MATHEMATICS O LEVEL(FORM ONE) NOTES – ALGEBRA

ALGEBRA

Algebra is a study which deals with situations whereby some values are unknown. Normally these unknowns 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 variable(s) in it.

Examples

• a + 2
• x + 3y + 9z
• 16p – qp
• a + b + c + d
• 40

An equation

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

Examples

• 2x – y = 16
• x + 2 = 6 – 5
• 3y + xy = 9

Each member of an equation or expression is called a term.

Coefficient

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

Example

What is the coefficient of the variables in the following?

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

Solutions

• Coefficients of a) x is 6, y is 1, p = -8
• b) k = -1, d = 3, = 1
• c) a = 2, b = 3, c = -1

Addition and subtraction of algebraic expressions

Addition and subtraction of algebraic expressions can be done by adding or subtracting like terms.

Like terms are those terms which have identical (same) variables.

Examples

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

Example: simplify the expression

3n – 7n + 12n

Solution:

• -4n + 12n
• 12n – 4n
• ∴ = 8n

Example: simplify

6m – 4 – 2m + 15

Solution:

• 6m – 2m – 4 + 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:

1. the number of terms
2. the coefficient of each of the terms

i) n + n + n + n + n + k + k + k + x + x = 5n + 3k + 2x

Solution:

• (a) There are 3 terms
• (b) Coefficient of “n” is 5
• Coefficients of “k” is 3
• Coefficients of “x” is 2

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) 3x + 7x – 3x =

Solution:

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

Simplify each of the expressions in numbers 2 – 6:

• 2. 12m + 13m = 25m
• 3. 5y + 7y – 4y = 8y
• 4. 24w – 28w = -4w
• 5. 15n – 9n = 6n
• 6. 4k – k + 3k = 6k
• 7. 8y – 3 – 7y + 4 = y + 1
• 8. 14x + 8 – 3x + 2 = 11x + 10
• 9. 3a – 5b – 7a + 6c + 7a + 8b = 3a + 3b + 6c
• 10. 4x – 6y + 7x + 2y = 11x – 4y
• 11. 3x + 4 + 8x – 4 – 11x = 0
• 12. 8m + 0.4m – 2 – 6m + 8 = 2.4m + 6

Multiplication and division of algebraic expressions

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

Solution:

(a – 2b + 6ab) × 12xy = 12axy – 24bxy + 72abxy

Example 2: Re-write without brackets

-16a (-2mn + 9xb – 3kbc)

Solution:

-16a (-2mn + 9xb – 3kbc) = (-16a × -2mn) + (-16a × 9xb) + (-16a × -3kbc)

= 32amn – 144axb + 48abck

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

ii) 8m ÷ 2 + 3mn ÷ n

Solution:

i) xy + yz + 2xy – zy ⇒ xy + 2xy + yz – 3zy = 3xy – 2yz

ii) 8m ÷ 2 + 3mn ÷ n = 4m + 3m = 7m

3. Simplify the following:

i) 5mn – 3mn = 2mn

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 = 26n – 3

iv) abc + bac – cab

Solution:

abc + abc – abc = abc – abc + abc = abc + 0 = abc

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

Solution:

10x + 6y + 9x + 6y = 19x + 12y

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

Solution:

2nm + 3m + 3mn + 4n = 5mn + 3m + 4n

vii) x (y – 5) + y (x + 2)

Solution:

xy – 5x + yx + 2y = xy + yx – 5x + 2y = 2xy – 5x + 2y

ix) Pq – 2qp + 3pq – 2qp

Solution:

Pq + 3pq – 2qp – 2qp = 4pq – 4pq = 0

x) (4x + 8y) ÷ 2 + (9xw + 4xy) ÷ w

Solution:

xi) Multiply 6a – 5b by 3x

Solution:

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

∴ = 18ax – 15bx

Equations

An equation is a mathematical statement which involves two expressions connected or joined by an equal sign.

So we define an equation also as a statement of equality e.g. 2y – 6 = 3x + 12.

The values of variables can be found in an equation if the number of equations is equal to the number of unknowns.

Formulation of an equation

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

1. Understand the problem/question, what it is asking for.
2. Let the unknown be represented by a variable.
3. Formulate the equation using the given information.

Signs, words or phrases 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, product.
• ÷ 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 of his sister, formulate an equation.

Solution:

Let y be the father’s age and x be the age of the daughter.

The age of the son = 3x

y = 3x + x

y = 4x

2. The sum of two numbers is 20. If one of the numbers is 8, formulate an equation.

Solution:

Let one of the numbers be x and the other number = 8.

x + 8 = 20

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

Solution:

A girl = 14 years

Let y be the girl’s age in x years time.

In years time = 14 + x

∴ y = 14 + x

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

Solution:

Let another number = x

∴ 24 – x = 16

Exercise 7.3

Formulate equations for each of the following:

1. Five times a number gives twenty.

Solution:

5x = 20

2. The difference between 123 and another number is 150.

Solution:

Let another number = x

Then x – 123 = 150

3. The sum of 21 and another number is 125.

Solution:

Let another number = y

Sum means (+)

21 + y = 125

4. When a certain number is increased by 15, the result is 88.

Solution:

Let the number be x

Then x + 15 = 88

5. When 99 is increased by a certain number the result is 63.

Solution:

Let the number = y

Then 99 + y = 63

6. The product of 12 and another number is the same as two times the sum of 12 and the number.

Solution:

Let the number be x

Then 12 × x = 2 × (12 + x)

∴ 12x = 24 + 2x

7. A number is such that when it is doubled 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 2x + 8 = 3x – 7

8. When 36 is added to a certain number, the result is the same as multiplying the number by 5.

Solution:

