ALGEBRA FRACTIONS

The term Algebra is derived from these two words; Alphabet and Numbers. It is a combination of both Alphabet (letters) and Numbers.

Therefore, we can say that Alphabet + Numbers = Algebra and Hence Alphabet + Numbers + Fractions = Algebra Fractions.

Types of Algebra Fraction

There are two types of Algebra fractions;

1. Monomial Fractions; These are fractions with only one value as denominator. Examples are; 

, etc…

2. Binomial Fractions; These are fractions with two values as denominator. Examples are;

 

, etc…

Rules for Solving Algebra Fractions

1. If the values are three digits, you factorise both numerator and denominator.

2. If the values are two digits, you look for their common factors.

3. If the values are single digit, just go ahead and simplify as there’s no need to factorise or look for common factors!

Types of Algebra Fractions

1. Simplification of Algebra Fractions

2. Multiplication & Division of Algebra Fractions

3. Substitution in Algebra Fractions

4. Equations in Algebra Fractions

1. Simplification of Algebra Fractions

Examples; Simplify the following Algebra;

Solution

Step 1

Note;

The example above contains three digit values, am I right? So we will factorise both the numerator and denominator before simplification;

NUMERATOR

a² - 5a + 6                +6a² (×)    (-3a & -2a)

                                   -5a (+)

a² – 3a – 2a + 6

a(a – 3) -2(a – 3)

(a – 3) (a – 2)

Step 2

DENOMINATOR

2 – 3a + a² (Rearrange)

a² – 3a + 2               +2a² (x)     (-2a & -1a)

                                   -3a (+)

a² – 2a – 1a + 2

a (a – 2) -1(a – 2)

(a – 2)(a – 1)

Step 3

Bring the Numerator and Denominator together;

Step 4

Cancellation Process; (a – 2) cancels (a – 2);

Step 1

The example 2 above contains two digit values, so we will look for common factors for both the Numerator and Denominator as well.

NUMERATOR;

6m²u² - 4mu³ ---The common factor is 2mu².

2mu²(3m – 2u)

Step 2

DENOMINATOR;

9m³u – 4mu³ ---The common factor is mu.

mu (9m² – 4u²) ….this can also be written thus as 3² gives 9 and 2² gives 4;

mu (3²m² – 2²u²)

mu (3m – 2u)² ….the “raise to power 2” has been brought out as the common factor;

mu (3m – 2u)(3m + 2u)

Step 3

Combining the Numerator and Denominator;

Step 4

Cancellation Process; mu goes in 2mu² to give 2u while (3m – 2u) cancels each other;

Step 1

This is a three digit value, remember you factorise when dealing with three digit values;

NUMERATOR;

a² + ax – 6x²         -6a²x² (×)    (+3ax & - 2ax)

                                       +ax (+)

a² + 3ax – 2ax – 6x²

a(a + 3x) – 2x(a + 3x)

(a + 3x)(a – 2x)

Step 2

DENOMINATOR;

2x² + ax – a²  (Rearrange)

- a² + ax + 2x²       -2a²x² (×)    (+2ax & - 1ax)

                                       +ax (+)

- a² + 2ax – 1ax + 2x²

-a(a – 2x) – x(a – 2x)

(a – 2x)(-a – x)

Step 3

Combining the Numerator and Denominator;

Step 4

Cancellation Process; (a - 2x) cancels each other;

Note; The minus sign was brought out as a common factor;

Note; The minus sign went up! This is a rule as minus sign must always remain with the numerator;

Step 1

NUMERATOR;

ab + ac  ----the common factor will be a;

a(b + c)

Step 2

DENOMINATOR;

ad + ae   -----the common factor here will also be a;

a(d + e)

Step 3

Combining the Numerator and Denominator;

Step 4

Cancellation Process; a cancels each other;

Step 1

Note; The values above are single digit, so there’s no need for factorization or common factors. Just go ahead and simplify!

 Step 2

Cancellation Process; 2 goes in 8 gives 4 and 2 goes in 10 gives 5 while x goes in x² gives x and z cancels z;

Series

A Series is the addition of successive terms in such a way that they are related to another according to a well defined rule. In other words, series is the addition of a sequence.

Examples of series are;

2 + 5 + 8 + 11 + 14 + ……..

