Remember that exam preparation also requires motivation and discipline, so try
to .... the two First term together, then the Outer terms(1st term of the first bracket
by ..... Step 2: Add the common difference to the 3rd term to get the 4th and so on:
.
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. Contact us We want to hear from you. So let us have your specific questions or just tell us what you think through any of the following: LearnXtra
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BROADAST SCHEDULE Grade 10 Catch up (Term 3) 10-Sep 11-Sep 12-Sep 13-Sep 15-Sep 17-Sep 18-Sep 19-Sep 20-Sep 22-Sep 24-Sep 25-Sep 26-Sep 27-Sep 29-Sep
17:00 -18:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 09:00 -10:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 09:00 -10:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 09:00 -10:00
10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
Algebraic Expressions
10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
Linear Functions
Algebraic Expressions Algebraic Expressions Exponents Exponents Numbers and Number Patterns Numbers and Number Patterns Linear Equations Quadratic Equations Linear & Quadratic Equations Literal Equations Inequalities Trigonometry Trigonometry Trigonometry
Grade 10 Exam Revision 15-Oct 16-Oct 17-Oct 18-Oct 20-Oct 22-Oct 25-Oct 27-Oct 29-Oct 30-Oct 2-Nov 8-Nov 9-Nov 10-Nov 12-Nov 13-Nov 14-Nov 15-Nov 16-Nov 17-Nov
17:00 -18:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 09:00 -10:00 17:00 -18:00 17:00 -18:00 09:00 -10:00 17:00 -18:00 17:00 -18:00 18:00 -19:00 17:00 -18:00 17:00 -18:00 09:00 -10:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 17:00 -18:00 17:00 -18:00
Quadratic Functions Hyperbolic Functions Exponential Functions Trig Functions Euclidean Geometry Euclidean Geometry Euclidean Geometry Finance Finance Statistics Analytical Geometry Measurement Probability Algebra Functions Finance Trigonometry Euclidean Geometry 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(1st term of the first bracket by the 2nd term of the second bracket) then the Inner terms (2nd term of the first bracket by the 1st 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+2)(x-2)
x(x + 1) x(x +2)
Step 2: Change the division sign and multiply by the reciprocal (x+1)(x-2) X (x+2)(x-2)
x(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)
(2 x 3 y)2 (3x 4 y)2
(4)
(b)
(9 x 4)(3x 2)(3x 2)
(2)
(c)
( x 2 y)( x2 2 xy 3 y 2 )
(3)
2
Radio Broadcast
11 Sept
17:00 -18:00
Questions for Discussion Factorise fully: (a) (b) (c) (d) (e)
15a4b6 3ab2 8a8 8b8 3x 2 3x 18 5x2 14 x 8 a3 3a2 a 3
Radio Broadcast
12 Sept
(2) (4) (2) (2) (4)
17:00 -18:00
Questions for Discussion Simplify: (a) (b) (c) (d)
1 2 1 2 3 2 xy 4x 3x x 1 x2 x 3 2 3 4 2 4 x 8x x 2 9x 12 x3 2 1 x x 3 3
1
(5) (5) (4) (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 . a n a m n
The bases are the same so you will then add the exponents For example
