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Anne Eadie & Gretel Lampe

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3-in-1
Grade 10 Mathematics 3-in-1 CAPS
CLASS TEXT & STUDY GUIDE

The Answer Series Grade 10 Maths 3-in-1 study guide uses simple, logical steps to explore the CAPS curriculum in great depth,
from first principles all the way up to final mastery. It addresses gaps in your memory from previous grades, before inviting
you to tackle new work through carefully selected graded exercises.

Key features:
• Comprehensive, explanatory notes and worked examples
• Graded exercises to promote logic and develop a technique for each topic
• Detailed solutions for all exercises
• An exam with fully explained solutions (paper 1 and paper 2) for thorough consolidation and final exam preparation.

This study guide has proven to be a great companion to Grade 10 Maths learners. Not only that, it builds confidence and
also lays the foundations for success in Grade 11 and 12.
 10
GRADE

Mathematics
CAPS Anne Eadie & Gretel Lampe

3-in-1

THIS CLASS TEXT & STUDY GUIDE INCLUDES

1 Comprehensive Notes

2 Exercises

3 Full Solutions eBook

available

Plus: Exam Papers and Memos

& Problem Solving Questions and Memos

2012 publication | 2021 edition | ISBN: 978-1-920297-79-4

 DETAILED CONTENTS
Amended Teaching Plan (2023/2024) 8 Analytical Geometry (Paper 2) 8.1
& A suggested November Exam ii
9 Finance & Growth (Paper 1) 9.1

1 Numbers & Number Patterns (Paper 1) 1.1

10 Statistics (Paper 2) 10.1

2 Exponents (Paper 1) 2.1

11 Measurement (Paper 2) 11.1

3 Algebraic Expressions (Paper 1) 3.1

12 Probability (Paper 1) 12.1

4 Algebraic Equations & Inequalities (Paper 1) 4.1

5 Trigonometry (Paper 2) 5.1 National Gr 10 Exemplars Questions Memos

Section 1: 'Pre-trig' Paper 1 E1 M1

Section 2: The Trigonometry of Acute Angles Paper 2 E3 M4
Section 3: Trigonometry Unlimited

Problem Solving PS1 PS3

6 Functions & Graphs (Paper 1) 6.1

Geometry Theorem Statements &

7 Euclidean Geometry (Paper 2) 7.1 Acceptable Reasons (at the back of the book)
i
 Simplify:
LAW 1 the PRODUCT of POWERS : am x an = am + n

(same bases) 5x + 2 - 5x + 1 3x - 3x - 2 2.2n + 1 - 2n

16. 17. 18.
5x + 1
x
8.3 3.2 n - 1
We know : a3 % a 2 = a % a % a % a % a = a 3 + 2 = a 5 . . . we add the exponents
These are EQUATIONS to be solved.
4
â a %a = ....? 2
a x % ay = . . . ?
2
ax  ax = . . . ? Solve for x:
19. 2 x + 1 = 23 20. 3 2x . 3 x = 9
2
am = an
Answers: a 6
; ax + y ; ax + x . . . not always that intuitive! 21. 7 4x
= 49 22. a .ax x+5
.a = 1 m=n

23. 2 x + 1 + 2 x + 2 x - 1 = 28 24. 5 x + 1 + 5 x - 1 = 26
Do you see that it becomes less and less intuitive? 25
Keep referring to the law to keep on track !
NOTE: There is no law for am + an – only for the PRODUCT of powers, am X an

& the law reversed : am + n = am x an

LAW 2 am = am - n
the QUOTIENT of POWERS:
an
i.e. 2 n + 1 = 2 n % 21 ; even 2 n - 1 = 2 n + (-1) = 2 n % 2 -1 (same bases)
The 4 factors
We know : 2 7 + 2 4 = 24 = 2  2  2  2  2  2  2
7
below cancel
2 2  2  2  2
EXERCISE 2.1 – An exercise on LAW 1 (Answers on page 2.9) = 23
4 of the
7 above.
27
So, = 27 - 4 . . . we subtract the exponents!
These are EXPRESSIONS to be simplified. 24
Complete : m
= an
5 n
& the law reversed : am - n
i.e. 2 5 - a = 2a ; 2 n - 1 = 2
1. x % x =
3 4
2. 2 p
% 2 =q
a 2 2
1 3
3. xa % xb = 4. 24  24 =
EXERCISE 2.2 – An exercise on LAW 2 (Answers on page 2.9)
5. 2 -3
% 2 -4
= 6. x % x =
n n

These are EXPRESSIONS to be simplified.

Complete :
7. am + n . am - n = 8. (x + y) 2 . (x + y) 3 =
11 1
23x
1. a5 + a2 = 2. x 4 + x4 = 3. p6 + p2 = 4. =
9. 3 2x + 4 . 3 - 2x - 3 = 10. 4 x + 1 . 7 x + 1 . 4 3 - x = 22x
3
71 - n a2
11. ab % cd = 12. 3 x2
%3 x
= 5. x16 + x4 = 6. = 7. a7 + a = 8. 1
=
7n -
EXPONENTS

a 2

 a b  p21
-1 b
10. ad = b
2 1 1
13. a b 3 2 3 2 = 14. 3
2 2 .4 5 =
9. = 11. =
p20 c b-2

Simplify :
Factorise :
7 x -2 y 5 15 4n a -3
15. 2 n + 3 = . . . . % . . . . ; 2 n - 5 = . . . . % . . . . (law 1 reversed) 12. 13. 14.
2 14 x -1 y 8 9 2n + 1
. 25 2n a -5
Copyright © The Answer Series: Photocopying of this material is illegal 2.4
Answers 24. 5 x + 1 + 5 x - 1 = 26
25
EXERCISE 2.1 – LAW 1 (Questions on page 2.4) â 5 x . 5 + 5 x . 5 -1 = 26
25
x7 2. 2 p + q 3. x a + b
 
