# MATHEMATICS FORM ONE TOPIC 5-ALGEBRA

MATHEMATICS FORM ONE TOPIC 5-ALGEBRA

UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU:0787237719

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## MATHEMATICS FORM ONE TOPIC 5-ALGEBRA

ALGEBRA

Algebra is a study which deals with situations whereby some values unknown. Normally these unknown are represented by letters. Those letters are also referred to as variables.

Algebraic expression

An expression – is a mathematical statement which consists of several variables. An expression can only be simplified, that is we cannot find values of the variables (s) on it.

Examples

1. a + 2
2. x + 3y + 9z

3.16p – qp

1. a + b + c + d
2. 40

An equation

An equation is formed when two expression are joined by an equal sign

E.g

1. i) 2x – y = 16
2. ii) x + 2 = 6 – 5

iii) 3y + xy = 9

Each member of an equation or expression is called Term

Coefficient

When a number is multiplied by a variable (s) that number is called coefficient of that variable

Example

What is the coefficient of the variables in the following?

1. a) 6x – 8p + y
2. b)– k + 3d
3. c) 2a + 3b – c

Solutions

Coefficients of a) x is 6

y is 1

P = -8

1. b) K = -1

d = 3,
= 1

1. c) a = 2

b = 3

c = -1

Addition and subtraction of algebraic expression

Addition and subtraction of algebraic expression can be done by adding or subtracting like term.

Like terms are those terms which has identical (same) variables

Examples

1. 2a + 4a = 6a
2. 5a + 16a = 21a
3. 2x + 10x – 3x = 9x

Examples: simplify the expression

3n – 7n + 12n

Solution

-4n + 12n

12n – 4n

= 8n

Examples: simplify

6m – 4 – 2m + 15

Soln

6m – 2m – 4n + 15

= 4m + 11

Example: simplify 4x + 6y – 3x + 5y

Solution

4x – 3x + 6y + 5y

= x + 11y

Coefficient: y = 11

x = 1

Number of terms = two

Exercise 7.1

1. Simplify each of the following expressions and after simplifying state
2. the number of terms
3. the coefficient of each of the terms
4. i) n + n + n + n + n + k + k + k + x + x = 5n + 3k + 2x

Solution

(a) There are3 terms

(b) Coefficient of “n” is 5

Coefficients of “y” is 3

Coefficients of “x” is 2

1. ii) 3x + 4y – 7z + 3x – 7y + 2z

Solution
a) There are 3 terms

6x – 3y – 5z
b) coefficients of x is 6

Coefficients of y is -3

Coefficients of z is -5

iii) 3 x + 7x –  x =

Solution

1. a) There is 1 term
b) Coefficient of x is 10

Simplify each of the expressions in numbers 2 – 6

1. 12m + 13m

12m
+ 13m

25m = 25m

1. 5y + 7y – 4y

12y – 4y

= 8y

1. 24w – 28w

-28w + 24w

= -4w

1. 15n – 9n

15n – 9n

= 6n

1. 4k – k + 3k

3k + 3k

= 6k

1. 8y – 3 – 7y + 4

8y – 7y – 3 + 4

y + 4 – 3

y + 1

= y+ 1

1. 14x + 8 – 3x + 2

14x – 3x + 8 + 2

=11x + 10

1. 3a – 5b – 7a + 6c + 7a + 8b

3a – 7a + 7a – 5b + 8b + 6c

=3a + 3b + 6c

1. 4x – 6y + 7x + 2y

4x + 7x – 6y + 2y

=11x – 4y

1. 3x + 4 + 8x – 4 – 11x

3x + 8x – 11x + 4 – 4

11x – 11x + 4 – 4

0 + 0

= 0

1. 8m + 0.4m – 2 – 6m + 8

8m + 0.4m – 6m – 2 + 8

8.4m – 6m + 6

= 2.4m + 6

Multiplication and division of algebraic expression

Example:1) Multiply a – 2b + 6ab by 12xy

Solution

(a – 2b + 6ab) x 12xy

= 12axy – 24bxy + 72abxy

Examples:2) Re – write without brackets

– 16a (-2mn + 9xb – 3kbc)

Solution

-16a (-2mn+9xb-3kbc) = (-16ax-2mn) + (-16a x 9xb) + (-16a x -3kbc)
= 32amn + -144axb + 48abck
= 32amn – 144axb +48abc

