Algebra

ALGEBRA

By : kasskassmaths​

​General Mathematics Syllabus

Section #6, Objectives ​1 - 21

TABLE OF CONTENTS

​Distributive Law - Slide 3

Changing the subject of the formula - Slide 4

Quadratics - Slide 6

​Completing the square - Slide 15

​Binary Operations - Slide 19

​Law of Indices - Slide 24

Operations involving algebraic fractions - Slide ​25​

​Solving simultaneously - Slide 27

Linear equ​alities - Slide 33

​Variation - Slide 38

Past Paper Questions to Practice - Slide ​41

DISTRIBUTIVE LAW

CHANGING THE SUBJECT OF THE FORMULA

continued...

Quadratics

m + n = 7 (b)

m x n = 10 (ac)

m and n are 2 & 5 ​

​m + n = -9

m x n = 14

m and n are -2 & -7​

​m + n = 1

m x n = -6

m and n are -2 & 3​

​Note : with the factorization in the 3rd line, there's a common factor before the 2nd factor can be accurately determined.

​m + n = 17 (b)

m x n = 16 (ac)

​m and n is 1 & 16

COMPLETING THE SQUARE

​Completing the square using perfect squares when a > 1.​

This involves a factoring of a common value from 2 terms before completing the square and then later multiplying the same value​ by two other terms, making use of the distributive law.

​

BINARY OPERATIONS

e.g 4

Consider a x (b + c)​ = (a x b) + (a x c)

Using a = 2, b = 3 , c = 4 , show that multiplication is distributive over addition.

​LHS

2 x ( 3 + 4)

= 2 x (7)

​= 14

Given that LHS = RHS, multiplication is distributive over addition.​

​

​RHS

(2 x 3) + (2 x 4)

= 6 + 8

= 14

LAW OF INDICES

​index(power)

​base

​number

​number

​index(power)

​base

Operations involving algebraic fractions

continued..​

SOLVING SIMULTANEOUSLY / Solving Equations simultaneously

There are two types : Elimination and Substitution.

​* Elimination is used when there is an identical term in each equation / the both equations have a common term. E.g Equ 1 contain a 3x and Equ 2 contain a 3x or -3x.

* Substitution is used when either one or both equations contain a term with a coefficient of 1.

E.g Equ 1 contain a 1x or Equ 2 contain 1y or both equations contain 1x or 1y.

​

​

Steps :

1) Number the equations. "(1)" and "(2)" ​

​2) Identify which type would be used to solve the equations

​3) If Elimination : Add or Subtract the equations & Solve for one unknown value.

If Substitution : Rearrange the equation that contain 1x or 1y and sub it into the other & solve for one unknown value.

​4) Solve for the next unknown value by subbing it into one of the equations.

​5) Checkback : sub both values into equations.

E.gs #1 - Elimination (addition)

​

​Note :

4y - 4y = 0

Check back :

2(3) + 4(2) = 14

4(3) - 4(2) = 4​

E.gs #2 - Elimination (subtraction)

​Note :

b - b = 0

Checkback :

6(2) + 6 = 18

4(2) + 6 = 14 ​

E.gs #3 - Elimination (different coefficients)

​Note : The equations needed to be adjusted to get 2 terms being the same for elimination to occur. Another option would have been (1) x 5 and (2) x 3​.

E.gs #3 - Substitution (one equation has a term with a coefficient of 1/-1)

​Note :

y was made the subject of the equation in eq 2

E.gs #4 - Substitution (both equations have a term with a coefficient of 1/-1)

Linear EQUALITIES

​On the graph or number line,

​: the more than or equal to and less than or equal to signs are represented by this (inclusive)

: the less ​than and more than sign are represented by this (exclusive)

E.gs

​

VARIATION

E.gs

​

Past Paper Questions to Practice

June 2019 No. 2 (a), (b) and (c)

July 2021 No. 2 (a) & (b)​

​May/June 2022 No. 2 (a) , (b) & (c)
May /June 2023 No. 2
May/June 2024 No. 2