1. 26
â 5x 5 + 1 =
1
+ 3
5 25
= 21 = 2 5. 2 -3 - 4 = 2 -7 6. x n + n = x 2n
â 5  26 
4. 24 4 x = 26
5 25
a m + n + m - n = a 2m 8. (x + y) 2 + 3 = (x + y) 5 9. 3(2x + 4) + (-2x - 3) = 31 = 3
26 
7. 5 26 5
x
% â 5 % =
25 26
4x + 1 + 3 - x . 7 x + 1 = 4 4 . 7 x + 1 11. cannot be simplified 12. 3 x % 3 x = 3 x +x
2 2
10. â 5x = 1
5
â 5 x = 5 -1
13.  a b  a
2
3
1
2
1
3
-1
b2  14.
3
22 . 4 5 =
3
22 . 210 15. 2n + 3 = 2n % 23 ;
â x = -1
3
2 1
+ 1 1
- = 212 2n - 5 = 2n % 2-5
= a3 3 . b2 2
1 0 = 24
= a b EXERCISE 2.2 – LAW 2 (Questions on page 2.4)
= 16
= a 1
1 -
1
1. a3 2. x 4 4 = x1 = x 3. p6 - 2 = p4

5 x. 52 - 5 x. 5 3 x - 3 x. 3 - 2 2 . 2n . 2 - 2n
16. 17. 18. 4. 23x - 2x = 2 x 5. x 16 - 4 = x 12 6. 7 1 - n - n = 7 1 - 2n
5 x. 5 8 . 3x 3 . 2n . 2 - 1
5 x (25 - 5) 3x 1 -  
1 2n (4 - 1) 7. a7 - 1 = a6 8. a2
3
 
- - 1
2
4
= a 2 = a2 9. p21 - 20 = p
= 9 =
5x. 5 = 2n .
3
8 . 3x 2 7 x- 2 y 5 1
10. cannot be simplified 11. b 1 - (- 2) = b3 12. =
= 20 14 x - 1 y8 2 x y3
5 = 8 % 1 = 3% 2 2
9 8 3
= 4 15 4n a-3 3
= 1 = 2 13. 14. = a -3 + 5 = a 2 15. 32 x = 3- x
9 9 2n + 1
. 252n a- 5 3
3 . 5 4n â 32 x = 33 + x
19. 2 x + 1 = 23 20. 32 x . 3 x = 9 21. 7 4 x = 49
3 2 2n + 1 . 52 2n â 2x = 3 + x
â x+1 = 3 â 32 x + x = 3 2 â 7 4x = 7 2 4n
3 4n
. 5 1 â x = 3
â x = 2 â 3x = 2 â 4x = 2 = 4n
= = 1
3 4n + 2
. 5 32 9
â x = 2 â x = 1
3 2
23
16. 24x = 2x - 2
17. 4 % 32 x = 9 % 2 2 x
x+x+5+1 x x x 2
22. a =1 23. 2 .2 + 2 + 2 .2 -1
= 28 2x
9
â 2 4 x = 23 - 2 x + 2  4  22 x  â 32 x =
â a 2x + 6
=1 
 2x 2 + 1 + 1
2  = 28 â 2 4 x = 25 - 2 x
2 4

2  2  b 
2x 2 n n

EXPONENTS
â 2x + 6 = 0 â 3 = 3 . . . an = a
â 2x = - 6
 2x  72  = 28 â 4x = 5 - 2x b
4 2
â 6x = 5 â 2x = 2
â x = -3  2x = 28 %
7 â x = 5 â x = 1 Law 5 reversed,
x 6 to be compared to law 2.
â 2 = 8
â 2 x = 23 n 3
18. 3n - 2 = 32 ; 53 - p = 5 p
â x = 3 3 5 2
2.9 Copyright © The Answer Series: Photocopying of this material is illegal
 FIVE FACTORISATION TESTS (approximately ½ hour each)
FACTORISING – SOME GOOD ADVICE
A
How many terms do I have? 1. x2 - x - 12 2. 8ax - 12ay - 10x + 15y 3. (x + 5)(x + 3) + k(3 + x) (2)(4)(2)

The number of terms determines my options. 4. 2

p - 14p - 32 5. 4m - pm + 8 - 2p 6. 12x2 - 19xy - 21y2 (2)(2)(2)

7. (x - y)3 - 3(x - y)2 8. 2a2 - 18 9. 28ab + 4a2 - 15b2 (3)(2)(2)

NO. OF
OPTIONS 10. ac + yd - ad - yc 11. 3k(2m - 3n) + 5t(3n - 2m) 12. (a - b)2 - 49 (2)(3)(2)
TERMS
13. 12x3 + 11x2 - x 14. x6 - 64y6 15. x(x - 1)(x - 2) - (x - 1)2 (3)(3)(4)
16
The difference of two squares OR 16. 1 - 16a 17. - 6m2 + 11m + 10 18. 4x - ax + ay - 4y (4)(2)(2)
Always look out for 19. Write down the simplest expression (in factorised form) into which the expressions in
2 terms
a common The sum of two cubes or the difference questions 6, 7 & 18 can divide (i.e. the lowest common multiple). (1)

factor of two cubes 20. Calculate the value of 1092 - 92 in the shortest possible way, without using a calculator. (3) [50]

FIRST! B
3 terms Trinomial 1. pa + pb + qa + qb 2. x2 + 5 x + 6 3. 4 x2 - 9 (2)(1)(1)

4. 5at + 9 + 3a + 15t 5. 4x2 - 20x + 25 6. 5 - 20a2 (3)(2)(2)

