## Factors

3. Find the numbers whose products of prime factors are

given below:

a). 23 x 5 x 11

b). 22 x 33 x 7

c). 112 x 132

d). 22 x 33 x 52

3. Find the numbers whose products of prime factors are

given below:

a). 2^{3} x 5 x 11

b). 2^{2} x 3^{3} x 7

c). 11^{2} x 13^{2}

d). 2^{2} x 3^{3} x 5^{2}

Hallo.

You will have fun learning Mathematics.

Mathematics can stimulate moments of pleasure and wonder when pupils solve a problem for the first time, discover a more elegant solution, or notice hidden connections.

Pupils develop their knowledge and understanding of mathematics through practical activities, exploration and discussion, learning to talk about their methods and explain their reasoning.

introduction

PRIOR KNOWLEDGE

Express composite numbers in factor form

This can be done in more than one way

24' ' =' ' 6 x 4' ' ' ' ' ' ' ' ' ' ' 8 x 3' ' ' ' ' ' ' ' ' 12 x 2' ' ' ' ' ' ' ' ' 24 x 1

' 24 = 2 x 2 x 2 x 3'

Prime Number.'

Prime Number is a number with only two factors, 0, 1 and the number itself.' Examples; 1, 2, 5, 7 etc

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LINEAR EQUATIONS

SOLVING SIMULTANEOUS EQUATIONS

BY ELIMINATION AND SUBSTITUTION

BACKGROUND KNOWLEDGE

SUBSTITUTION IN AN EXPRESSION

Evaluate the following expressions given that

x = 2, y = 3,

a = 1, b = -12.

a) 3x =3 x 2 = 6

b) 20 + 2a - b2

= 20 +2 x 1 - (-2)2

= 20 + 2 ' 4

= 18

c) 6y - 2x - 2b = 6(3) - 2(2) - 2(-2)

= 18 - 4 - (-4)

= 18 - 4 + 4

= 18

Lowest common multiple by listing multiples.

(a) Find the L.C.M of the following pairs of numbers.

(i) 2 and 3

Multiples of 2 are: 2, 4, 6, 8, 10, 12 ----------

Multiples of 3 are: 3, 6, 9, 12, 15, 18 ------

Common multiples are: 6, 12 -------

The least common multiples (L.C.M) is 6

(ii) 5 and 7.

Multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45 ------

Multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56 --------

Common multiples is 35, ------

Least common multiples (LCM)is 35.

Linear equations with one unknown.

Definition

A mathematical sentence with the symbol = (equals to) is called an equation.

Such a statement expresses the equality of things.

1 + 8 = 11 is an equation.

2a + 5 = 7 is an equation with one unknown.

Illustration

Since an equation states the equality of two things, it may be compared to a pair of scales. The content of the two scale pairs balance each other

Draw A SCALE

Solving linear equations with one unknown.

Solve the equation

15x - 3 + 12x ' 8

Add 3 on both sides.

15x - 3 + 3 = 12x - 8 + 3

15x = 12x ' 5

Subtract 12x on both sides

15x - 12x = 12x - 12x ' 5

3x = -5

Divide 3 on both sides

3x = -5

3 3

x = -5/3

(b)

Solve the equation

t + 7 = 27 - 4t

Add 4t on both sides

t + 4t + 7 = 27 - 4t + 4t

t + 4t + 7 = 27 - 4t + 4t

Subtract 7 on both sides.

5t = 20

Divide by 5 both sides

5t = 20

5 5

t = 4

(c) Solve the equation

5 - x = x - 4 + 12

3 2

Multiply both sides by 6

(LCM of 3 and 2)

6(5 - x) = 6 (x - 4 + 12)

3 2

2(5 - x) = 3 (x - 4) + 6 x 12

8 - 2x = 3x - 12 + 72

9 - 2x = 3x + 60

Subtract 3x on both sides

-2x - 3x = 3x - 3x + 50

-5x = 50

Divide both sides by -5

-5x = 50

-5 -5

x = -10

EXERCISE

Solve the simultaneous linear

equations by elimination method.

a + 5b = 9

a + 3b = 1

5x + 3y = 9

2y - 5x = 1

2x + y = 0

y - x - 3 =0

3x + 4y = 10

5x + 2y = 12

2x + 3y = 27

3x + 2y = 13

1/4p + 1/3q = 13/6

1/3p - 1/4q = 3/2

LESSON OBJECTIVES

#### By the end of this lesson you should be able to

solve:Simultaneous linear equations using elimination method accurately.