Example:3) divide 36xyz – 48xwz – 24xz by 12z

Solution
(36xyz – 48xwz – 24xz) ÷ 12z

Exercise 7.2

1. Complete the following

60xy – 30y + 90z = 30 ( )
Solution

60xy – 30y + 90z = 30 (2xy – y +3x)

2. Simplify i) xy + yz + 2xy – 3zy

1. ii) 8m ÷ 2 + 3mn ÷ n

Solution

1. i) xy + yz + 2xy – zyxy+yz +2xy – zy = xy + 2xy +yz – 3zy
= 3xy – 2yz
2. ii) 8m ÷ 2 + 3mn ÷ n

4m + 3m

(4 + 3) m

=7m

1. Simplify the following
2. i) 5mn – 3mn

= 2mn

1. ii) xyz + 3xy + 4zx – zyx

= xyz – zyx + 3xy + 4zx

= 0 + 3xy + 4zx

= 3xy + 4zx

iii) 3 (2n + 3) + 4 (5n – 3)

Solution

6n + 9 + 20n – 12

6n + 20n + 9 – 12

=26n – 3

1. iv) abc + bac – cab

Solution

abc + abc – abc

abc – abc + abc

abc + 0

= abc

1. v) 2 (5x + 3y) + 3(3x + 2y)

Solution

10x + 6y + 9x + 6y

10x + 9x + 6y + 6y

= 19x + 12y

1. vi) m (2n + 3) + n (3m + 4)

Solution

2nm + 3m + 3mn + 4n

2nm + 3mn + 3m + 4n

= 5mn + 3m + 4n

vii) x (y – 5) + y (x + 2)

Solution

xy – 5x + yx + 2y

xy + yx – 5x + 2y

= 2xy – 5x + 2y

1. ix) Pq -2qp + 3pq – 2qp

Solution

Pq + 3pq – 2qp – 2qp

4pq – 4pq

= 0

1. x) (4x + 8y) ÷ 2 + (9xw + 4xy) ÷ w

solution

1. xi) Multiply 6a – 5b by 3x

Solution

3x (6a – 5b) = 3x x 6a – 3x x 5b
= 18ax – 15bx

= 18ax – 15bx

Equations

An equations is a mathematical statement which involves two expression connected or joined by an equal sign
So we define an equation also as statement of equality e.g. 2y – 6 = 3x + 12
The values of variables can be found in equation if the number of equations is equal to the number of unknown.

FORMULATION OF AN EQUATION

There are three steps to follow when formulating an equation which are;

1. i)Understand the problem/question, what it is asking for
2. ii) Let the unknown be represented by a variable

iii) Formulate the equation using the given information

Signs, words or phrase used when formulating an equation:-

+ Addition, sum of, increase by, greater than, plus, taller than, more than

– Difference, subtract, decrease, less than, shorter than.

× Multiplication, times, products.

÷ Division, divided, Quotient.

= Equals, is, given, result.

Example 01

1. The age of the father is equal to the sum of the ages of his son and daughter. If the son’s age is thrice the age his sister, formulate an equation.

Solution

Let y be the father age

And x be the age of the daughter

The age of son = 3x

y = 3x + x

y = 4x

1. The sum of two numbers is 20. If one of thenumber is 8 formulate an equation.

Solution

Let one of the number be x

And the other number = 8

x + 8 = 20

1. A girl is 14 years old, how old will she be in x years time

Solution

A girl = 14 years

Let “y” be a girls age in x years time.

In years time = + x

y = 14 + x

1. The difference between 24 and another number is 16, form an equation

Soln

Let another number = x

24 – x = 16

Exercise 7.3

Formulate equations for each of the following

1. Five times a number gives twenty

Soln

x= 20

5x = 20

1. The differencebetween123 and another number is 150

Solution

let another number = x

Then x – 123 = 150

x – 123 = 150

1. The sum of 21andanother number is 125

solution

let another number = y

Sum means (+)

21 + y = 125

21 + y = 125

1. When a certainnumberis increased by 15, the result is 88

solution

Let the number be x

Then x + 15 = 88

x + 15 = 88

1. When 99 is increased by a certainnumberthe result is 63

Solution

Let the number = y

Then 99 + y = 63

99 + y = 63

1. The product of 12andanother number is the same as two times the sum of 12 and the number

Solution

Let the number be x

Then 12 x x = 2 x (12 + x)