2 - 2 – for common 'brackets', or
7. 3ac + 2bc - 2bd - 3ad 8. 3y2 + 15y - 108 9. ac + 6b - 3ab - 2c (2)(2)(3)
4 terms 3 - 1 or
3
10. p - 8 11. 522 - 502 (evaluate) 12. 132x2 + 96xy - 36y2 (2)(3)(3)
1 - 3 – leading to difference of squares
Grouping 13. 9 x2 - 5y - 3x - 25y2 14. 8m2 - 50mn + 33n2 15. x4 - x3 +x-1 (3)(3)(3)
3 - 2 or
5 terms ... watch out for 16. 12mb + 9a2 - 4m2 - 9b2 17. x2 - 2 xy - a2 + y2 18. k4 - 37k2 + 36 (3)(3)(3)
2 - 3 – for common 'brackets' 2
switchrounds ! 19. 16(2a + b) - 9(a - 2b)2 (6) [50]
3 - 3 – for common 'brackets' or C
6 terms difference of squares or 1. x2 - 25xy + 144y2 2. x2 - 24xy + 144y2 3. 2ac - 3ad - 2bc + 3bd (2)(2)(3)
2 - 2 - 2 – for common 'brackets' 4. k(a - b) + n(b - a) 5. 1222 - 1202 (evaluate) 6. 2a2 - 2a - 12 (2)(3)(3)
1
7. ax - b2 - bx + ab 8. a2x2 + 5ax - 24 9. x2 -2 (3)(2)(3)
And then, two good habits . . . 4
10. 20m2n + 62mn2 - 28n3 11. ax2 + 3by2 - 3bxy - axy 12. 125x3 + y3 (3)(3)(2)
ALGEBRAIC EXPRESSIONS

Once you've factorised:

13. 20 x2 - 45y2 14. 3a3 + 12a2b + 9ab2 15. x2 - 2x + 2y - y2 (3)(3)(3)
● CHECK your factors by multiplying out.
16. Factorise a2 - b2 and then write down the factors of (2x2 - 4x + 1)2 - (x2 - 3x + 3)2
Remember: factorising w multiplying out . . . REVERSE PROCESSES ! in the simplest form. (10) [50]

● Ask yourself : 'Have I finished?' D

Double-check for any simplification or any further factorisation 1. 2 x2 - 8 2. a 2 - b2 + a - b 3. x2 - 12x + 36 (3)(3)(2)
2 2
(e.g. common factor or difference of squares). 4.
 1  1
r +  - r -  5. 10x2 + 38xy - 8y2 6. 2 x3 - x2 + 4 - 8 x (3)(3)(3)
4 4  r  r
e.g. (1) 2x(a + b) + 4(a + b) (2) x - y
7. 40ap2 + 82a2 p + 40a3 8. 12ab + 8b2 - 6af - 4bf 9. - 3x2 + 21x - 30 (3)(3)(3)
= (a + b)(2x + 4) = (x2 + y2)(x2 - y2)
16
10. 3 - 3(x - y)2 11. x2 + 8 + 12. k(x3 - 1) - k(x - 1)3 (3)(2)(5)
= 2(a + b)(x + 2) = (x2 + y2)(x + y)(x - y) x2
13. 4p2 (3p - 1) - 5p 14. x2 - y2 + 4x + 4 15. 3a3 - 24b3 (4)(4)(3)
Knowing that there are guidelines shifts the mind into a more confident way of thinking.
3 16. If P = 3x + 2 and Q = 2x - 1, express P2 - 2PQ + Q2 in terms of x. (3) [50]

Copyright © The Answer Series: Photocopying of this material is illegal 3.14

E B
1. 9a2 - 49b2 2. xy + 6x + 2y + 12 3. 6a2 - 5ab - 6b2 (2)(2)(2) 1. p(a + b) + q(a + b) 2. (x + 2)(x + 3) 3. (2x + 3)(2x - 3)
= (a + b)(p + q)
4. 18a2 - 8b2 5. - 6 x3 + 5 x2 + 25x 6. x4 + 24x2y2 + 108y4 (3)(3)(2)

7. 2
21a + 26a - 15 8. - 0,81 + c2 9. 4b3 - 8b2 - ab + 2a (2)(2)(4) 4. 5at + 15t + 3a + 9 5. (2x - 5)2 6. 5(1 - 4a2)
= 5t(a + 3) + 3(a + 3) = 5(1 + 2a)(1 - 2a)
3
10. - 3 x2 11. y3 - y2 + y - 1 12. x3 + 4x2y + 3xy2 (3)(3)(3)
= (a + 3)(5t + 3)
4

13. a2 + c2 - b2 - 2ac 14. (x2 - 2x - 8)2 + 5(x2 - 2x - 8) (3)(4) 7. 3ac + 2bc - 3ad - 2bd 8. 3(y2 + 5y - 36) 9. ac - 3ab - 2c + 6b

15. 3a2 + 6ab + 3b2 + 9a2y + 18aby + 9b2y 16. 2,72 - 2,32 (evaluate) (5)(2) = c(3a + 2b) - d(3a + 2b) = 3(y + 9)(y - 4) = a(c - 3b) - 2(c - 3b)
= (3a + 2b)(c - d) = (c - 3b)(a - 2)
17. Factorise x3 + 13 and then determine the value of this expression if x + 1 = 2. (5) [50]
x x 10. (p - 2)(p2 + 2p + 4) 11. (52 + 50)(52 - 50) 12. 12(11x2 + 8xy - 3y2)
= (102)(2) = 12(11x - 3y)(x + y)
Answers to Factorisation Tests
= 204
A
13. (9x2 - 25y2) - 3x - 5y 14. (4m - 3n)(2m - 11n) 15. x3(x - 1) + (x - 1)
1. (x - 4)(x + 3) 2. 4a(2x - 3y) - 5(2x - 3y) 3. (x + 3)(x + 5 + k)
= (2x - 3y)(4a - 5) = (3x + 5y)(3x - 5y) - (3x + 5y) = (x - 1)(x3 + 1)
= (3x + 5y)(3x - 5y - 1) = (x - 1)(x + 1)(x2 - x + 1)
4. (p - 16)(p + 2) 5. m(4 - p) + 2(4 - p) 6. (4x + 3y)(3x - 7y)
= (4 - p)(m + 2) 16. 9a2 - 4m2 + 12mb - 9b2 17. x2 - 2xy + y2 - a2 18. (k2 - 36)(k2 - 1)
= 9a2 - (4m2 - 12mb + 9b2) = (x - y)2 - a2 = (k + 6)(k - 6)(k + 1)(k - 1)
7. (x - y)2 [(x - y) - 3] 8. 2(a2 - 9) 9. 4a2 + 28ab - 15b2
2 2
2 = (3a) - (2m - 3b) = (x - y + a)(x - y - a)
= (x - y) (x - y - 3) = 2(a + 3)(a - 3) = (2a - b)(2a + 15b)
= [3a + (2m - 3b)][3a - (2m - 3b)]
10. ac - ad - yc + yd 11. 3k(2m - 3n) - 5t(2m - 3n) 12. (a - b + 7)(a - b - 7)
= (3a + 2m - 3b)(3a - 2m + 3b)
= a(c - d) - y(c - d) = (2m - 3n)(3k - 5t)
= (c - d)(a - y) 19. [4(2a + b) + 3(a - 2b)][4(2a + b) - 3(a - 2b)]
= (8a + 4b + 3a - 6b)(8a + 4b - 3a + 6b)
13. x(12x2 + 11x - 1) 14. (x3 ) 2 - (8y3 ) 2 . . . the difference of
= (11a - 2b)(5a + 10b)
= x(12x - 1)(x + 1) two squares
= (x3 + 8y3)(x3 - 8y3) = 5(11a - 2b)(a + 2b)
= (x + 2y)(x2 - 2xy + 4y2)(x - 2y)(x2 + 2xy + 4y2)
C

