MTN Learn. Mathematics. Grade 10. radio support notes

MTN Learn

Mathematics

Grade 10

radio support notes

Contents

INTRODUCTION ... 2

GETTING THE MOST FROM MINDSET LEARN XTRA RADIO REVISION ... 3

BROADAST SCHEDULE ... 4

ALGEBRAIC EXPRESSIONS ... 5

EXPONENTS ... 9

NUMBER PATTERNS ... 18

LINEAR EQUATIONS ... 23

QUADRACTIC EQUATIONS ... 27

LINEAR AND QUADRATIC EQUATIONS ... 30

LITERAL EQUATIONS ... 32

INEQUALITIES ... 34

TRIGONOMETRY ... 37

INTRODUCTION

Have you heard about Mindset? Mindset Network, a South African non-profit organisation, was founded in 2002. Through our Mindset Learn programme, we develop and distribute high quality curriculum aligned educational resources for Grade 10 - 12. We make these materials available on TV (Dstv and Toptv channels 319), the Internet

(www.mindset.co.za/learn) and as DVDs and books.

At Mindset we are committed to helping South African learners succeed. This is why Mindset Learn is proud to offer Mindset Learn Xtra especially for Grade 10 – 12 learners. Learn Xtra offers you hundreds of hours of video and print support, live television shows between 4pm and 7pm every Monday to Thursday and a free Helpdesk where our expert teachers are on standby to help you. You can find out more about Mindset Learn Xtra at

www.learnxtra.co.za.

Learn Xtra also offers specific exam revision support. Every year we run Winter, Spring, Exam and Supplementary Schools to help you ace your exams. And now, Mindset is proud to announce a powerful partnership with MTN and the Department of Basic Education to bring you Mindset Learn Xtra Radio Revision powered by MTN – a full 3 months of radio programmes dedicated to supporting Grade 10 and 12 Mathematics, Physical Sciences and English First Additional Language.

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GETTING THE MOST FROM MINDSET LEARN XTRA RADIO REVISION

In the Grade 10 Mathematics radio programme, we will focus on questions that come from recent previous exam and test papers. This booklet contains diagrams taken from the exam and test papers so that you will be able to follow what is said during the broadcast.

Before you listen to the show, read through the questions for the show and try to answer them without looking up the solutions. If you have a problem and can’t answer any of the questions, don’t panic. Make a note of any questions you need answered. When listening to the show, compare your approach to the teacher’s. Don’t just copy the answers down but take note of the method used.

Make sure you keep this booklet. You can re-do the questions you did not get totally correct and mark your own work.

Remember that exam preparation also requires motivation and discipline, so try to stay positive, even when the work appears to be difficult. Every little bit of studying, revision and exam practice will pay off. You might benefit from working with a friend or a small study group as long as everyone is as committed as you are. Mindset believes that Mindset Learn Xtra Radio Revision will help you achieve the results you want.

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BROADAST SCHEDULE

Grade 10 Catch up (Term 3)

10-Sep 17:00 -18:00 10

Algebraic Expressions

11-Sep 17:00 -18:00 10

Algebraic Expressions

12-Sep 17:00 -18:00 10

Algebraic Expressions

13-Sep 17:00 -18:00 10

^Exponents

15-Sep 09:00 -10:00 10

^Exponents

17-Sep 17:00 -18:00 10

Numbers and Number Patterns

18-Sep 17:00 -18:00 10

Numbers and Number Patterns

19-Sep 17:00 -18:00 10

Linear Equations

20-Sep 17:00 -18:00 10

Quadratic Equations

22-Sep 09:00 -10:00 10

Linear & Quadratic Equations

24-Sep 17:00 -18:00 10

Literal Equations

25-Sep 17:00 -18:00 10

Inequalities

26-Sep 17:00 -18:00 10

Trigonometry

27-Sep 17:00 -18:00 10

Trigonometry

29-Sep 09:00 -10:00 10

Trigonometry

Grade 10 Exam Revision

15-Oct 17:00 -18:00 10

Linear Functions

16-Oct 17:00 -18:00 10

Quadratic Functions

17-Oct 17:00 -18:00 10

Hyperbolic Functions

18-Oct 17:00 -18:00 10

Exponential Functions

20-Oct 09:00 -10:00 10

Trig Functions

22-Oct 17:00 -18:00 10

Euclidean Geometry

25-Oct 17:00 -18:00 10

Euclidean Geometry

27-Oct 09:00 -10:00 10

Euclidean Geometry

29-Oct 17:00 -18:00 10

^Finance

30-Oct 17:00 -18:00 10

^Finance

2-Nov 18:00 -19:00 10

Statistics

8-Nov 17:00 -18:00 10

Analytical Geometry

9-Nov 17:00 -18:00 10

Measurement

10-Nov 09:00 -10:00 10

Probability

12-Nov 17:00 -18:00 10

^Algebra

13-Nov 17:00 -18:00 10

^Functions

14-Nov 17:00 -18:00 10

^Finance

15-Nov 17:00 -18:00 10

Trigonometry

16-Nov 17:00 -18:00 10

Euclidean Geometry

17-Nov 17:00 -18:00 10

Maths (Helpdesk Questions)

For more information, free downloads and all the schedules, visit www.mtnlearning.co.za.