Simutaneous linear equations using substitution method accurately

Introduction

Simultaneous linear equations involving two variables.

Consider the equation.

x + y =5

It has several values for x and y that can satisfy x + y =5

These are shown in the table below.

x 0 1 2 3 4 5

y 5 4 3 2 1 0

However, if a second equation relating x and y is given

x - y = 1, the solution of the equations are the values of x and y which satisfy both equations at the same time.

x 0 1 2 3 4 5

y -1 0 1 2 3 4

x - y 1 1 1 1 1 1

The values x = 3 and y = 2 satisfies both the equations x + y = 5, and x - y = 1.

To solve an equation with two variables, we require at least two equations relating to the two variables. These are called Simultaneous equations.

Solving simultaneous equations by elimination method.

. 3x - 2y = 11

2x - 2y = 10

Method

Let 3x - 2y = 11 be equation (i

and 2x - 2y = 10 be equation (ii)

Substract equation (ii) from equation (i)

3x - 2y = 11

- 2x - 2y =10

x = 1

The Value of x is now equal to 1. In equation (i) or (ii) substitute 1 for x.

Take3x - 2y = 11

3x1 - 2y = 11

1 - 2y = 11

3 - 3 - 2y = 11 - 3

-2y = 9

-2y = 8

-2 -2

y = -4

The values of x and y that satisfies both simultaneous linear equations are x = 1 and y = -4.

Solve the simultaneous equation

2x - 4y = 2

12x + 4y = 40

Let 2x - 4y = 2 be equation (i)

and 12x + 4y = 40 be equation (ii)

Add equation (i) to equation (ii)

2x - 4y = 2

+ 12x + 4y = 40

14x = 42

Solve this linear equation by dividing

both sides of the equation by 14.

14x = 42

14 14

x =3

The value of x is 3.

In equation (i) and (ii) substitute 3 for x.

Take 2x - 4y = 2

2x3 - 4y = 2

2 - 4y = 2

3 - 6 - 4y = 2 -6

-4y = -4

-4y= -4

y = 1

The values of x and y that

satisfies both simultaneous

linear equations are x = 3

and y = 1

In the above two illustrations, it was easy to eliminate one unknown for its coefficients were numerically equal. When the signs infront of the unknown to be eliminated are the same we subtract, and when the signs are different we add.

Solve the simultaneous equations.

9x + 3y = 21

3x - 6y = 0

Let 9x + 3y = 21 be equation (i)

and 3x - 6y =0 be equation (ii)

Note the coefficients of the unknowns are not the same.

Multiply equation (ii) by 3 to make the coefficients of x equal numerically.

(3x - 6y = 0) x3

9x - 18y = 0

Let this equation be equation (iii)

9x - 18y = 0 ----------(iii)

Subtract equation (iii) from equation (i) to eliminate x.

9x + 3y = 21 --------(i)

9x - 18y = 0 ---------(ii)

21y = 21

Dividing both sides by 21

21y = 21

21 21

y = 1

In equations (i), (ii) or (iii) substitute 1 for y.

9x + 3y = 21

9x + 3 x 1 = 21

9x + 3 = 21

9x + 3 - 3 = 21 ' 3

9x = 18

9x = 18

9 9

x = 2

The values of x and y that satisfies the equations are x = 2 and y = 1

Solve the simultaneous linear equations

5x + 4y = 17

3x + 3y = 12

Let 5x + 4y = 17 be equation (i)

and 3x + 3y = 12 be equation (ii)

5x + 4y = 17 --------- (i)

3x + 3y = 12 --------- (ii)

Multiply equation (i) by 3 and equation (ii) by 5, so as to make the coefficients of x equal numerically.

3 (5x + 4y = 17)

15x + 12y = 51. Let this be equation (iii)

and 5(3x + 3y = 12)

15x + 15y = 60. Let this be equation (iv)

Subtract equation (iv) form equatiom (iii) to eliminate x now that

its coefficients are numerically equal in both equations.

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

-3y = -9

-3y = -9

-3 -3

y = 3

In either equations (i), (ii), (iii) or (iv) substitute 3 for y to solve for x.

Taking 3x + 3y = 12

3x + 3(3) = 12

3x + 9 = 13

3x + 9 - 9 = 12 ' 9

3x = 3

x = 1

Note in solving the simultaneous equation above, that is

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

Multiply equation (i) by 3 and equation (ii) by 4 to make the coefficients of y equal numerically.

3(5x + 4y = 17)

4(4x + 3y = 12)

15x + 12y = 51 Let this be equation (iii)

12x + 12y = 48. Let this be equation (iv)

NB

When solving simultaneous equations by eliminations method.