ALGEBRAIC EXPRESSIONS
OR (x2 ) 3 - (4y2 ) 3 . . . the difference of
1. (x - 16y)(x - 9y) 2. (x - 12y)2 3. a(2c - 3d) - b(2c - 3d)
= (x2 - 4y2)(x4 + 4x2y2 + 16y 4 ) two cubes
= (2c - 3d)(a - b)
= (x + 2y)(x - 2y)(x4 + 4x2y2 + 16y 4)
4. k(a - b) - n(a - b) 5. (122 + 120)(122 - 120) 6. 2(a2 - a - 6)
15. (x - 1)[x(x - 2) - (x - 1)] 16. (1 + 4a8)(1 - 4a8) 17. - (6m2 - 11m - 10)
= (a - b)(k - n) = (242)(2) = 484 = 2(a - 3)(a + 2)
= (x - 1)(x2 - 2x - x + 1) = (1 + 4a8)(1 + 2a4)(1 - 2a4) = - (2m - 5)(3m + 2)
= (x - 1)(x2 - 3x + 1) OR 10 + 11m - 6m2 7. ax - bx + ab - b2 8. (ax + 8)(ax - 3) 9. x2 - 9
4
= (5 - 2m)(2 + 3m) = x(a - b) + b(a - b)
  
= (a - b)(x + b) =  x + 3  x - 3 
 2  2
18. 4x - ax - 4y + ay 20. 1092 - 92 = (109 + 9)(109 - 9)
= x(4 - a) - y(4 - a) = (118)(100) 10. 2n(10m2 + 31mn - 14n2) 11. ax2 - axy - 3bxy + 3by2 12. (5x + y)(25x2 - 5xy + y2)
= (4 - a)(x - y) = 11 800 = 2n(5m - 2n)(2m + 7n) = ax(x - y) - 3by(x - y)
2
19. (4x + 3y)(3x - 7y)(x - y) (x - y - 3)(4 - a) = (x - y)(ax - 3by) 3
3.15 Copyright © The Answer Series: Photocopying of this material is illegal
 In other words, are there values of x and y that will make both equations true ?
SIMULTANEOUS LINEAR EQUATIONS
In words, let us ask ourselves :
Can you think of two numbers which add up to ten AND have a difference of two ?
This refers to solving problems where you have TWO equations and TWO unknowns.
To build up our understanding of how to solve these, let us start with ONE equation
Hint : Look at the list of possible solutions for each equation above – see A & B
and TWO unknowns.

K Single Equation, Two Unknowns – Infinite Solutions! K Solving Simultaneous Equations

Until now we have dealt with equations with only ONE unknown
If we are able to find the answer (6; 4) by just looking at the two equations and
e.g. 3x + 8 = 14
thinking about it, this is termed: solving the equations BY INSPECTION.
â x = 2
Now we will learn to find the answers to simultaneous equations algebraically,
Linear equations with ONE unknown have ONE solution (mostly) so that we can solve more complex problems.

 Now consider a linear equation with TWO unknowns : Equations can be added or subtracted . . .
e.g. x + y = 10 ... A RELATIONSHIP between x and y LOGIC is
because . . . if a = b
If x = 1, y=....? or (1; . . . ?) and c = d essential
If x = 6, y=....? or (6; . . . ?) in Maths !
then a + c = b + d or a-c = b-d
If x = 2,96, y=....? or (2,96; . . . ?)
If x = - 2, y=....? or (- 2; . . . ?) In our example above we had : x + y = 10 . . . (1)
and x - y = 2 . . . (2)
â Possible solutions : (1; 9), (6; 4), (2,96; 7,04), (- 2; 12), etc. A

ALGEBRAIC EQUATIONS & INEQUALITIES

` These are all 'solutions' to this equation since, in each case, the PAIR of values of Let us add equation (1) to equation (2) to find the answer algebraically . . .
x & y makes the equation true. And we can just keep going, i.e. there appears to be (1) + (2) : â x + y + x - y = 10 + 2 Notice that by
an INFINITE set of answers !
adding the
We always explain on the left â 2x = 12 ...
A linear equation with TWO unknowns has an INFINITE number of solutions ! equations we
which operation we are doing. â x = 6 have
 Now consider : x -- y = 2 Substitute x = 6 in (1) : â 6 + y = 10 eliminated
This equation also has an infinite set of solutions of which some examples are: (lost) y !
â y = 4
Possible solutions: (20; 18), (12; 10), (6; 4), (- 4; - 6), etc. B
â The two numbers are 6 and 4 