ALGEBRAIC EXPRESSIONS

STUDY NOTES Expanding

When multiplying a binomial or trinomial by a single term we place the binomial or trinomial in brackets. We use the Distributive Law of multiplication to expand term by term. This means that we multiply each term inside a bracket by the term outside the brackets Simplify: 2a(a - 1) - 3(a² - 1).

Solution:

2a(a - 1) - 3(a² - 1) = 2a(a) + 2a(-1) + (-3)(a²) + (-3)(-1) = 2a2 - 2a - 3a² + 3

= -a² - 2a + 3

When multiplying two binomials together we apply the same Distributive Law. We multiple the two First term together, then the Outer terms(1^st term of the first bracket by the 2^nd term of the second bracket) then the Inner terms (2^nd term of the first bracket by the 1^st term of the second bracket) and then Last terms. Remember FOIL

Find the product: (3x - 2)(5x + 8) Solution:

(3x - 2)(5x + 8) = (3x)(5x) + (3x)(8) + (-2)(5x) + (-2)(8) = 15x² + 24x - 10x - 16

= 15x² + 14x - 16

Multiplying a Binomial by Trinomial

Find the product: (x - 1)(x² - 2x + 1) Solution:

Step 1: Expand the bracket

(x - 1)(x² - 2x + 1) = x(x² - 2x + 1) - 1(x² - 2x + 1) = x³ - 2x² + x - x² + 2x - 1 Step 2: Simplify

= x³ - 3x² + 3x - 1

Factorising

Methods of Factorising

 taking out a common factor

 difference of two squares

 trinomials to binomials

 sum and difference of cubes

 grouping

Example: Factorise: 5(a - 2) - b(2 - a) Solution:

(a-2) and (2-a) are not common factors. The terms have different signs. If we take out a factor of -1 from (2-a), we can change the signs

2 - a = - (a - 2)

Now we have a common factor in both terms. Be careful of the signs.

5(a - 2) - b(2 - a) = 5(a - 2) - [-b(a - 2)]

= 5(a - 2) + b(a - 2) = (a - 2)(5 + b) Factorising trinomials

 Recognise the pattern. The process is the opposite of FOIL.

 The factors of the first term of a trinomial become the first terms of each of the binomials which we place in brackets.

 The factors of the last term of a trinomial become the second term of each of the binomials in the brackets.

 We select the factors so that when we expand the multiplication of inner and outer terms gives us the middle term of the trinomial.

 Always check that you have the correct factors by expanding the factors using FOIL.

Difference and sum of two cubes

You need to recognise the pattern. Notice that the product of a binomial and trinomial gives us the sum of two cubes.

(x + y)(x² - xy + y²) = x(x² - xy + y²) + y(x² - xy + y²)

= [x(x²) + x(- xy) + x(y²)] + [y(x²) + y( - xy) + y(y²)]

= x³ - x²y + xy² + x²y - xy² + y³ = x³ + y³

Also this product gives the difference of two cubes:

(x - y)(x² + xy + y²) = x(x² + xy + y²) - y(x² + xy + y²)

= [x(x²) + x(xy) + x(y²)] - [y(x²) + y(xy) + y(y²)]

= x3 + x²y + xy² - x²y - xy² - y³ = x³ - y³

So to factorise the sum of cubes, we use the pattern. The terms in the first bracket are added together and the middle term in the second bracket is subtracted.

x³ + y³ = (x + y)(x² - xy + y²)

For the difference of cubes, the terms in the first bracket are subtracted and the middle term in the second bracket is added.

x³ - y³ = (x - y)(x² + xy + y²)

Algebraic Fractions

Simplifying algebraic fractions

We working with algebraic fractions in the same way we work with fractions of numbers.

Multiplication and division Simplify:

x² – x – 2 ÷ x² + x (x ≠ 0; x≠ ±2) x² - 4 x² + 2x

Solution:

Step 1: Factorise the numerator and denominator

(x+1)(x-2) ÷ x(x + 1) (x+2)(x-2) x(x +2)

Step 2: Change the division sign and multiply by the reciprocal

(x+1)(x-2) X x(x + 2) (x+2)(x-2) x(x +1)

Step 3: Write the final answer

= 1

Radio Broadcast 10 Sept 17:00 -18:00

Questions for Discussion

Expand and simplify the following:

(a) (2x3 )y ²(3x4 )y ² (4)

(b) (9x²4)(3x2)(3x2) (2)

(c) (x2 )(y x²2xy3y²) (3)

Radio Broadcast 11 Sept 17:00 -18:00

Questions for Discussion

Factorise fully:

(a)

15

a b^{4 6}

 3

ab² (2)

(b)

8

a⁸

 8

b⁸ (4)

(c) 3x²3x18 (2)

(d) 5x²14x8 (2)

(e) a³

 3

a²

 

a

3

(4)

Radio Broadcast 12 Sept 17:00 -18:00

Questions for Discussion

Simplify:

(a) 2 3

1 2 1

1  4x  3x  2xy (5)

(b) 1 2 3

2 3 4

x x x

  ⁽⁵⁾

(c)

2 3

4 8 2

12 9

x x x

x x

 

 ⁽⁴⁾

(d) 2 1

3 3

    

  

x x  (4)

EXPONENTS

STUDY NOTES

Exponential notation is a short way of writing the same number multiplied by itself many times. For any real number a and natural number n, we can write a multiplied by itself n times as aⁿ. a is called the base, n is called the exponent or index.