x4 . x x4 . x1 x41 x5
(same bases, add exponents)
23.23 26 64
(same bases, add exponents)
am a
n
a mn
The bases are the same so you will then subtract the exponents For example
x4 x4 1 x 41 x3 x x 212 10
(same base, subtract exponents)
21210 22 4
(same base, subtract exponents)
2
(a m )n amn
With this rule, you will need to multiply the exponents For example
(3x4 )2 (31 x4 )2 312. x42 9 x8
xn
1 x
n
1
and
x
n
xn
With this rule, you make the exponents positive For example
1 1 32 2 9 3
1 2
32 9
3
a0 1 Any number raised to an exponent of 0 equals 1 For example
30 1 4 x 0 4 1 4
a b
n
b a
n
For example
3 4
2
2
4 16 9 3
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 9 is called an exponential equation because the unknown 2 variable is the exponent. Clearly, the solution to this equation is x 2 because 3 9 . x
Getting the bases on both sides of the equation to be the same and then equating the exponents will solve the equation as follows:
3x 9 3x 32 x 2 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: 3x 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)
(3x4 )2 (2 x2 )3 4(2 x)2 3x0 2 2 3
(b) (c)
2(a b ) (ab) (2b6 )3 ( 2)3. 218 (28 ) 2 . 2
(5)
6
(5) (5)
Radio Broadcast
15 Sept
09:00 -10:00
Questions for Discussion (a) 3x = (b) 2m+2 = (0,5)3+2m
(c) (d)
2 x 0,125 1 16 x. 42 x 4
(3) (6)
SOLUTIONS TO QUESTIONS Radio Broadcast (a)
10 Sept
17:00 -18:00
(2 x 3 y ) 2 (3 x 4 y) 2 4 x 2 12 xy 9 y 2 (9 x 2 24 xy 16 y 2 ) 4 x 2 12 xy 9 y 2 9 x 2 24 xy 16 y 2 5 x 2 36 xy 7 y 2
(b)
(9 x 2 4)(3x 2)(3x 2) (9 x 2 4)(9 x 2 4) 81x 4 16
(c)
( x 2 y )( x 2 2 xy 3 y 2 ) x( x 2 2 xy 3 y 2 ) 2 y ( x 2 2 xy 3 y 2 ) x3 2 x 2 y 3xy 2 2 x 2 y 4 xy 2 6 y 3 x3 4 x 2 y 7 xy 2 6 y 3
Radio Broadcast (a)
11 Sept
17:00 -18:00
15a 4b6 3ab 2 3ab 2 5a3b 4 3ab 2 1 3ab 2 (5a3b 4 1)
(b)
8a8 8b8 8(a8 b8 ) 8(a 4 b 4 )(a 4 b 4 ) 8(a 4 b 4 )(a 2 b 2 )(a 2 b 2 ) 8(a 4 b 4 )(a 2 b 2 )(a b)(a b)
(c)
3x 2 3x 18 3( x 2 x 6) 3( x 3)( x 2)
(d)
5 x 2 14 x 8 (5 x 4)( x 2)
(e)
a3 3a 2 a 3 a 2 (a 3) (a 3) (a 3)(a 2 1) (a 3)(a 1)(a 1)
Radio Broadcast (a)
1
12 Sept
17:00 -18:00
1 2 1 2 3 2 xy 4x 3x
1 12 x3 y 1 3xy 2 4y 1 6 x2 1 12 x3 y 2 xy 6 x 2 4 x 2 3xy 3x3 4 y
12 x3 y 3xy 8y 6x2 12 x3 y 12 x3 y 12 x3 y 12 x3 y
12 x3 y 3xy 8 y 6 x 2 12 x3 y x 1 x2 x 3 2 3 4 x 1 6 x2 4 x 3 3 2 6 3 4 4 3 6( x 1) 4( x 2) 3( x 3) 12 12 12 6( x 1) 4( x 2) 3( x 3) 12 6 x 6 4 x 8 3x 9 12 7x 5 12
(b)
LCD 12
(c)
(d)
4x2 8x
x2 9x 12 x3 4 x( x 2) 9x 3 ( x 2) 12 x ( x 2) 9x 2 ( x 2) 3x 3 x 2 1 x x 3 3 1 2 2 x2 x x 3 3 9 2 9 x 3x 6 x 2 9 2 9 x 3x 2 9
Radio Broadcast
13 Sept
17:00 -18:00
(a) (3x 4 ) 2 (2 x 2 )3 ( x 2 ) 4(2 x) 2 3 x0 (9 x8 ) (8 x 6 )( x 2 ) 4(4 x 2 ) 3(1) 9 x8 8 x8 16 x 2 3 x8 16 x 2 3
(b) 2(a 2b 2 ) 3 (ab) 6 (2b 6 )3
2a 6b 6 a 6b 6 8b 18
2a 0b 12 8b 18
2(1)b18 8b12
2b18 8b12
b6 4 (c)
( 2)3. 218 (28 ) 2 . 2
( 2) 2 ( 2). 218 216. 2 2( 2). 218 216. 2 2 . 218 216 219 216
23 8
Radio Broadcast (a)
(b) 2m+2 = (0,5)3+2m
(c)
