# How to do Algebra

Algebra is often regarded by many primary and secondary school students as a challenging topic to learn in mathematics. In these examples, we’ll do the 4 parts to Patterns and Algebra by using simple exam questions as illustrations.

‘Simplify’ is commonly used in many exam questions. However, it may require you to find the value of an ‘unknowm’ by applying one of the fours below:

1. Factorising
2. Expanding
3. Substituting
4. Solving

Perhaps, it is important to note that learning how to factorise, expand, substitute and solve algebra early at Grade 7 and 8 is crucial.

## 1. Factorising

Factorise the following expressions (note that expanding and factorising are opposites)

a. 2a + 10 =
Highest Common Factor (HCF) of 2a and 10 is 2

2(a + 10)

b. 4 + 6x =
HCF = 2

2(2 + 3x)

c. 3x – 9=
3 (x – 3)

d. 2x^2 + 4x

HCF = 2x
2x (x + 2)

## 2. Substitution

a. Write down the value of abc when a=10, b=2 and c=0

abc = 10x2x0=0

Maths knowledge: any number x 0 is 0

b. Work out the value of 1/2x – 3y when x= 10 and y= 2

5 – 6 = – 1 (many students write 1 instead)
Concept tested: Addition and subtraction of -ve and +ve numbers. (Reinforce that differences that 6-5=1and 5-6=-1)

c. Find the value of 3x + 2y when x = 4 and y = 5

12 – 10 = 2 …………. 🙂

## 3. Expanding Brackets

(Note that expanding and factorising are opposites)

Expand the following expressions

a. 3(2y – 5) =

6y – 15 ( many student forget to do 3 x – 15)

b. 4(2m + 3n) =
8m + 12n

c. x(x – 10) =

x^2 – 10x

## 4. Solving equation

Solve the following equations to find the value of x

a. 4x = 20
x = 5

b. 3x – 7 = 8
3x = 8 + 7 ( it is important to get the order 8 + 7 right and not 7 + 8: even the answer is same, the answer is NOT be the same when subtracting, see the example below)

3x = 15
x = 5

c. 8(x + 12) = 100

8x + 96 = 100 ………………Expand the brackets
8x = 100 – 96………………. (subtract 96 on both sides (remember balancing equations?)
x = 4/8 (Why divide 8? In order to find the value of x, you must divide LHS and RHS by 8)
x = 1/2 (or 0.5)

Solve the following to find the value of y

a. y/3 = 9

y = 27 ………….multiply 3 x 9 (now, this is important as you can use this to solve complex equations that have a divisor)

b. 2y/5 = 4
2y = 20………. ( 20 = 5 x 4)
y = 10

c. 2y + 3 / 2 = 5

2y + 3 = 10………. ( 10 = 2 x 5)
2y = 10 – 3 ………..(subtracting 3 on both sides of the equation)
2y = 7
y = 7/2
y = 3.5