When doing calculations with exponents we use the following Law of Exponents:

 a^m

.

aⁿ



a^{m n}

The bases are the same so you will then add the exponents For example

4 4 1 4 1 5

. .

x x



x x



x ^



x (same bases, add exponents)

3 3 6

2 .2  2  64

(same bases, add exponents)

 ^{m m n}

n

a a

a

 

The bases are the same so you will then subtract the exponents

For example

4 4

4 1 3

1

x x

x x

x x

    (same base, subtract exponents)

12 12 10 2

10

2 2 2 4

2

    (same base, subtract exponents)



(

a^{m n}

) 

a^{m n}

With this rule, you will need to multiply the exponents

For example

4 2 1 4 2 1 2 4 2 8

(3

x

)  (3

x

)  3

^

.

x ^

 9

x

 _n

1

x n

x





and 1 _n

n x x^



With this rule, you make the exponents positive

For example 2

2

1 1

3 3 9

   ²

2

1 3 9

3

^

 

 a⁰

 1

Any number raised to an exponent of 0 equals 1 For example

3

0

 1

4

x0

   4 1 4



n n

a b

b a

 



  

   

   

For example

2 2

3 4 16

4 3 9

 



   

   

   

Simplifying Exponents

Change the bases to products of their prime factors and simply using the laws.

Example 1:

Simplify:

Solution:

Step 1: Change the bases to prime numbers

Step 2: Simplify the exponents

Example 2:

Simplify:

Step 1: Look for common factors

Here is a common factor in both numerator and denominator Step 2: Take out the common factor

Step 3: Cancel the common factor

Step 4: Simplify the fraction

Solving exponential equations

In an exponential equation, the exponent is the unknown.

For example, the equation

3

^x

 9

is called an exponential equation because the unknown variable is the exponent. Clearly, the solution to this equation is x2 because

3

2

 9

. Getting the bases on both sides of the equation to be the same and then equating the exponents will solve the equation as follows:

2

3 9

3 3 2

x x

x



 

 

If a^x a^y then we can equate the exponents, since the bases are the same.

Therefore we can say that x



y.

Example 1

Solve for x:

2

^x

 16

Step 1:

Change the bases to the same prime numbers

Step 2: Bases are the same so equate the exponents

Example 2

Solve for x: 3^x 3 100 64

Step 1:

Simplify the equation

Step 2: Bases are the same so equate the exponents

Radio Broadcast 13 Sept 17:00 -18:00

Questions for Discussion

Simplify the following without using a calculator:

(a)  ( 3x^{4 2})  ( 2x^{2 3}) 4(2 )x ²3x⁰ (5) (b)

2 2 3 6

6 3

2( ) ( )

(2 )

a b ab

b

  



 (5)

(c)

3 18 8 2

( 2) . 2 (2 ) . 2

(5)

Radio Broadcast 15 Sept 09:00 -10:00

Questions for Discussion

(a) 3^x =

(b) 2^m+2 = (0,5)^3+2m

(c)

2^x0,125

(3)

(d)

16 . 4^{2 1} 4

x x

(6)

SOLUTIONS TO QUESTIONS

Radio Broadcast 10 Sept 17:00 -18:00

(a) 2 2

2 2 2 2

2 2 2 2

2 2

(2 3 ) (3 4 )

4 12 9 (9 24 16 )

4 12 9 9 24 16

5 36 7

x y x y

x xy y x xy y

x xy y x xy y

x xy y

  

     

     

   

(b) ²

2 2

4

(9 4)(3 2)(3 2)

(9 4)(9 4)

81 16

x x x

x x

x

  

  

 

(c) 2 2

2 2 2 2

3 2 2 2 2 3

3 2 2 3

( 2 )( 2 3 )

( 2 3 ) 2 ( 2 3 )

2 3 2 4 6

4 7 6

x y x xy y

x x xy y y x xy y

x x y xy x y xy y

x x y xy y

  

     

     

   

Radio Broadcast 11 Sept 17:00 -18:00

(a) 4 6 2

2 3 4 2

2 3 4

15 3

3 5 3 1

3 (5 1)

a b ab

ab a b ab

ab a b



   

 

(b) ^{8 8}

8 8

4 4 4 4

4 4 2 2 2 2

4 4 2 2

8 8

8( )

8( )( )

8( )( )( )

8( )( )( )( )

a b

a b

a b a b

a b a b a b

a b a b a b a b



 

  

   

    

(c) 2 2

3 3 18

3( 6)

3( 3)( 2)

x x

x x

x x

 

  

  

(d) 2

5 14 8

(5 4)( 2)

x x

x x

 

  