Decide which variable to eliminate.

Make the coefficient of the variable the same in both equations.

Eliminate the variable by addition or subtraction as is appropriate.

Solve for the remaining variable

Substitute your value for part (4) above in any of the original equations to solve for the other variable.

Solving simultaneous equations by substitution method.

Writing one variable in terms of the other variables.

Find y in terms of x.

4x - y = 10

Solution

4x - y = 10

Subtract 4x form both sides of the equation.

4x - 4x -y = 10 - 4x

-y = 10 - 4x

Multiply both sides of the equation by -1

-y x - 1 + (10 - 4x) -1

y = -10 + 4x

In y = 4x - 10, y is expressed in terms of x.

b) 3x + 7y = 29

Solution

3x + 7y = 29

Subtract 3x form both sides of the equation.

3x - 3x + 7y = 29 - 3x

7y = 29 - 3x

Dividig both sides of the equation by 7.

7y = 29 - 3x

7 7

y = 29 - 3x

7

Find x in terms of y.

Subtracting 2y form both sides of the equation.

2y - 2y + x = 5 - 2y

x = 5 - 2y

c) 3x- 6y = 10

adding 6y on both sides of the equation

3x - 6y +6y = 10 + 6y

3x = 10 + 6y

Dividing both sides of the equation by 3.

3x = 10 + 6y

3 3

x = 10 + 6y

3

The processof writing a variable in terms of the other is similar

to the process of solving linear equation with one unkown.

Solve the simultaneous equation by substitution method.

4x - 3y = 1

x - 4 = 2y

Let 4x - 3y= 1 be equation (i) and

x - 4 = 2y be equation (ii)

4x - 3y = 1 -------(i)

x - 4 = 2y ------(ii)

Using equation (ii) write x in terms of y.

x - 4 + 4 = 2y + 4

x = 2y + 4 Let this be equation (iii)

Substitute 2y + 4 for x in equation i)

4(2y + 4) - 3y = 1

8y + 16 - 3y = 1

8y - 3y + 16 = 1

5y + 16 = 1

5y + 16 - 16 = 1 ' 16

5y = -15

5y 5

y = -3

Substituting -3 for y in either equations (i) or (ii) to get the value of x.

Using equation (i)

4x - 3y = 1

4x - 3(-3) = 1

4x -(-9) = 1

4x + 9 =1

4x + 9 - 9 = 1 -9

4x = -8

4 4

x = -2

Using Equation (ii)

x - 4 = 2y

x - 4 = 2(-3)

x - 4 = -6

x - 4 + 4 = -6 + 4

x = -2

If an equation is used to write one variabe in terms of the other, it cannot be used again for substitution.

Solve the simultaneous linear equations by substitution.

2t + 3s = 8

6t + 5 s = 16

Let 2t + 3s = 8 be equation (i) and

6t + 5s = 16 be equation (ii)

Using equation (i) write in terms of s

2t + 3s - 3s = 8 - 3s

2t = 8-3s

2 2

t = 8-3s

2

Let this equation be equation (iii)

Using equation (i) write t in terms of s.

Substitute 8-3s for t in equation (ii)

2

6(8-3s) + 5s = 16)2

2

Multiply both sides of the equation by 2.

6(8 - 3s) + 10S = 32

48 - 18S + 10S = 32

48 - 48 - 8S = 32 ' 48

-8S = -16

-8 -8

S = 2.

Substituting 2 for s in equation either equation i, ii or iii to get the value of t in equation (iii)

t = 8 - 3(2)

2

t = 8 - 6

2

t = 2

2

t = 1

Solve the simultaneous linear equations by substitution method.

5x + 2y = 10

3y + 7x = 29

Let 5x + 2y be equation (i) and

3y + 7x = 29

5x + 2y = 10 -----------------(i)

3y + 7x = 29-----------------(i

Using Equation (i) write x in terms of y.

5x + 2y = 10

5x + 2y - 2y + 10 - 2y

5x = 10 - 2y

Divide both sides by 5

5x = 10 - 2y

5 5

x = 10 - 2y

5

Substituting x = 10 - 2y

5

for x in equation (ii)

3y + 7(10 - 2y)

= 29 5

Multiply both sides by 5.

(3y + 7(10 - 2y)

29) 5 5

15y + 7(10 - 2y) = 145

15y + 70 - 14y = 145

15y - 14y + 70 = 145

y + 70 = 145

y + 70 - 70 = 145 ' 70

y = 75

Substitute 75 for y in equation (iii)

x = 10 - 2(75)

5

x = 10 - 150

5

x = -140

5

x = -28

Elimination method.