4 + 16 + 64 + 256 + 1024 + ………

-1 + 0 + 7 + 14 + 23 + 34 + ……..

Note

Just like sequence you will be given what you’ll know as nth term formula to find the terms of series and the formula always differ per questions.

Examples;

1. Find the series of the first six terms of 2ⁿ + 4n².

Solution

Note

2ⁿ + 4n² stands as a formula that you’ll use to solve for the first six terms mentioned in the question. Finding the first six terms means solving for T_1, T_2, T_3, T_4, T₅ and T₆ respectively_.

Step 1

T_n = 2ⁿ + 4n² ---The nth term formula.

T₁ = 2¹ + 4(1)²

T₁ = 2 + 4(1²)

T₁ = 2 + 4(1)

T₁ = 2 + 4

T₁ = 6

Step 2

T₂ = 2² + 4(2)²

T₂ = 4 + 4(2²)

T₂ = 4 + 4(2 × 2)

T₂ = 4 + 4(4)

T₂ = 4 + 4 × 4

T₂ = 4 + 16

T₂ = 20

Step 3

T₃ = 2³ + 4(3)²

T₃ = 8 + 4(3²)

T₃ = 8 + 4(3 × 3)

T₃ = 8 + 4(9)

T₃ = 8 + 36

T₃ = 44

Step 4

T₄ = 2⁴ + 4(4)²

T₄ = 16 + 4(4²)

T₄ = 16 + 4(4 × 4)

T₄ = 16 + 4(16)

T₄ = 16 + 64

T₄ = 80

Step 5

T₅ = 2⁵ + 4(5)²

T₅ = 32 + 4(5²)

T₅ = 32 + 4(5 × 5)

T₅ = 32 + 4(25)

T₅ = 32 + 100

T₅ = 132

Step 6

T₆ = 2⁶ + 4(6)²

T₆ = 64 + 4(6²)

T₆ = 64 + 4(6 × 6)

T₆ = 64 + 4(36)

T₆ = 64 + 144

T₆ = 208

 The series of the first six terms will be; 6 + 20 + 44 + 80 + 132 + 208 +…

2. Find the sum of the series n² + 5n up to the 4^th term.

Solution

Step 1

T_n = n² + 5n---Formula

T₁ = 1² + 5(1)

T₁ = 1 + 5

T₁ = 6

Step 2

T₂ = 2² + 5(2)

T₂ = 4 + 5 × 2

T₂ = 4 + 10

T₂ = 14

Step 3

T₃ = 3² + 5(3)

T₃ = 9 + 5 × 3

T₃ = 9 + 15

T₃ = 24

Step 4

T₄ = 4² + 5(4)

T₄ = 16 + 5 × 4

T₄ = 16 + 20

T₄ = 36

 The sum of the series will be; 6 + 14 + 24 + 36 = 80.

 

Practice these questions below;

1. What is the sum of the eight series of an nth term equals 2ⁿ + 5n?

2. Find the seven series of the term (n + 1)² – 2n.

3. Write out the series of the term 3ⁿ(n – 1).

4. Find the series of the first five terms of n(2 + 3n).

5. Find the sum of the series 5(3^{n - 1}) up to the 4^th term.

Linear Inequalities

An inequality is just like linear equation which has two sides and an unknown variable which are always represented with alphabets (a-z) but deals with a lot of signs.

Signs of inequalities

> Greater than

< Less than

≥ Greater or equal to

≤ Less than or equal to

Solve the following inequalities;

Example 1; Simplify 2x – 4 > 8

Solution

Note;

In order to do this, you must consider the two sides of this inequality.

2x – 4 > 8 (Make x the subject of the equation)

2x > 8 + 4

2x > 12 (Divide both sides by 2)

x > 6.

Example 2; Solve the inequalities 3x – 8 ≤ 10 + 5x

Solution

3x – 8 ≤ 10 + 5x (Collect like terms)

3x – 5x ≤ 10 + 8

- 2x ≤ 18 (Divide both sides by the co-efficient of x which is -2)

x ≥ - 9.

Note; the inequality sign must change when dividing with a minus sign).

Practice these questions below;

1. Simplify 4x – ¹/₃   ≥   ²/_3x + 2

3. Simplify 6x – 2 < 2x + 8

4. Simplify ¹/_3y  >  ¹/_2y  +  ¹/₄

5. Simplify ⁶/_x  ≥  2