(e) ^{3 2}

2

2

3 3

( 3) ( 3) ( 3)( 1) ( 3)( 1)( 1)

a a a

a a a

a a

a a a

  

   

  

   

Radio Broadcast 12 Sept 17:00 -18:00 (a)

2 3

3 2

3 2 3 2

3 2

3 3 3 3

3 2

3

1 2 1

1 4 3 2

1 12 1 3 2 4 1 6

1 12 4 3 3 4 2 6

12 3 8 6

12 12 12 12

12 3 8 6

12 x x xy

x y xy y x

xy y xy

x y x x x

x y xy y x

x y x y x y x y

x y xy y x

x y

  

       

   

  



(b)

1 2 3

2 3 4

1 6 2 4 3 3

LCD 12

2 6 3 4 4 3

6( 1) 4( 2) 3( 3)

12 12 12

6( 1) 4( 2) 3( 3) 12

6 6 4 8 3 9

12

7 5

12

x x x

x x x

x x x

x x x

x x x

x

  

 

  

      

  

  

    



    



 

(c)

²

3

3

2

4 8 2

12 9

4 ( 2) 9

( 2) 12

( 2) 9

( 2) 3

3

x x x

x x

x x x

x x

x x

x x

x

  

  



  





(d)

2

2

2

2 1

3 3

1 2 2

3 3 9

9 3 6 2

9

9 3 2

9

x x

x x x

x x x

x x

    

  

  

   

  



 



Radio Broadcast 13 Sept 17:00 -18:00 (a)

4 2 2 3 2 2 0

8 6 2 2

8 8 2

8 2

( 3 ) ( 2 ) ( ) 4(2 ) 3 (9 ) ( 8 )( ) 4(4 ) 3(1)

9 8 16 3

16 3

x x x x x

x x x x

x x x

x x

     

     

    

   

(b)

2 2 3 6

6 3

6 6 6 6

18 0 12

18 18 12 18 12 6

2( ) ( )

(2 ) 2

8 2

8 2(1)

8 2 8

4

a b ab

b a b a b

b a b

b b b b b b

  



  



















(c)

3 18 8 2

2 18

16 18 16

18 16 19 16 3

( 2) . 2 (2 ) . 2

( 2) ( 2). 2 2 . 2 2( 2). 2

2 . 2 2 . 2

2 2 2 2 8













Radio Broadcast 15 Sept 09:00 -10:00

(a)

(b) 2^m+2 = (0,5)^3+2m

(c)

3

2 0,125

2 125

1000 2 1

8

2 2

3

x x

x

x

x





 

 

 

  

(d)

2

4 2 2

2

4 4

2 4 4

2

8 2

16 . 4 1 4

(2 ) . (2 ) 1 2 2 . 2 1

2 2 1

2 2 2 8 2

2 8 1 4

x x

x x

x x

x x

x

x x

x







 

 

 

 

  

  

  

NUMBER PATTERNS

STUDY NOTES

Linear number patterns

A linear or arithmetic number pattern has a constant difference between consecutive terms.

Example 1: Consider the pattern:

2;5;8;11;...

Each term of the pattern is obtained by adding 3. The difference between the second and first term is therefore 3. The difference between the third and the second term is also 3.

We say that:

2 (first term)

3 (constant difference) a

d





We can continue the pattern as follows:

2;5;8;11;14;17;...

We define the first term as T₁2, the second term as T₂ 5, the third term as T₃ 8 and so forth. It is easy to now find the sixth term, which is 17. However, to find the 1000^th term would take too long.

We can develop a rule or formula to help us find the 1000^th term as follows:

1 2 3 4 5 10

T 2

T 2 3 2 (1)3 T 2 3 3 2 (2)3 T 2 3 3 3 2 (3)3 T 2 3 3 3 3 2 (4)3 T 2 (10 1)3

T 2 ( 1)3

T 2 3 3

T 3 1

n n n

n n n



   

    

     

      

  

  

   

  

The rule (general or nth term) can be used to determine any specific term of the pattern.

For example, the 4^th term can be calculated as follows:

4

T 3 1

T 3(4) 1 11

n  n

    The 1000^th term is:

1000

T 3 1

T 3(1000) 1 2999

n  n

   

Example 2

Consider the sequence:

3 ; 5 ; 7 ; 9 ; ...

   

a.) What type of sequence is this?

b.) Find the nth term of the sequence.

c.) What is the 10^th term of the sequence?

d.) Will the number -201 be part of this sequence? If so what term will it be?

Solution:

a.) Is there a common difference?

Start with the last term and subtract the previous term d= T4 – T3 = -9 – (-7) = -2

d =T3 – T2 = -7 – (-5) = -2 d =T2 – T₁ = -5 – (-3) = -2

There is the same difference between each of the terms.