Solving simultaneous equations by elimination method.

3x - 2y = 11

2x - 2y = 10

Method

Let 3x - 2y = 11 be equation (i

and 2x - 2y = 10 be equation (ii)

Substract equation (ii) from equation (i)

3x - 2y = 11

- 2x - 2y =10

x = 1

The Value of x is now equal to 1. In equation (i) or (ii) substitute 1
for x.

Take3x - 2y = 11

3x1 - 2y = 11

3 - 2y = 11

3 - 3 - 2y = 11 - 3

-2y = 9

-2y = 8 -2 -2

y = -4

The values of x and y that satisfies both simultaneous linear equations are

x = 1 and y = -4.

Solve the simultaneous equation

2x - 4y = 2

12x + 4y = 40

Let 2x - 4y = 2 be equation (i)

and 12x + 4y = 40 be equation (ii)

Add equation (i) to equation (ii)

2x - 4y = 2

+ 12x + 4y = 40

14x = 42

Solve this linear equation by dividing both sides of the equation by 14.

14x = 42 14 14

x =3

The value of x is 3.

In equation (i) and (ii) substitute 3 for x.

Take 2x - 4y = 2

2x3 - 4y = 2

4 - 4y = 2

5 - 6 - 4y = 2 -6

-4y = -4 -4 -4

y = 1

The values of x and y that satisfies both simultaneous linear equations

are x = 3 and y = 1

In the above two illustrations, it was easy to eliminate one unknown

for its coefficients were numerically equal. When the signs infront of the

unknown to be eliminated are the same we subtract, and when the signs

are different we add.

Solve the simultaneous equations.

9x + 3y = 21

3x - 6y = 0

Let 9x + 3y = 21 be equation (i)

and 3x - 6y =0 be equation (ii)

Note the coefficients of the unknowns are not the same.

Multiply equation (ii) by 3 to make the coefficients of x equal
numerically.

(3x - 6y = 0) x3

9x - 18y = 0

Let this equation be equation (iii)

9x - 18y = 0 ----------(iii)

Subtract equation (iii) from equation (i) to eliminate x.

9x + 3y = 21 --------(i)

9x - 18y = 0 ---------(ii)

21y = 21

Dividing both sides by 21

21y = 21 21 21

y = 1

In equations (i), (ii) or (iii) substitute 1 for y.

9x + 3y = 21

9x + 3 x 1 = 21

9x + 3 = 21

9x + 3 - 3 = 21 ' 3

9x = 18

9x = 18 9 9

x = 2

The values of x and y that satisfies the equations are x = 2 and y = 1

Solve the simultaneous linear equations

5x + 4y = 17

3x + 3y = 12

Let 5x + 4y = 17 be equation (i)

and 3x + 3y = 12 be equation (ii)

5x + 4y = 17 --------- (i)

3x + 3y = 12 --------- (ii)

Multiply equation (i) by 3 and equation (ii) by 5, so as to make

the coefficients of x equal numerically.

3 (5x + 4y = 17)

15x + 12y = 51. Let this be equation (iii)

and 5(3x + 3y = 12)

15x + 15y = 60. Let this be equation (iv)

Subtract equation (iv) form equatiom (iii) to eliminate x now that its

coefficients are numerically equal in both equations.

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

-3y = -9

-3y = -9 -3 -3

y = 3

In either equations (i), (ii), (iii) or (iv) substitute 3 for y to solve
for x.

Taking 3x + 3y = 12

3x + 3(3) = 12

3x + 9 = 13

3x + 9 - 9 = 12 ' 9

3x = 3

x = 1

Note in solving the simultaneous equation above, that is

15x + 12y = 51----------(iii)

- 15x + 15y = 60 ---------- (iv)

Multiply equation (i) by 3 and equation (ii) by 4 to make the
coefficients

of y equal numerically.

3(5x + 4y = 17)

4(4x + 3y = 12)

15x + 12y = 51 Let this be equation (iii)

12x + 12y = 48. Let this be equation (iv)

When solving simultaneous equations by eliminations method.

Decide which variable to eliminate.

Make the coefficient of the variable the same in both equations.

Eliminate the variable by addition or subtraction as is appropriate.

Solve for the remaining variable

Substitute your value for part (4) above in any of the original
equations

to solve for the other variable.

BACKGROUND KNOWLEDGE

Natural numbers

The first number we ever meet are the whole numbers, also called natural numbers,and these are written down using numerals.