K The Meaning of 'Simultaneous' Equations We also say: (6; 4) is the SOLUTION to the simultaneous linear equations
– Two Equations, Two Unknowns x + y = 10 and x - y = 2
Simultaneous means TOGETHER or AT THE SAME TIME. So when we have as these values of x & y make both the equations true.
'simultaneous equations', it means we need to find a solution that solves
TWO equations AT THE SAME TIME.
Sometimes one can't eliminate a variable by adding or subtracting
Let us take the two equations we have just been looking at, simultaneously . . . straight away . . .
x + y = 10 . . . (1) We always number the What do we mean by that?
x -- y = 2 . . . (2) equations on the right
Can both of these equations be true at the same time? 4
4.9 Copyright © The Answer Series: Photocopying of this material is illegal
 Worked example Can you solve these equations?
Solve for x and y : 2x + 3y = -1 . . . (1) ar 6 = 162 . . . (1)
ar 2 = 2 . . . (2)
3x - 6y = -12 . . . (2)
Let us try addition . . .
We first need to alter equation number (1) so that adding it to or subtracting it (1) + (2) : ar 6 + ar 2 = 162 + 2
from equation number (2) will eliminate one of the variables : a(r 6 + r 2) = 164 and now what ? ? ? – no good!
(1) % 2 : 4x + 6y = - 2 . . . (3) Now, observe the coefficient
Addition and subtraction don't work, do they! What else can we try?
of y in (2) and (3)
Equations can also be multiplied or divided :
(2) + (3) : â 7x = -14 . . . y has been eliminated ! because, if a = b
â x = -2 and c = d
OR : Eliminate x
a b
– see below then a.c = b.d and =
Subst. x = - 2 in (1) : â - 4 + 3y = - 1 c d
â 3y = 3 Which of these is best to use in our example to eliminate one of the variables?
â y = 1
This is an ORDERED PAIR
Answer
â Solution : (- 2; 1)  .... – x first, then y second!
ar 6
– a good way to give the solution (1)  (2) : = 162 . . . Division seems to be the way to eliminate a
ar 2 2
of 2 equations with 2 unknowns. â r 4 = 81 Even though this sum
â r = 3 looks different (and is !)
CHECK THIS ANSWER BY substituting the values into the Subst. r =  3 in (2) : a % 9 = 2 the LOGIC is the same !
equations (1) and (2) to see whether they hold true ! 2
ALGEBRAIC EQUATIONS & INEQUALITIES

â a = 
9
We could've gone for eliminating x (instead of y) :
(1) % 3 : 6x + 9y = - 3 . . . (3) Remember :
EXERCISE 4.6
EXERCISE 4.6
It is possible to check your answers !
(2) % 2 : 6x - 12y = - 24 . . . (4)
(3) - (4) : â 21y = 21 ... x has been eliminated Solve the following pairs of simultaneous equations :
â y = 1 1. x + y = 12 Do this one by inspection 2. a + 7b = 49
Subst. y = 1 in (1) : 2x + 3 = - 1 x-y = 4 first and then algebraically. a + 3b = 9
â 2x = - 4
3. p + 2q = 1 4. 2x + 3y = 8 5. 2x = 3y - 4
â x = - 2, etc.
3p - q = 10 3x + 4y = 11 y = 3-x

y x x+y 2x - y
6. +1= 7. = 7- ...–
2 5 2 3
1 x-y x+y
and x+ 1 = 1y and - +4
1
=0 ... —
4 2 3 4 3 2

8. The length of a rectangle is a mm and the breadth is b mm. The area of the
rectangle is unchanged if the length is increased by 6 mm and the breadth is
diminished by 2 mm. The area is also unchanged if the length is decreased by
6 mm and the breadth is increased by 3 mm. Find the length and breadth of the
4 original rectangle.
Copyright © The Answer Series: Photocopying of this material is illegal 4.10
Answers (1) % 3 : 6x - 15y = 30 . . . (3)
1. 'By inspection' : The sum of 2 numbers is 12 and their difference is 4. (2) % 2: 6x - 8y = -12 . . . (4)
What are the numbers ? (3) - (4) : â -7y = 42
 (-7) : â y = -6
Algebraically :
From (1) : 2x = 5y + 10
x + y = 12 . . . (1) 2. a + 7b = 49 . . . (1)
= 5(- 6) + 10
x - y = 4 . . . (2) a + 3b = 9 . . . (2)
= - 30 + 10
(1) + (2) : 2x = 16 (1) - (2) : 4b = 40 = - 20
â x = 8 â b = 10 â x = -10 and y = - 6
(1) : 8 + y = 12 (2) : a + 30 = 9 â Solution : (-10; - 6)
â y = 4 â a = - 21
â Solution : (8; 4) â Solution : (- 21; 10) 7. (1) % 6 : 3(x + y) = 42 - 2(2x - y)
Did you get this by inspection? â 3x + 3y = 42 - 4x + 2y
â 7x + y = 42 . . . (3)
2x + 3y = 8 . . . (1)
3. p + 2q = 1 . . . (1) 4. (2) % 12 : 3(x - y) - 4(x + y) + 54 = 0
3x + 4y = 11 . . . (2)
3p - q = 10 . . . (2) â 3x - 3y - 4x - 4y + 54 = 0
(2) % 2 : 6p - 2q = 20 . . . (3) (1) % 3 : 6x + 9y = 24 . . . (3)
â - x - 7y = - 54
(1) + (3) : 7p = 21 (2) % 2 : 6x + 8y = 22 . . . (4) % (-1) : â x + 7y = 54 . . . (4)
â p = 3 (3) - (4) : â y = 2 (3) % 7 : â 49x + 7y = 294 . . . (5)
(2) : 9 - q = 10 (3) : â 6x + 18 = 24
q = -1 (4) - (5) : â - 48x = - 240
â 6x = 6
â x = 5