12x = 24 + 2x

7.A number is such that when it is double and 8 added to it, the result is the same as multiplying the number by 3 and subtracting 7.

solution

Let the number be x

Then x + x + 8 = x x 3 – 7

2x + 8 = 3x – 7

1. When 36 is added to a certainnumber, theresult is the same as multiplying the number by 5.

solution

Let x be the number

Then x + 36 = x x 5

x + 36 = 5x

1. If John is n years old and is 6 years older that James older, write an expression of the sum of their ages.

Solution

Let “J” be john, and “Q” be James and “N” be the year

Let Q = q years

J = n + 6 years

The sum of their age = q + n + 6

= q + n + 6

1. When the sum of nand(n + 3) is multiplied by 5 the result half the product of the two numbers.

Write the expression of this statement:-

Solution

(n + (n +3) x 5 = ½ (n + (n +3) )

(2n + 3) x 5 = ½(2n + 3)

SOLVING FOR EQUATIONS

Solving means finding the value of the unknown in the equation

Example 1

1. x + 5 = 8

Solution

x + 5 = 8

x + 5 – 5 = 8 – 5

x + 0 = 3

x = 3

1. x – 8 = 15

x – 8 + 8 = 15 + 8

x = 23

1. 3x – 5 = 7

3x – 5 + 5 = 7 + 5

3x = 12

x = 4

4. + 3 = 12

solution;

multiply 2 both side

1. (3x – 2) = 10

Solution

1. = 2

Solution

=

1 = (3x – 2)

8 = 6x – 4

8 + 4 = 6x – 4 + 4

12 = 6x

=

x = 2

1. –  = 4

Solution

2m = 4 x 15

2m = 60

m= 30

1.  = 5

Solution

10x = 5 x 8

10x = 40

x = 4

1. 2x – 5 = 3x – 8

Solution

3x-8=2x-5

3x-2x=8-5

x = 3

1. 4 – 3t = 0.3t – 5.9

Solution

4 + 5.9 = 0.3t + 3t

9.9 = 3.3t

9.9 = 3.3t
3.3 3.3

t = 3

Solution

Multiply by 8 both side

1.  –  solve for x

Solution

=  –

7

14x-7 = 9x

14x -9x =7

5x = 7

x =

### MATHEMATICS FORM ONE TOPIC 5-ALGEBRA

EXERCISE 7.4

Solve the following equations

1. x + 12 = 25

Solution

x = 25 – 12

x = 13

x = 5

1. 2x + 12 = 25

Solution

2x + 12 – 12 = 25 – 12

2x + 0 = 13

1. x – 8 = 8

Solution

x– 8 + 8 = 8 + 8

x = 16

1. x = 55

Solution

X = 5 5

X = 25

1. 2x – 8 = 8

Solution

2x =8+8

2x = 16

x = 8

1. 3x – 3 = 15

Solution

3x – 3 + 3 = 15 + 3

x =6

1. – 3 = 5

Solution

1. 0.2x + 7 = 9

Solution

0.2x + 7 – 7 = 9 – 7

x = 10

1. 0.6x – 5 = 7

Solution

0.6x – 5 + 5 = 7 + 5

x = 20

1. + 3 = 5

Solution

1. 4x – 7= 7

Solution

4x = 7 + 7

4x = 14

=

1. = 14

Solution

Solution

1. = 6

Solution

= 6 x 5

=

x = 10

Solution

3x = 25 + 1

=

1. = 10

5 = 10x

1. 18.10

Solution

5 x 1 = 10 (x + 1)

5 = 10x + 10

5 – 10 = 10x

solution

1 (x + 5) = 3 (x – 1)

x+ 5 = 3x – 3

3 + 5 = 3x – x

x = 4

Solution

1 (x + 5) = 5 (x – 1)

x + 5 = 5x – 5

5 + 5 = 5x – x

Solving word problems

E.g. 1

If John has hundred shillings, how many oranges can be buy if orange costs 50 shillings?

Solution

Let k be the number of oranges John can buy but one orange costs 50shs.

50 x k = 200

K = 4

John can buy 4 oranges

Example 2:

A father age is 4 times the age of his son. If the sum of there is fifty years Find the age of the son.