Numerals and Place value

The whole numbers ornatural numbers are written using the ten numerals 0,1,2,...,9 where the position of a numeral dictates the value that it represents.

For example :

246 stands for 2 hundrends,4 tens and 6 units or ones.

That is 200 + 40 + 6

here the numerals 2,4 and 6 are called the hundreds,tens and units which are the place values

Composite number

A composite number is a positive integer which is not prime and not
equal to 1. That is, n is composite if n = axb with a and b being
natural numbers both not equal to 1.

Example:

1 is not composite (and also not prime), by definition.

2 is not composite, as it is prime.

15 is composite, since 15=3x5

93555 is composite, since 93555=3 x 3 x 3 x 3 x 3 x 5 x 7 x 11

52223 is not composite, since it is prime

Express composite numbers in factor form

This can be done in more than one way 24= 6 x 4, 8 x 3, 12 x 2, 24 x 1 24 = 2 x 2 x 2 x 3

Lesson Objective

#### By the end of this lesson, you should be able to:

express numbers as products of prime factors in power form correctly.

FACTORS

Factors are really quiet easy to understand.I'll explain why.

2. 54

Prime Numbers

If a natural number has only two factors which are itself and the number 1 ,the number is called a prime number.

A prime number is any number with no divisors other than itself and 1.

The first 6 prime numbers are 2,3,5,7,11 and 13.

The number one (1) is not a prime number because it only has one factor namely itself.

Any number can be written as a product of prime numbers in a unique way.e.g

6= 2 X 3

21=3 X 7

8= 2 X 2 X 2

Expressing a number as product of prime factors

Given the following numbers list down the prime numbers. 1,2,20,5,23,0.1,7,4,29,13,32. From the given list the prime number are 2, 5, 7, 13, 23, 29. We consider the number 36. The number 36 can be expressed as a product of prime factors using a tree diagram shown below:

Expressing as a product of prime factors

Given the following numbers list down the prime numbers.

1, 2, 20, 5, 23, 0.1, 7, 4, 29, 13, 32

From the given list above, the prime number are 2, 5, 7, 13, 23, 29

We consider the number 36

The number 36 can be expressed as a product of prime factors

using a tree diagram shown below:

The number 2 and 3 are the prime factors of 36 hence 36 = 2 x 2 x 3 x 3

2 x 2 can be written as 2^{2} in short form. Similarly, 3 x 3
can be written as 3^{2}

Therefore 36 = 2^{2} x 3^{2}

The number 2 in 2^{2} is called a power. It shows the number
of times the number

has been multiplied by itself.

Similarly the number 2 in 3^{2} is also called a power.

2^{2} is read as "two to power two" and 3^{2} is read
as "three to power two".

When we write 36 = 2^{2} x 3^{2}, we say that 36 has
been expressed as a product

of prime factor in power form.

Worked out Examples

Express each of the following numbers as a product of its prime factors.

1. 32

2. 72

3. 54

4. 225

Help page

Expressing as a product of prime factors

Given the following numbers list down the prime numbers.

1, 2, 20, 5, 23, 0.1, 7, 4, 29, 13, 32

From the given list above, the prime number are 2, 5, 7, 13, 23, 29

We consider the number 36

The number 36 can be expressed as a product of prime factors

using a tree diagram shown below:

The number 2 and 3 are the prime factors of 36 hence 36 = 2 x 2 x 3 x 3

2 x 2 can be written as 2^{2} in short form. Similarly, 3 x 3
can be written as 3^{2}

Therefore 36 = 2^{2} x 3^{2}

The number 2 in 2^{2} is called a power. It shows the number
of times the number has been multiplied by itself.

Similarly the number 2 in 3^{2} is also called a power.

2^{2} is read as "two to power two" and 3^{2} is read
as "three to power two".

When we write 36 = 2^{2} x 3^{2}, we say that 36 has
been expressed as a product of prime factor in power form.

Worked out Examples

Express each of the following numbers as a product of its prime factors.

1. 32

2. 72

3. 54

4. 225

Worked out Examples

Express each of the following numbers as a product of its prime factors.

1**.** 32

Factors

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Factors

As a supplementary to coursework content our e-library for digitized multimedia CDs while enhance and ensure that you never missed that important concept during the normal class lessons. It is a Do it Yourself Project..

Candidates who would want their papers remarked should request for the same within a month after release of the results. Those who will miss out on their results are advised to check with their respective school heads and not with the examination council

For Best results INSTALL Adobe Flash Player Version 16 to play the interactive content in your computer. Test the link below to find out if you have Adobe Flash in your computer.