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Unit 4:
Mathematics for Engineering Technicians

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Aims
Introduce the unit contents and learning outcomes.
Appreciate algebraic methods.
Objectives
Identify the learning outcomes of the unit.
Identify why is there a need for Algebra.
Finding the subject in a formula.
Expanding brackets.

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This unit aims to give learners a strong foundation in mathematical skills.
These skills will help them to successfully complete many of the other units within the qualification.
One of the main responsibilities of engineers is to solve problems quickly and effectively. This unit will enable learners to solve mathematical, scientific and associated engineering problems at technician level

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Learning outcomes ……….
On completion of this unit a learner should:
1. Be able to use algebraic methods
2. Be able to use trigonometric methods and standard formula to determine areas
3. Be able to use statistical methods to display data
4. Be able to use elementary calculus techniques.

5. Introduction To Algebra
Uned 4 mathemateg ar gyfer Technegwyr Peirianneg
Introduction To Algebra
Which country does the word Algebra come from? Where is it used in Engineering? Power = Head x Flow x Gravity
Head (H) = 12 meters
Flow (F) = 20 Litres per Second
Gravity = 9.81
What is the Power (P) ?
P=H x F x G
The word, Algebra is an Arabic word meaning bringing together broken parts. This useful tool of mathematics was invented by the 9th century Arab mathematician, Mohammed Ibn Musa Al-Khwarizmi (from who’s Europeanised name comes the word, algorithm).
Algebra is used in every aspect of every type of engineering. If I were designing a water canal, I would have to consider slope, cross sectional area, flow rates.
Many different disciplines of engineering exist today such as mechanical, petroleum and civil. Engineering utilizes math---more specifically, calculus and algebra---to solve physical problems such as how to build a bridge or design an airplane. Take designing a rocket going to the moon, for example: An engineer must use algebra to solve for flight trajectory, how long to burn each thruster at what intensity and at what angle to lift off. Though a very difficult, math-heavy discipline, engineering provides a very rewarding career both in achievement and pay.
POWER = 2354 WATTS
IF WE REQUIRE WATTS WHAT SHOULD THE Head be ? (500) First we n

6. Introduction To Algebra
Uned 4 mathemateg ar gyfer Technegwyr Peirianneg
Introduction To Algebra
The formula is P= H x F x G
What if we require watts
How would we work out the Head?
(500) We need to make H the subject so divide each side by F x G to get H = P/F x G now put in the figures and we get 500 meters

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RULES FOR REARRANGING THE FORMULA
Add the same amount to both sides.
Subtract the same amount from both sides.
Multiply both sides by the same amount.
Divide both sides by the same amount.
Square the whole of both sides.
Square root the whole of both sides.
It is often useful to change the subject of a formula. We should follow the same rule, do the same to both sides, as we follow with an equation. That is we can:

8. Changing the subject of the formula
Uned 4 mathemateg ar gyfer Technegwyr Peirianneg
Changing the subject of the formula
a + b = c
C is the
subject
Rearrange the formula to make b the subject so that
The
method
is the
same as
solving
equations
b = ……... -a -a
-a from both sides
a + b = c
Equal = Hafal
b = c - a

9. Make b the subject a = bx + c
Uned 4 mathemateg ar gyfer Technegwyr Peirianneg
Make b the subject a = bx + c
rearrange so that b is on the right
bx + c = a
-c from both sides -c -c
bx + c = a
bx = a - c
÷ both sides by x
Equal = Hafal ÷x ÷x
bx = a - c
b = a – c x

10. Make s the subject n = m -3s
n + 3s = m
add 3s to both sides
3s = m - n
- n from both sides
3s = m - n
divide both sides by 3 3

11. reverse to get r on the right
Make r the subject
p = q + r s
q + r = p s
reverse to get r on the right
q + r = sp
multiply both sides by s
r = sp – q
- q from both sides

12. divide both sides by 3 s = 3(a + b) 3(a + b) = s 3a + 3b = s
Make b the subject
s = 3(a + b)
3(a + b) = s
reverse to get b on left
3a + 3b = s
multiply out the bracket
3b = s - 3a
- 3a from both sides
3b = s - 3a 3
divide both sides by 3

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Exercise:
Make the letter in brackets the subject.
1. S = t + a (t)
2. PV = T (V)
3. 2A = PQ (Q)
4. V2 = 4gh (h)
5. a = p/q (p)
6. v = u + at (u)
T=s-a
V=t/p
Q=2a/p
H=v2/4g
P=aq
U=v-at

14. 3(x + 2) 4(2x + 3) x(x + 3) 3(x – 2) + 2(x – 3) (x + 3)(x – 2)
This means ‘get rid’ of the brackets in cases like:
3(x + 2)
4(2x + 3)
x(x + 3)
3(x – 2) + 2(x – 3)
(x + 3)(x – 2)

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4(2x – 3)
p(q – 2p)
5(2y – 3)
x(2x +y)
4(x + 5) + 3(x – 7)
5(3p + 2) – 2(5p – 3)
1) 7n-21
2) 8x-12
3) pq-2p²
4)10y-15
5) 2x² + xy
4x x = 7x – 12
15p p = 5p +16
t² - 2t + 4t = t² + 2t -8
x² + 2xy + 3xy + 6y² = x² + 5xy + 6y²

16. + x + = + + x - = - - x + = - - x - = + -12 (6) 4 + – 3 = 17
(2) –2 x –5= 10
(7) 4 – – 3 =
- x - = +
-14
-12
(3) - 7 x 2 =
(8) – 5 – + 7 = 48
(4) 6 x 8 =
(9) 6 + – 3 = 3
(5) – 6 x – 9= 54
(10) – 5 – – 4 = -1

17. .
( 2x – 7 ) ( 5x – 4)
(3x – 4 ) ( 6x – 7)
( 2x + 4 ) ( 3x – 8 )

18. (x + 2)(x + 3)
(a + b)(c + 3)
(y − 3)(y + 2)
d) (2x + 1)(3x − 2)
e) (3x − 1)(3x + 1)
f) (5x − 1)(x − 5)
g) (2p + 3q)(5p − 2q)
h) (x + 2)(2x2 − x − 1)