The common difference d = -2. So this sequence is a linear sequence.

b.) Draw up a table to find the pattern:

Number of terms (n) Term (T_n)

Pattern

1 T₁ = -3 -3

2 T₂ = -5 -3 -2 = -3 + (1)(-2)

3 T₃ = -7 -3 -2 -2 = -3 + (2)(-2)

4 T₄ = -9 -3 -2-2-2 = -3 + (3)(-2)

n T_n -3 -2 …..-2 =-3 +(n-1)(-2)

c.) To find the 10^th term substitute into the formula for the nth term:

Let n =10

d.) Look at the terms of the sequence and notice that they are consecutive negative odd numbers. The pattern starts at -3. Since -201 is a negative odd number less than -3 it will be part of the sequence. It can be written in the form of the general term

-201 = -2(100) -1

This means that -201 is the 100^th term of the sequence.

Example 3

Write down the next three terms in each of the following sequences:

-8; -3; 2; ...

Solution

Step 1: Identify the common difference:

d= 2 –(-3) = -3 – (-8) = 5

This sequence is a linear sequence.

Step 2: Add the common difference to the 3^rd term to get the 4^th and so on:

T₄ = T₃ + d = 2 + 5 = 7 T₅ = T₄ + d = 7 + 5 = 12 T₆ = T₅ + d = 12 + 5 = 17

Notice: In this example the sequence is increasing in value Example 4

30; 27; 24; ...

Solution

Step 1: Identify the common difference:

d= 24 –27 = 27 – 30 = -3

This sequence is a linear sequence.

Step 2: Add the common difference to the 3^rd term to get the 4^th and so on:

T4 = T3 + d = 24 + (-3) = 21 T₅ = T₄ + d = 21 + (-3) = 18 T₆ = T₅ + d = 18 + (-3) = 15

Notice: In this example the sequence is decreasing in value

Radio Broadcast 17 Sept 17:00 -18:00

Questions for discussion Question 1

Identify the pattern in each of the following sequences:

(a) -2; -7; -12; -17;...

(b) 3; 6; 12; 24;...

(c) 0; 3; 8; 15;...

Solution

(a) Linear pattern with a common difference of -5

(b) Not linear – each new term is found by multiplying the last by 2 (c) Not linear – the terms are found by following the pattern: n² – 1

Question 2

Determine the nth term and 16^th term:

(a) 5; ; ; 4; ……. (3)

(b) -2,5; -0,5; ; 2,5; …… (3)

Solutions (a)

Number of terms (n) Term (T_n)

Pattern

1 T1 = 5 5

2 T₂ = 8 5 + 3 = 5 + (1)(3)

3 T₃ = 11 5 + 3 + 3 =5 + (2)(3)

4 T4 = 14 5 + 3 + 3 + 3= 5 + (3)(3)

N T_n 5 + 3 …. + 3= 5 +(n-1)(3)

T_n = 3n + 2

T₁₆ = 3(16) + 2 = 48 + 2 = 50

(b)

Number of terms (n) Term (T_n)

Pattern

1 T₁ = -2,5 -2,5

2 T₂ = -0,5 -2,5 + 1,5 = -2,5 + (1)(1,5)

3 T₃ = 1 -2,5 + 1,5 + 1,5 =-2,5 + (2)(1,5)

4 T₄ = 2,5 -2,5 + 1,5 + 1,5 + 1,5 = -2,5 + (3)( 1,5)

n T_n -2,5 + ,5 …. + ,5= -2,5 +(n-1)(1,5)

T_n = 1,5n -3,5

T₁₆ = 1,5(16) -3,5 = 48 + 2 = 20,5

Radio Broadcast 18 Sept 17:00 -18:00

Questions for discussion

Chains of squares can be built with matchsticks as follows:

(a) Use matches and create a chain of 4 squares. How many matches were used?

(b) Create a chain of 5 squares. How many matches were used?

(c) Determine a conjecture for calculating the number of matches in a chain of

n

squares.

(d) Now determine how many matches will be needed to build a chain of 100 squares.

Solution

Number of terms (n) Term (T_n)

Pattern

1 T₁ = 4 4

2 T2 = 7 4 + 3 = 4 + (1)(3)

3 T₃ = 10 4 + 3 + 3 = 4+ (2)(3)

4 T₄ = 13 4 + 3 + 3 + 3 = 4+ (3)( 3)

5 T₅ = 16 4 + 3 + 3 + 3 = 4+ (3)( 3)

N T_n 4 + 3…. + 3= 4 +(n-1)(3)

(a) 13 matches (b) 16 matches

(c) For a conjecture we need to describe a pattern for the nth term T_n = 3n + 1

(d) T100 = 3 (100) +1 = 301

LINEAR EQUATIONS

STUDY NOTES

Linear Equation: an equation in which can be written in the form of binomial equal to zero.

Variable has a maximum power of 1.

Variable: a number or set of numbers, represented by a letter. Also called an unknown in an equation.

Solving linear equations

Step 1: Group like terms

Step 2: Collect numbers on the one side of the equation and variables on the other Remember: If a term is subtracted on one side of the equation, we add the term to both sides of the equation to collect like terms

If a term is added on one side of the equation, we subtract the term from both sides of the equation to collect like terms

To simplify a variable multiplied by a number, we divide both sides of the equation by the number.

To simplify a variable divided by a number, we multiply both sides of the equation by the number.