ALGEBRAIC EQUATIONS & INEQUALITIES

â Solution: (3; -1) â x = 1
â Solution : (1; 2) (3) : â y = 42 - 7x
= 42 - 35
5. 2x = 3y - 4 . . . (1) â y = 7
y = 3 - x . . . (2)
â Solution : (5; 7)
(2) in (1) : 2x = 3(3 - x) - 4
â 2x = 9 - 3x - 4
â 2x + 3 x = 9 - 4
8. (a + 6)(b - 2) = ab and (a - 6)(b + 3) = ab
â 5x = 5
â ab - 2a + 6b - 12 = ab â ab + 3a - 6b - 18 = ab
â x = 1
â - 2a + 6b = 12 â 3a - 6b = 18
(2) : y = 3 - 1  2) â - a + 3b = 6 . . . (1)  3) â a - 2b = 6 . . . (2)
â y = 2
(1) + (2) : b = 12
â Solution : (1; 2)
From (2) : a = 2(b) + 6
y x x+ 1 = y = 2(12) + 6
6. +1=
2 5 4 2 3 = 24 + 6
% 10) 5y + 10 = 2x % 12) 3x + 6 = 4y = 30
â 5y - 2x = -10 â 3x - 4y = - 6 . . . (2) â length = 30 mm and breadth = 12 mm
% (-1) : 2x - 5y = 10 . . . (1) â Solution: (30; 12)
4
4.11 Copyright © The Answer Series: Photocopying of this material is illegal
 The SIGNS of the trig ratios IN A FLASH! The trig ratios of 90º and multiples of 90º
y sin  Use this procedure to find the trig ratios of 90º ; 180º ; 270º & 360º (& 0º)
sin  = and y is positive in I and II
r II I y 5 y 0
 sin  is POSITIVE in quadrants 1 & 2 (0; 5) sin 90º =
r
= = 1 180º sin 180º =
r
= = 0
5 r=6 6
(and negative in 3 & 4) 'SUNRISE FOR SINE' r=5
90º cos 90º =
x
=
0
= 0
(- 6; 0)
O cos 180º =
x
=
-6
= -1
r 5 r 6
O y 5 y 0
cos  =
x  and x is positive in I and IV cos  tan 90º =
x
=
0
=  tan 180º =
x
=
-6
=0
r I
x=0; y=5; r=5 x = -6 ; y=0; r=6
 cos  is POSITIVE in quadrants 1 & 4 IV
(and negative in 2 & 3) 270º y y
'PULL A CURTAIN FOR COS' sin 270º = =
-4
= -1 sin 360º = =
0
= 0
O r 4 r 5
x 0 r=5 x 5
y
r=4 cos 270º =
r
=
4
= 0 O (5; 0)
cos 360º =
r
=
5
= 1
tan  = tan  y -4 360º tan 360º =
y
=
0
=0
x  I (0; - 4) tan 270º = = ∞ x
x 0 5
and x & y have the same sign in I and III III x=0; y = -4 ; r = 4 x=5; y=0; r=5
 tan  is POSITIVE in quadrants 1 & 3
Apparently, this is THE BMW SYMBOL!!! Note : The results for 0º and 360º are the same.
(and negative in 2 & 4)
SUMMARY
Learn these easy PICTURES so that you know
the SIGNS of your trig ratios IN A FLASH ! : 0 90 180 270 360

sin  : 0 1 0 -1 0
NO MORE CAST RULE!!!
cos  : 1 0 -1 0 1
tan  : 0 ∞ 0 ∞ 0
The 4 steps to find the trig ratios of any angle:
→
1. Place the ø in STANDARD POSITION (starting at OX ) . . . end arm Trigonometric graphs

O X We will learn how to sketch the graphs y = sin , y = cos  and y = tan  for 0º ≤  ≤ 360º.
2. Pick a point (x; y) on the end arm of the ø beginning arm We will use the critical values of these ratios to make it easy. But first, some terminology . . .
– we'll call its distance from the origin r Terminology
TRIGONOMETRY

(hypotenuse)
The sine and cosine graphs are WAVE-shaped.
3. Write down x = y = r = (x; y)  The amplitude of a WAVE is the deviation from its centre line :
r y
(opposite)  The period of a graph is the number of degrees spanning a FULL WAVE.
4. Apply the DEFINITIONS   The range is the set of all the possible y-values.
O
x Our investigations of the trig ratios have shown us that the range of values
y x y (adjacent) of sines and cosines is very small - only between - 1 and 1.
sin  = cos  = tan  =
r r x We write: -1  sin   1 and -1  cos   1 for all values of !
5 By contrast, the range of tan values is from - ∞ to + ∞ !
-1 0 1

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 A Summary of Algebraic Functions Exponential functions : y = abx + q

a>0 y y a<0
Straight lines : y = ax + q  6 x-intercept (y = 0)
0<b<1 b>1
a>0 y y a<0  x-intercept (y = 0) & y-intercept (x = 0) only if a > 0 & q < 0 or
O x O x a<0&q>0
q  Domain : x ∈ R
q
O x O x  Range : y ∈ R b>1
 y-intercept (x = 0)
0<b<1
² (0; a) if q = 0
(No axes of symmetry; no asymptotes) y y
otherwise ² (0; a + q)
0<b<1 b>1

Parabolas : y = ax2 + q y=q  Domain : x ∈ R

(q > 0)
6 6
O x O x
a>0 y y a<0  x-intercept(s) (y = 0) & y-intercept (x = 0)  Range :
0<b<1 b>1 for a > 0, y > q &
q  Domain : x ∈ R
y y for a < 0, y < q
O x O x  Range : y  q if a > 0
y  q if a < 0 0<b<1 b>1
q 6
O 6
x O x  Asymptote : y = q
 Axis of symmetry: x = 0 (the y-axis) y=q
(q < 0)  Axes of symmetry :
(No asymptotes)
there are no axes of symmetry
0<b<1 b>1
Hyperbolas : y = k + q
x
 x-intercept (y = 0) only when q ≠ 0 EXERCISE 6.9
EXERCISE 6.9 -–Algebraic
Algebraicgraphs
graphs (Answers on page 6.25)
k>0 y y k<0
 no y-intercept (x ≠ 0) 1. Given :
 Domain : x ∈ R, x ≠ 0 y = ax + q ; y = ax2 + q ; y= a +q; y = abx + q (b > 0)
O x O x x

