600 = 0∙025
Yes, the slope does meet the regulations because 0∙025 < 0∙027.
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1_ExpressionsFormulae.indd 14-15
DON’T FORGET
When you are asked to
give a reason for your
answer you must answer by
comparing both numerical
values or by stating the
difference between them
as well as reaching a
conclusion. Answers such as
‘Yes because it is less’ are
unsatisfactory.
DON’T FORGET
If you are calculating gradients
which involve coordinates,
while you can still use the
same formula, watch out
for lines which may have a
negative gradient due to the
way they slope.
Vertical height
3
Gradient of AB = Horizontal distance = –3
7 = – 7
600 m
Does this slope meet the safety regulations?
EE NOTE
THINGS TO DO AND THINK ABOUT
1. According to regulations, the gradient on a cycle track should be less than 0·05.
A slope on a cycle track is shown below.
12 m
180 m
Does this slope meet the safety regulations? Give a reason for your answer.
2. Find the gradient of the straight line joining the points A (3, 1) and B (9, 4).
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