Example 1: Solve for a 2 -3(a + 3) = 2a +3 Solution

Step 1: Expand brackets 2 -3a -9 = 2a +3

Step 2: Collect like terms and solve for a -3a -7 = 2a + 3

-7 -3 = 2a + 3a 5a = -10 a = -2

Solving linear simultaneous equations

There are two methods to solve linear equations that have two unknowns (variables) By elimination

In this method, we add or subtract the equations from each other to eliminate one of the variables. You can multiply or divide all terms of one of the equations by a number to get a simple solution. Once the value of one variable is known, we substitute back into one of the original equations to solve for the second variable.

Example 2

Solve for

x

and

y

: 3x y 10...A

6

x y ...B

Multiply each term of equation B by

 3

and call the new equation C:

3x y 10...A 3x 3y 18...C

   

The reason for multiplying equation B by

 3

is to be able to eliminate the terms in

x

. These terms will be eliminated if they differ in sign.

3x y 10...A 3x 3y 18...C

   

4y 8...(add the like terms of A and C)  y 2

Now substitute

y  2

into either A, B or C to get

y

: 3x(2) 10 (equation A was used)

3 12 4

x x

 

 

By substitution

In this method, we simplify one of the equations so that one of the unknowns (variables) is equal to an expression that includes the other unknown. Next we substitute this new equation into the second equation and solve for the unknown. Once the value of one variable is known, we substitute back into one of the original equations to solve for the second variable.

Example 3

Solve for

x

and

y

: 3x y 10

6 x y

Solution

Label each equation as follows:

3x y 10...A 6

x y ...B

Now pick either one of the equations and solve for one of the variables. Let’s solve for the variable

y

in equation A:

3 10

3 10

3 10...C x y

y x

y x

 

   

  

Now replace the variable

y

in equation B with

3

x

 10

and solve for

x

:

6

(3 10) 6 4 10 6 4 16

4

x y

x x

x x x

 

   

  

 

 

Now substitute x

 4

into either equation A, B or C to get

y

:

3(4) 10 (equation C was chosen)

2

y

y

 

 

Radio Broadcast 19 Sept 17:00 -18:00

Questions for discussion

Solve the following equations:

(a) 1 2

2 3 1

x x  (4)

(b)

2(

x

 1)

²

 (2

x

 3)(

x

  1) 0

(4)

SOLUTIONS

(a) 1 2

2 3 1 ( 2) 2 3 1

3 2( 2) 6

3 2 4 6

2 x x

x x

x x

x x

x

  

   

   

   

 



LCD  6

 3x2(x2)6



3

x

 2

x

  4 6

x

 2

(4)

(b) ²

2 2

2 2

2 2

2 2

2( 1) (2 3)( 1) 0

2( 2 1) (2 2 3 3) 0

2( 2 1) (2 3) 0

2 4 2 2 3 0

2 2 4 2 3 0

3 5

3 5

3 3

5 3

x x x

x x x x x

x x x x

x x x x

x x x x

x x

x

    

       

      

      

      

  

 

 

 

 

 2x²4x2

 2x² x 3

 2x²  x 3 0

 5

x3

(4)

QUADRACTIC EQUATIONS

STUDY NOTES

Quadratic equation: equation in which can be written in the form of trinomial equal to zero.

Variable has a maximum power of 2.

Solving quadratic equations

Step 1: Expand and group like terms so that the equation is equal to zero Step 2: Factorise the trinomial

Step 3: Let the terms in both brackets equal zero

Step 4: Solve the equations to find two possible values of the variable.

Step 5: Check that the solution values are valid. Values that result in division by zero are not real and are excluded as solutions

Example 1

Solve for x

Step 1: Expand and group like terms so that the equation is equal to zero

Step 2: Factorise the trinomial

Step 3: Let the terms in both brackets equal zero

Step 4: Solve the equations to find two possible values of the variable.

Step 5: Check that the solution values are valid.

(10)(10 -5) = 50 (-5)(-5-5) = 50 Both solutions are valid

Radio Broadcast 20 Sept 17:00 -18:00

Questions for discussion

Solve the following:

(a) x²

 16

(3)

(b)

12

x²

 3

x (4)

(c)

2

x²

 10

x

 12  0

(3)

(d) (a2)(2a 1) 0 (2)

(e) (a2)(2a 1) 25 ⁽⁵⁾

Solutions

(a) ²

2

16 16 0 ( 4)( 4) 0

4 or 4 x

x

x x

x x



  

   

   

 x²160

 (x4)(x4)0

 x

  4 or

x

 4

(3)

(b) 12x²3x

12 2 3 0 3 (4 1) 0 3 0 or 4 1 0

0 or 4 1 1

4

x x

x x

x x

x x

x

  

  

   

  



 12x²3x0

 3 (4x x 1) 0

 x

 0

 1

x 4

(4)

(c) ²

2

2 10 12 0

5 6 0

( 6)( 1) 0 6 or 1

x x

x x

x x

x x

  

   

   

   

 x²5x 6 0

 (x6)(x 1) 0

 x

 6 or

x

  1

(3)

(d) ( 2)(2 1) 0

2 or 1 2

a a

a a

  

   

 a

  2

 1

a2

(2)