FUNCTIONS & GRAPHS

 Range : y ∈ R, y ≠ q 1.1 Choose which one of the above equations suits each graph best.
Give a reason for your answer.
 Axes of symmetry : y y y y
y y k
for y = ² y = x & y = -x A B C D
x
k O x x O x O x
y=q & for y = +q ² y=x+q O
O x O x x
& y = -x + q
 Asymptotes : x = 0 (y-axis) & y = q
1.2 State in each case if a > 0 or a < 0. Motivate your answer.
a > 0 means : a is positive 1.3 State in each case if q > 0; q < 0 or q = 0.  Note : q is not necessarily
a < 0 means : a is negative Motivate your answer. the y-intercept ! 6
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 Investigating quadrilaterals, using diagonals:
QUADRILATERALS
fig. 1 : Use a diagonal to determine the sum of How would you find the sum
the interior angles of a quadrilateral. of the interior angles of
K Revision of Properties of Quadrilaterals fig. 2 : Use a diagonal to find the area of a pentagon? A hexagon?
 Recall all the quadrilaterals ... (kite, trapezium, parallelogram, rectangle, rhombus, square). a trapezium.
 What properties do they have? fig. 3 - 6 : Which of these quadrilaterals have their areas bisected by
the diagonal?
 Equal sides?
fig. 3 - 6 : Draw in the second diagonal. For each figure, establish whether the
diagonals are :
 equal  bisect each other
 intersect at right angles  bisect the angles of the quadrilateral
kite rectangle parallelogram rhombus square
fig. 6 : Find the area of a kite in terms of its diagonals.
Could this formula apply to a rhombus? A square?

'any' quadrilateral rhombus square

K Defining Quadrilaterals
kite
 A trapezium
 Parallel sides?
rectangle Definition : A trapezium is a quadrilateral with
'any' ONE PAIR OF OPPOSITE SIDES parallel.
square
quadrilateral trapezium parallelogram
 A parallelogram
rhombus
We have observed the properties of a parallelogram :
 both pairs of opposite sides parallel
 both pairs of opposite sides equal We will, however, define the
 Angles? parallelogram in terms of
 both pairs of opposite angles equal
Which are equal? Which are supplementary? Which are right angles? its parallel lines.
 diagonals which bisect one another.

 Diagonals? 4a
EUCLIDEAN GEOMETRY

Definition : A parallelogram is a quadrilateral with

TWO PAIRS OF OPPOSITE SIDES parallel.
5
1 2 3
Observe the progression of quadrilaterals below as we discuss further definitions :
rectangle
4b
'any' square
quadrilateral trapezium parallelogram
6a 6b
A diagonal of a
quadrilateral is rhombus
or a line joining
7 opposite vertices.

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  Venn Diagrams Remember !
A graphical method to visually represent the outcomes of two or more different  n(event) = number of favourable outcomes in the event
events (by means of circles) ; together with common elements of the events (by
means of overlapping circles) ; as well as the sample space of all the events (by  P(event) = probability of the number of favourable outcomes n(E) divided
means of a rectangle). by the total number of elements in the sample space n(S)
the number of favourable outcomes that exist n(E)
 Sample Space & Events i.e. P(E) =
the total number of possible outcomes n(S)

Possible Venn Diagram layout :  The sum of the probabilities in the Venn diagram must be 1 or 100%
Sample Space
S – the set of all possible outcomes
A  Mutually Exclusive / Disjoint Events
Event B
B
– an event which contains all S
the possible outcomes of B Event B
A B
– contains all the possible
Common elements (shaded) outcomes of B
Event A – from event A and event B
– an event which contains all [should they exist] Events A & B are . . .
the possible outcomes of A
Event A Mutually exclusive
– contains all the – the events do not overlap
Worked Example 1
possible outcomes of A  they have no common elements
Set up a Venn Diagram to illustrate the following :
 A sample space from 1 to 10 (whole numbers only)
 Event A : the factors of 8
Worked Example 2

 Event B : the multiples of 2 Given that S = {1; 2; 3; 4; 5; 6} and that event A is all the even numbers and event B
is all the odd numbers :
Answer (a) Draw a Venn Diagram to illustrate the situation.
S
(b) Are these events A and B mutually exclusive? Give a reason for your answer.
Event A : Factors of 8 = {1; 2; 4; 8}
A B
6  n(A) = 4 elements
PROBABILITY

1 Answer s
3 2 5
4 Event B : Multiples of 2 = {2; 4; 6; 8; 10} (a) S (b) Yes, these events are mutually
8 10
9  n(B) = 5 elements exclusive as an even number
7
A B can never be an odd number.
Common elements = {2; 4; 8} 246
135

12
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 TRIGONOMETRY [36] QUESTION 5 QUESTION 6

QUESTION 4 5.1 Solve for x, correct to ONE decimal place, 6.1 Consider the function y = 2 tan x.
in each of the following equations where
4.1 In the diagram below, ΔABC is right-angled at B. 0º ≤ x < 90º. 6.1.1 Make a neat sketch of y = 2 tan x for
0º ≤ x ≤ 360º on the axes provided below.
A 5.1.1 5 cos x = 3 (2) Clearly indicate on your sketch the
intercepts with the axes and the asymptotes.
5.1.2 tan 2 x = 1,19 (3)
y
5.1.3 4 sec x - 3 = 5 (4)
B C 6-

5.2 An aeroplane at J is flying directly over a 5-

Complete the following statements : point D on the ground at a height of 5 kilometres.
4-
AB
It is heading to land at point K. The angle of
4.1.1 sin C = (1) depression from J to K is 8º. S is a point
... 3-
along the route from D to K.
AB 2-
4.1.2 . . . A = (1)
BC
J 1-

4.2 Without using a calculator, determine the 8º

0 x
sin 60º . tan 30º 90º 180º 270º 360º
value of : (4) -1 -
sec 45º
5 km
-2 -
4.3 In the diagram, P(- 5; 12) is a point in the
ˆ = θ. -3 -
Cartesian plane and ROP
-4 -
y D S K
-5 -
P(- 5; 12)
5.2.1 ˆ .
Write down the size of JKD (1) -6 -