15 Sept
09:00 -10:00
2 x 0,125 125 1000 1 2x 8 2x
2 x 23 x 3
(d)
16 x. 42 x
1 4
(24 ) x . (22 ) 2 x 24 x. 24 x 24 x 4 x 28 x 22 8 x 2 2 8 1 x 4 x
1 22 1
22
1 22
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: a2
(first term)
d 3
(constant difference)
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 1000th term would take too long. We can develop a rule or formula to help us find the 1000th term as follows:
T1 2 T2 2 3 2 (1)3 T3 2 3 3 2 (2)3 T4 2 3 3 3 2 (3)3 T5 2 3 3 3 3 2 (4)3 T10 2 (10 1)3 Tn 2 (n 1)3 Tn 2 3n 3 Tn 3n 1 The rule (general or nth term) can be used to determine any specific term of the pattern. For example, the 4th term can be calculated as follows:
Tn 3n 1 T4 3(4) 1 11 The 1000th term is:
Tn 3n 1 T1000 3(1000) 1 2999
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 10th 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) 1 2 3 4
Term (Tn) T1 = -3 T2 = -5 T3 = -7 T4 = -9
Pattern -3 -3 -2 = -3 + (1)(-2) -3 -2 -2 = -3 + (2)(-2) -3 -2-2-2 = -3 + (3)(-2)
n
Tn
-3 -2 …..-2 =-3 +(n-1)(-2)
c.) To find the 10th 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 100th 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 3rd term to get the 4th 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 3rd term to get the 4th 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 16th term: (a) (b)
5; ; ; 4; ……. -2,5; -0,5; ; 2,5; ……
Solutions (a) Number of terms (n)
(3) (3)
1 2 3 4
Term (Tn) T1 = 5 T2 = 8 T3 = 11 T4 = 14
Pattern 5 5 + 3 = 5 + (1)(3) 5 + 3 + 3 =5 + (2)(3) 5 + 3 + 3 + 3= 5 + (3)(3)
N
Tn
5 + 3 …. + 3= 5 +(n-1)(3)
Tn = 3n + 2 T16 = 3(16) + 2 = 48 + 2 = 50 (b) Number of terms (n) 1 2 3 4
Term (Tn) T1 = -2,5 T2 = -0,5 T3 = 1 T4 = 2,5
-2,5 -2,5 + 1,5 = -2,5 + (1)(1,5) -2,5 + 1,5 + 1,5 =-2,5 + (2)(1,5) -2,5 + 1,5 + 1,5 + 1,5 = -2,5 + (3)( 1,5)
n
Tn
-2,5 + ,5 …. + ,5= -2,5 +(n-1)(1,5)
Tn = 1,5n -3,5 T16 = 1,5(16) -3,5 = 48 + 2 = 20,5
Pattern
Radio Broadcast
18 Sept
17:00 -18:00
Questions for discussion Chains of squares can be built with matchsticks as follows:
(a) (b) (c) (d)
Use matches and create a chain of 4 squares. How many matches were used? Create a chain of 5 squares. How many matches were used? Determine a conjecture for calculating the number of matches in a chain of n squares. Now determine how many matches will be needed to build a chain of 100 squares.
Solution Number of terms (n) 1 2 3 4 5
Term (Tn) T1 = 4 T2 = 7 T3 = 10 T4 = 13 T5 = 16
Pattern 4 4 + 3 = 4 + (1)(3) 4 + 3 + 3 = 4+ (2)(3) 4 + 3 + 3 + 3 = 4+ (3)( 3) 4 + 3 + 3 + 3 = 4+ (3)( 3)
N
Tn
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 Tn = 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 x y 6 ......................B Multiply each term of equation B by 3 and call the new equation C:
3x y 10 ..................A 3x 3 y 18............C The reason for multiplying equation B by 3 is to be able to eliminate the terms in terms will be eliminated if they differ in sign.