Solution

Let the age of father be y

Let the age of the son be x

Therefore the age of the father is y = 4x

Their sum = 4x + x = 50
5x = 50

The son’s age is 10years old

Example 3:

The sum of 2 consecutive numbers is 31. Find the smaller numbers

Solution

Let the smaller number be x

Let the bigger number be x + 1

x+ x + 1 = 31
2x + 1 = 31
2x = 31 – 1
2x = 30

The smaller number is 15

Exercise 7.5

1. If 4 is added to anumberand the sum is multiplied by 3 the result is 27. Find the number.

Solution

Let the number be ‘b’

(b + 4) x 3 = 27

12 + 3b = 27

3b = 15

b= 5

1. Okwi’s age is six times uli’s age.15 yearshenceOkwi will be three times as old as Uli. Find their ages.

Solution

Let the age of Uli be x

Okwi = 6x

Okwi Uli

6x x

6x + 15 x + 5

6x + 15 = 3x + 45

fifteen years to come
6x + 15 15 + x
Then 6x + 15 = 3(x + 15)
6x + 15 = 3x + 45
6x – 3x = 45 – 15
3x = 30

x = 10

Okwi = 60 years

Uli = 10 years

3 . The sum of two consecutive odd numbers is 88. Find the numbers

Solution

Let the number be n

n + 2, n + 4

n + 2 + n + 4 = 88

2n + 6 = 88

n = 41

The smaller number = 41 + 2 = 43

The bigger number = 41 + 4 = 45

1. Obi’s age is twice Oba’s age. 4 years ago Obi was three times as old as Oba. Find their ages.

Solution

Oba’s age let it be x

Obi Oba

2x x

2x – 4 x – 4

2x – 4 = (x – 4) 3

2x – 4 = 3x – 12

8 = x

Obi = 16 years old.

Oba = 8 years old.

Inequalities in one unknown

The following rules are useful when solving inequalities

1. i) Adding or subtracting equal amounts from each side does not change the inequalities sign

Example : solve x – 2 ≤ 4

Solution

X – 2 + 2 ≤ 4 + 2

X ≤ 6

Example 2: 2x + 4 ≥ 16

Solution

2x + 4 – 4 ≥ 16 – 4

≥

X ≥ 6

1. ii) Multiplying or dividing by samepositive number each side change the inequality sign

Example: solve 3y + 16 < 50

Solution

3y + 16 – 16 < 50 – 16

,

Example 2: (2x – 4) ≥ 9

(2x – 4) ≥ 9 x 3

2x – 4 + 4 ≥ 29 x 3

≥

X ≥ 3

iii) Multiplying or dividing each side by negative number CHANGES the inequality sign.

Example. Solve the inequality

((4 – 3x) < 4

Solution

2 (4 – 3x) < 4 x 3

2 () <

4 – 3x < 6

-3x < 6 – 4

The sign changes

Examples 1: Solve -4x + 3≥

-4x + 3≥
-4x ≥ -3

-4x ≥ –

÷ -4

x ≤

Examples 2. solve

Find their L.C.M

>  x 4

3 (2x – 6) 4 (3 – 2x)

6x – 18 > 12 – 8x

6x + 8x > 12 + 18

X >

BINARY OPERATIONS

Is an operation denoted by *, which describe the formula of a given variables.

if P * q = 5pq – p: Find

1. i) 2 * 3 =

p = 2 and q = 3

2 * 3 = 5 (2) (3) – 2

= 30 – 2 = 28

2 * 3 = 28

1. ii) (1* 2) * 3

Solution

(1 * 2) = p = 1 and q = 2

In (1 * 2) = 5 (1) (2) -1 = 9
=10-1
=9
12=9

9 * 3 = p * q

9 * 3 = 5 (9) (3) – 9

= 135 – 9

= 126

(1 * 2) * 3 = 126

1. iv) (2 * 1) * (3 * 2)

Solution

2 * 1 = p = 2 and q = 1

5 (2) (1) – 2

2 * 1 = 10 – 2

= 8

3 * 2 = p = 3 and q = 2

5 (3) (2) – 3

15 x 2 – 3

3 * 2 = 30 – 3

32= 27

2 * 1 = 8

3 * 2 = 27

8 * 27

8 * 27 = p = 8 and q = 27

5 (8) (27) – 8

40 x 27 – 8

1080 – 8

Then: 8 * 27 = 1072

(2 * 1) * (3 8 2) = 1072

1. if (t * 5) = 50 find t

Solution

t * 5 = p = t and q = 5

t * 5 = 5 (t) (5) – t

15t – t

24t=50

t * 5 = 24t

=

t =