(e) (a2)(2a 1) 25 _ 2a²3a 2 25

 2a²3a270

 (2a9)(a 3) 0

 9

a 2

 a

 3

(5)

2 2 2

2 4 2 25 2 3 2 25 2 3 27 0 (2 9)( 3) 0 2 9 0 or 3 0 2 9 or 3

9 2

a a a

a a

a a

a a

a a

a a

a

    

   

   

   

    

   

  

LINEAR AND QUADRATIC EQUATIONS

Radio Broadcast 22 Sept 09:00 -10:00

Questions for discussion

1. Solve the following linear equations:

(a.) 2 6 1

4 3 2

m m  (4)

(b)

(

x

 3)(

x

  4) (3

x

 2)

²

 

x

8 (

x x

 1)

(5) 2. Solve the following quadratic equations:

(a.)

(

p

 2)

²

 16

(4)

(b) 20 ( x 1)(x2)0 ⁽⁶⁾

3. Solving the following system of linear equations simultaneously:

2 and 2 10

x y x y ⁽⁴⁾

Solutions

1(a)

2 6 1

4 3 2

3( 2) 4( 6) 6

3 6 4 24 6

24 24

m m

m m

m m

m m

   

    

    

  

 



LCD 12 

 3(m 2) 4(m 6) 6



3

m

  6 4

m

 24  6

m

 24

(4)

1(b) ²

2 2 2

2 2 2

2 2

( 3)( 4) (3 2) 8 ( 1)

12 (9 12 4) 8 8

12 9 12 4 9 8

8 13 16 9 8

4 16 4

x x x x x x

x x x x x x x

x x x x x x

x x x x

x x

      

        

       

    

 

 

 x² x 12

 9x²12x4

 8x²8x

9x²12x4

 x

 4

(5)

2(a) ²

2 2

( 2) 16

4 4 16

4 12 0

( 6)( 2) 0

6 or 2 p

p p

p p

p p

p p

 

   

   

   

   

 p²

 4

p

 4

 p²

 4

p

 12  0

 (p6)(p2)0

 p6 or p 2 (4)

2(b)

2 2 2 2

20 ( 1)( 2) 0

20 ( 3 2) 0

20 3 2 0

3 18 0 3 18 0 ( 6)( 3) 0

6 or 3

x x

x x

x x

x x

x x

x x

x x

   

    

    

   

   

   

   



20 ( 

x²

 3

x

  2) 0



20 

x²

   3

x

2 0



   

x²

3

x

18 0

 x²

 3

x

  18 0

 (x6)(x 3) 0

 x

  6 or

x

 3

(6)

3.

2

2( 2) 10

2 4 10

3 6

2 2 2 4

x y

y y

y y

y y x x

 

   

   

 

 

  

 

OR

2

2 10

x y x y

 

 

3 12

4

4 2

2 2 x x

y y y





  

  

 

 x y 2

 2(y  2) y 10

 y2

 x

 4

(4)



3

x

 12

 x

 4

 y2

LITERAL EQUATIONS

STUDY NOTES

Literal equation: An equation which has different letters which represent variables Changing the subject of formulae

Literal equations are a way of showing the relationship between variables and are used as formula in every day calculations. However, the way the literal equation is written may not be in the form we require. We often need to rearrange a literal equation so that a particular variable is isolated on one side of the equation and set equal to an expression of the other variables on the other side of the equal sign. We call this process, changing the subject of the formula.

Step 1: Identify the variable that needs to be the subject of the formula Step 2: Identify what operations involve this variable

Step 3: Perform the opposite operation to both sides of the equation

Start by adding or subtracting terms that do not include the variable Simplify then do the next operations

Divide or multiply both sides of the equation to isolate the variable

If the variable is squared, take the square root of both sides of the equation.

If the unknown variable is in the denominator, we multiply both sides by the lowest common denominator (LCD) and then continue to solve.

Example 1: Make s the subject of the formula for the equation:

v² = u² + 2as Step 1: v² = u² + 2as

(Highlight the variable the needs to be the subject of formula: Here it is s)

Step 2: The variable on the right hand side of the equation is multiplied by 2a and the u² is added.

Step 3: First subtract u² from both sides of the equation and simplify v² = u² + 2as

v² - u² = u² -u² + 2as v² - u² = 2as

Next divide both sides by 2a and simplify v² - u² 2as

2a 2a v² - u² 2a

Re-write with the variable on the left hand side on the equation v² - u²

2a

s =

= s

=

Radio Broadcast 24 Sept 17:00 -18:00

Questions for discussion

Question 1:

(a) Solve for I: P = VI

(b) Make m the subject of the formula: E = mc² (c) Solve for t: v = u + at

(d) Solve for c:

(e) Make f the subject of the formula:

INEQUALITIES

STUDY NOTES

Inequality: A mathematical expression for comparing the size (magnitude) of a variable or variables

Linear inequality: A mathematical expression that is similar in form to a linear equation but may include the following signs:

Greater than and equal to: ≥

Greater than (not equal): >

Less than and equal to: ≤ Less than (not equal): <

Solving Linear Inequalities

We solve linear inequalities in the same way we solve linear equations, except that when multiplying or dividing both sides of an inequality by a negative number the sign changes from greater than to less than or from less than to greater than.