(4)
EXAM PAPERS: PAPER 2

θ 5.2.2 Calculate the distance DK, correct to

the nearest metre. (3)
O x 6.1.2 If the graph of y = 2 tan x is reflected
R
5.2.3 If the distance SK is 8 kilometres, about the x-axis, write down the
calculate the distance DS. (1) equation of the new graph obtained
by this reflection. (1)
Determine the value of : 5.2.4 Calculate the angle of elevation
from point S to J, correct to
4.3.1 cos θ (3)
ONE decimal place. (2) [16]
2
4.3.2 cosec θ + 1 (3) [12]

Copyright © The Answer Series: Photocopying of this material is illegal E4

 2 12 13 5.2.3 DS = DK – SK
3.2 CD2 = (x - 1)2 + (5 + 2)2 = ( 53 ) 4.3.2 sin θ =  cosec θ =
y 13 12 = 35,58 km – 8 km
D(x; 5) â (x - 1)2 + 49 = 53
13 2 169
â (x - 1)2 = 4 â cosec 2 θ + 1 = ⎛⎜ ⎞⎟ + 1 = +1 = 27,58 km 
⎝ 12 ⎠ 144
x â x-1 = ±2 5
169 + 144 313 ˆ =
5.2.4 tan JSD
Note : x must = =  ⎛ = 2 25  ⎞
â x = 3 or -1 ⎜ ⎟ 27,58
be negative. C(1; - 2) 144 144 ⎝ 144 ⎠
ˆ l 10,3º  . . . tan -1 ⎛ 5 ⎞ =
â JSD ⎜ 27,58 ⎟
⎝ ⎠
But x < 0 in the second quadrant
5.1.1 5 cos x = 3 correct to 1 dec. place
â x = -1  . . . only the neg. value of x is valid
3
÷ 5) â cos x = (= 0,6)
5 6.1.1
AB y
4.1.1 sin C = -1 ⎛ 3 ⎞
AC  â x l 53,1º  . . . cos ⎜ ⎟ =
6-
⎝5⎠
AB
4.1.2 cot A = 5-
BC 5.1.2 tan 2 x = 1,19
â 2x = 49,958…º . . . tan -1 1,19 = 4-
BC 1
Note : tan A = ; cot A =
AB tan A ÷ 2) â x l 25,0º 
3-

4.2 The expression 5.1.3 4 sec x – 3 = 5 2-

3 1 30º + 3) â 4 sec x = 8 1-
× 2
2 3 3 ÷ 4) â sec x = 2 x
= 2
45º
2 1 (0º; 0) 90º (180º; 0) 270º (360º; 0)
60º 1 â cos x =
1 45º 2 -1 -
1 1
= % 1
2 2
1 x = 60º  . . . cos -1 ⎛⎜ ⎞⎟ = -2 -
⎝2⎠
1 2
= x . . . The denominator must -3 -
2 2 2 be rationalised ˆ = 8º 
5.2.1 JKD . . . alternate ø's; || lines
-4 -
2
=  ... 2 % 2 = 2
4 DK 1 -5 -
5.2.2 In ΔJDK : = cot 8º ... =
5 tan 8º
EXAM MEMOS: PAPER 2

4.3.1 OP = 13 units . . . 5 : 12 : 13  ; Pythagoras -6 -

5
% 5) â DK =
y
tan 8º x = 90º x = 270º
P(- 5; 12)
= 35,5768 . . . km
6.1.2 y = - 2 tan x 
13
12 θ = 35 576,8 metres
g(x) = a sin x  g(90º) = a sin 90º
l 35 577 metres  6.2.1 a = 4   4 = a
O x
-5 R . . . correct to the nearest metre
6.2.2 The range of h :
-5 5 x -2 ≤ y ≤ 6  . . . the values of y
â cos θ = = -  . . . cos  =
13 13 r
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 EUCLIDEAN GEOMETRY:
THEOREM STATEMENTS & ACCEPTABLE REASONS

LINES TRIANGLES
EUCLIDEAN GEOMETRY: THEOREM STATEMENTS & ACCEPTABLE REASONS

The adjacent angles on a straight line are ø sum in Δ OR sum of øs in Δ

øs on a str line The interior angles of a triangle are supplementary.
supplementary. OR int øs in Δ

If the adjacent angles are supplementary, the outer The exterior angle of a triangle is equal to the
adj øs supp ext ø of Δ
arms of these angles form a straight line. sum of the two interior opposite angles.

The angles opposite the equal sides in an

øs around a pt OR øs opp equal sides
The adjacent angles in a revolution add up to 360º. isosceles triangle are equal.
øs in a rev
The sides opposite the equal angles in an
sides opp equal øs
isosceles triangle are equal.
Vertically opposite angles are equal. vert opp øs
In a right-angled triangle, the square of the
Pythagoras OR
hypotenuse is equal to the sum of the squares
If AB || CD, then the alternate angles are equal. alt øs; AB || CD Theorem of Pythagoras
of the other two sides.

If AB || CD, then the corresponding angles If the square of the longest side in a triangle is Converse Pythagoras
corresp øs; AB || CD equal to the sum of the squares of the other two OR Converse Theorem of
are equal.
sides then the triangle is right-angled. Pythagoras
If AB || CD, then the co-interior angles are
co-int øs; AB || CD If three sides of one triangle are respectively
supplementary.
equal to three sides of another triangle, the SSS
triangles are congruent.
If the alternate angles between two lines are equal,
alt øs =
then the lines are parallel. If two sides and an included angle of one triangle
are respectively equal to two sides and an
If the corresponding angles between two lines are SAS OR SøS
corresp øs = included angle of another triangle, the triangles
equal, then the lines are parallel. are congruent.

If the co-interior angles between two lines are If two angles and one side of one triangle are
co-int øs supp respectively equal to two angles and the
supplementary, then the lines are parallel. AAS OR øøS
corresponding side in another triangle, the
triangles are congruent.
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