Let x be the number

Then x + 36 = 5x

9. If John is n years old and is 6 years older than James, 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 ages = q + n + 6

   ∴ = q + n + 6

10. When the sum of n and (n + 3) is multiplied by 5, the result is half the product of the two numbers.

Write the expression of this statement:

Solution:

(n + (n + 3)) × 5 = ½ (n × (n + 3))

(2n + 3) × 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

2. x – 8 = 15

x – 8 + 8 = 15 + 8

x = 23

3. 3x – 5 = 7

3x – 5 + 5 = 7 + 5

3x = 12

∴ x = 4

4. 2x + 3 = 12

Solution:

Multiply both sides by 2



5. (3x – 2) = 10

Solution:



6. = 2

Solution:

=

8 × 1 = (3x – 2)

8 = 6x – 4

8 + 4 = 6x – 4 + 4

12 = 6x

∴ x = 2

7. – = 4

Solution:



2m = 4 × 15

2m = 60

m = 30

8. – 3 = 5

Solution:



9. 0.2x + 7 = 9

Solution:

0.2x + 7 – 7 = 9 – 7

∴ x = 10

10. 0.6x – 5 = 7

Solution:

0.6x – 5 + 5 = 7 + 5

∴ x = 20

11. + 3 = 5

Solution:



12. 4x – 7= 7

Solution:

4x = 7 + 7

4x = 14

∴ x = 3.5

Exercise 7.4

Solve the following equations:

1. x + 12 = 25

Solution:

x = 25 – 12

x = 13

2. =

x = 5

3. 2x + 12 = 25

Solution:

2x + 12 – 12 = 25 – 12

2x = 13

x = 6.5

4. x – 8 = 8

Solution:

x – 8 + 8 = 8 + 8

x = 16

5. x = 5

Solution:

x = 5

6. 2x – 8 = 8

Solution:

2x = 8 + 8

2x = 16

x = 8

7. 3x – 3 = 15

Solution:

3x – 3 + 3 = 15 + 3

3x = 18

x = 6

8. – 3 = 5

Solution:



9. 0.2x + 7 = 9

Solution:

0.2x + 7 – 7 = 9 – 7

x = 10

10. 0.6x – 5 = 7

Solution:

0.6x – 5 + 5 = 7 + 5

x = 20

11. + 3 = 5

Solution:



12. 4x – 7= 7

Solution:

4x = 7 + 7

4x = 14

x = 3.5

Solving word problems

Example 1:

If John has 200 shillings, how many oranges can he buy if one orange costs 50 shillings?

Solution:

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

50 × k = 200

k = 4

∴ John can buy 4 oranges.

Example 2:

A father’s age is 4 times the age of his son. If the sum of their ages 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 10 years old.

Example 3:

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

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 a number and the sum is multiplied by 3, the result is 27. Find the number.

Solution:

Let the number be ‘b’

(b + 4) × 3 = 27

3b + 12 = 27

3b = 15

b = 5

2. Okwi’s age is six times Uli’s age. 15 years hence Okwi will be three times as old as Uli. Find their ages.

Solution:

Let the age of Uli be x

Okwi = 6x

15 years to come:

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

2n = 82

n = 41

The smaller number = 41 + 2 = 43

The bigger number = 41 + 4 = 45

4. 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 = 2x

4 years ago:

2x – 4 = 3(x – 4)

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. Adding or subtracting equal amounts from each side does not change the inequality 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

2. Multiplying or dividing by the same positive number each side does not change the inequality sign.

Example: solve 3y + 16 < 50

Solution:

3y + 16 – 16 < 50 – 16

3y < 34

Example 2: (2x – 4) ≥ 9

Solution:

2x – 4 + 4 ≥ 9 + 4

2x ≥ 13

x ≥ 6.5

3. Multiplying or dividing each side by a negative number CHANGES the inequality sign.

Example: Solve the inequality (4 – 3x) < 4

Solution:

–3x < 0

Multiply both sides by –1 and reverse the inequality:

3x > 0

Examples 1: Solve -4x + 3 ≥

Solution:

-4x + 3 ≥

-4x ≥ -3

Divide both sides by -4 and reverse the inequality:

x ≤ ¾

Examples 2: solve

Find their L.C.M

3 (2x – 6) > 4 (3 – 2x)

6x – 18 > 12 – 8x

6x + 8x > 12 + 18

14x > 30

x > 30/14

Binary operations

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

If P * q = 5pq – p: Find

i) 2 * 3 =

p = 2 and q = 3

2 * 3 = 5 × 2 × 3 – 2 = 30 – 2 = 28

2 * 3 = 28

ii) (1 * 2) * 3

Solution:

(1 * 2) = p = 1 and q = 2

1 * 2 = 5 × 1 × 2 – 1 = 10 – 1 = 9

∴ 1 * 2 = 9

9 * 3 = p * q

9 * 3 = 5 × 9 × 3 – 9 = 135 – 9 = 126

(1 * 2) * 3 = 126

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

Solution:

2 * 1 = p = 2 and q = 1

2 * 1 = 5 × 2 × 1 – 2 = 10 – 2 = 8

3 * 2 = p = 3 and q = 2

3 * 2 = 5 × 3 × 2 – 3 = 30 – 3 = 27

8 * 27 = p = 8 and q = 27

8 * 27 = 5 × 8 × 27 – 8 = 1080 – 8 = 1072

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

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

Solution:

t * 5 = p = t and q = 5

t * 5 = 5 × t × 5 – t = 25t – t = 24t

24t = 50

t = 50/24