The solution to an inequality gives a set of numbers that satisfy the conditions of the inequality. The solution can be expressed in three ways:

 An inequality

We use the inequality signs to indicate all the values that form the solution:

e.g greater than and equal to 2: x ≥ 2 less than 2: : x < 2

greater than and equal to 2 but less than and equal to 5: 2 ≤ x ≤ 5

 A number line

We show the points included in the solution with a filled in circle and a solid line with an arrow above a number line. Eg: greater than and equal to 2:

If a point is not included, we draw a circle that is not filled in on the point followed by an arrow above a number line:

E.g. less than 2:

 An interval

We use square brackets separated by a semi-colon to include all real number values in a solution. This is called a closed interval.

e.g: greater than and equal to 2 but less than and equal to 5: [2; 5]

We use round bracket to show when a number is not included in the interval. We use the sign ∞ to represent infinity. We always use a round bracket with infinity because it is not a fixed number.

e.g greater than and equal to 2: [2 ; ∞)

less than 2: (- ∞; 2)

Solving Linear Inequalities

You solve a linear inequality by following the same steps as you do for a linear equation except that when you multiply or divide both sides by a negative number, the direction of the inequality changes – less than becomes greater than.

Example: Solve for y:

1 (3 6) 6 3( 2) 3 y    y 

Solution:

Step 1: Simplify both sides by expanding the brackets y – 2 ≤ 6 + 3 y + 6

y – 2 ≤ 3 y + 12

Step 2: Group like terms, variables on the left hand side of the equation and numbers on the right

y – 3 y ≤ 12 + 2 (Add 2 and subtract 3 y to both sides of the equation and simplify) – 2 y ≤ 14

Step 3: Divide both sides of the equation by -2 and change the sign y ≥ 7

Step 4: Check that the solution is valid. Select a number greater than 7 say 10:

Substitute 10 into left hand side of original inequality:

3(3(10) – 6) =₃( 24) = 8

Substitute 10 into right hand side of original inequality:

6 + 3(10 + 2) =6 + 3(12) = 6 + 36 = 42

Radio Broadcast 25 Sept 17:00 -18:00

Questions for discussion

Solve the following inequalities:

(a) 4(2x 1) 5x2 ⁽³⁾

(b) 2 1

2 1

3

x  x (3)

TRIGONOMETRY

STUDY NOTES

Defining the trig ratios in a right angled triangle:

Defining trigonometric ratios

Defining the Reciprocal Ratios

Special Angles

Instead of relying on a calculator there are a few special angles you need to remember or know how to work out by using the two triangles shown below:

Defining Trigonometric Ratios on the Cartesian Plane

A line, OP, drawn from the origin, will form an angle A with the x-axis. We measure angle A from the x-axis in an anti-clockwise direction.

We can construct a right and triangle OPX, with OP as the hypothenuse. Since the line OP can rotate about O, we think of it as a radius and call its length r. The x co-ordinates of P, gives the size of the side adjacent to A and the y co-ordinates of P, gives the size of the side opposite to A. Now we can re-define the trigonometric ratios on the Cartesian plane as follows:

These definitions of the trig ratio are used to:

 define the ratios for angles greater than 90^o

 add the values of 0⁰ and 90⁰ as special angles

Trig Ratios for angles greater than 90

^o

The sign of the different ratios are all positive in the 1^st Quadrant in Quadrants II, III and IV the sign changes. In Quadrant II the sine ratio is positive, in Quadrant III the tan ratio is positive and in Quadrant IV the Cosine ratio is positive.

Special Angles

The table below summarises all the special angle values you must know:

Radio Broadcast 26 Sept 17:00 -18:00

Questions for discussion

Question 1

Use your calculator to determine the value of the following ratios (correct to 2 decimal places):

a) b) c.)

Use your calculator to find the angle theta for a right angled triangle given the following trig ratios:

d.) sin θ = 0,707 e.) 3 tan θ = ,244 Question 2

Calculate the value of the following without the use of a calculator a.

b.

c. Show that:

Radio Broadcast 27 Sept 17:00 -18:00

Questions for discussion

Question 1:

If sin θ = 0,4 and θ is an obtuse angle, determine:

a.) cos θ b.) √2 tan θ Question 2:

Given tan θ = p /2 , where 0^o ≤ θ ≤ 90^o. Determine the following in terms of p:

a.) sec θ b.) cot θ c.) cos² θ

d.) tan² θ - sec² θ

Radio Broadcast 29 Sept 09:00 -10:00

Questions for discussion

Question 1

Consider the triangle PQS below:

a.) Calculate the length of SR b.) Find the size of the angle RPQ

Question 2

The distance, YA, from a block of flats to a radio mast is 350m. Someone stands at point X.

They measure the angle from X to the top of the mast (B) to be 32˚ (the angle of elevation).

They then measure the angle from X to the bottom of the mast A to be 56˚ (the angle of depression). What is the height of the radio mast AB (correct to the nearest meter)?

X

A B

56^o 32^o

Y 350m