x . These
3x y 10 ..................A 3x 3 y 18............C 4 y 8 ..............(add the like terms of A and C) y 2 Now substitute y 2 into either A, B or C to get
3x (2) 10 3x 12
y:
(equation A was used)
x 4 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 x y 6
Solution Label each equation as follows:
3x y 10 ...................A x y 6 ......................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:
3x y 10 y 3x 10 y 3x 10................C Now replace the variable
y in equation B with 3x 10 and solve for x :
x y 6 x (3x 10) 6 4 x 10 6 4 x 16 x 4 Now substitute x 4 into either equation A, B or C to get
y 3(4) 10
y:
(equation C was chosen)
y 2 Radio Broadcast
19 Sept
17:00 -18:00
Questions for discussion Solve the following equations: (a) (b)
1 x2 x 1 2 3 2( x 1)2 (2 x 3)( x 1) 0
SOLUTIONS (a) 1 x2
2
x
1
3 x ( x 2) 1 2 3 3 x 2( x 2) 6 3x 2 x 4 6 x 2
(4) (4) LCD 6 3x 2( x 2) 6 3x 2 x 4 6 x2 (4)
(b)
2( x 1) 2 (2 x 3)( x 1) 0 2( x 2 2 x 1) (2 x 2 2 x 3 x 3) 0 2( x 2 2 x 1) (2 x 2 x 3) 0 2x 4x 2 2x x 3 0 2
2
2 x2 2 x2 4 x x 2 3 0 3 x 5 3 x 5 3 3 5 x 3
2 x2 4 x 2 2 x2 x 3 2 x2 x 3 0 x
5 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 Both solutions are valid
(-5)(-5-5) = 50
Radio Broadcast
20 Sept
17:00 -18:00
Questions for discussion Solve the following: (a) (b) (c) (d) (e)
x 2 16 12 x 2 3x 2 x2 10 x 12 0 (a 2)(2a 1) 0 (a 2)(2a 1) 25
(3) (4) (3) (2) (5)
Solutions (a)
x2 16 0 ( x 4)( x 4) 0 x 4 or x 4
x 2 16 x 2 16 0 ( x 4)( x 4) 0 x 4
(b)
(3)
or x 4
12 x 2 3x 12 x 2 3x 0
3 x(4 x 1) 0
12 x2 3x 0 3x(4 x 1) 0 x0
3 x 0 or 4 x 1 0
x
x 0 or 4 x 1
x2 5x 6 0 ( x 6)( x 1) 0 x 6 or x 1
2 x 2 10 x 12 0 x2 5x 6 0 ( x 6)( x 1) 0
(d)
x 6 or x 1 (a 2)(2a 1) 0
a 2 or a
(e)
(4)
1 4
x (c)
1 4
1 2
(a 2)(2a 1) 25
(3) a 2 a
1 2 (2)
2a2 3a 2 25 2a2 3a 27 0 (2a 9)(a 3) 0 a a3
9 2 (5)
2a 2 a 4a 2 25 2a 2 3a 2 25 2a 2 3a 27 0 (2a 9)(a 3) 0 2a 9 0 2a 9 a
9 2
or a 3 0 or a 3
LINEAR AND QUADRATIC EQUATIONS Radio Broadcast
22 Sept
09:00 -10:00
Questions for discussion 1. Solve the following linear equations: (a.)
m2 m6 1 4 3 2
(4)
(b)
( x 3)( x 4) (3x 2)2 x 8x( 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:
x y 2 and 2 x y 10
(4)
Solutions 1(a)
m2 m6 1 4 3 2 3(m 2) 4(m 6) 6 3m 6 4m 24 6 m 24
LCD 12
3(m 2) 4(m 6) 6 3m 6 4m 24 6 m 24 (4)
m 24 1(b)
( x 3)( x 4) (3x 2) 2 x 8 x( x 1) x 2 x 12 (9 x 2 12 x 4) x 8 x 2 8 x x 2 x 12 9 x 2 12 x 4 9 x 8 x 2 8 x 2 13x 16 9 x 8 x 2
x2 x 12 9 x2 12 x 4 8x 2 8x 9 x2 12 x 4 x4
4 x 16
(5)
x 4 2(a)
( p 2) 2 16
p 4p 4
p 2 4 p 4 16
p 4 p 12 0
p 2 4 p 12 0 ( p 6)( p 2) 0 p 6 or p 2
2
2
( p 6)( p 2) 0
p 6 or p 2 (4)
2(b)
20 ( x 1)( x 2) 0 20 ( x 2 3 x 2) 0 20 x 2 3 x 2 0 x 2 3 x 18 0 x 2 3 x 18 0 ( x 6)( x 3) 0
3.
x 6 or x 3 x y2
2( y 2) y 10 2 y 4 y 10 3 y 6
2 20 ( x 3x 2) 0
20 x 3x 2 0 2
x 3x 18 0 2
x 3x 18 0 2
( x 6)( x 3) 0
x 6 or x 3 (6) x y2 2( y 2) y 10 y2 x4 (4)
y 2 x 2 2 x 4 OR
x y 2
2 x y 10 3 x 12 x4 4 y 2 y 2 y 2
3x 12 x4 y2
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: Step 2: Step 3:
Identify the variable that needs to be the subject of the formula Identify what operations involve this variable 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 2 Step 1: v = 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 u 2 is added. Step 3: First subtract u2 from both sides of the equation and simplify v2 = u2 + 2as 2 v - u2 = u2 -u2 + 2as v2 - u2 = 2as Next divide both sides by 2a and simplify v2 - u2 2as = 2a 2a v2 - u2 =s 2a Re-write with the variable on the left hand side on the equation 2 2 s= v -u
2a
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) (e)
Solve for c: 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):
