Important terms Variable: This is the letter in a term or an expression whose value can vary. Some of the letters most used for variables are x and y. Term: This is an algebraic quantity which contains only a letter and may contain a number. e.g. 3n means 3 multiplied by the variable n n 2

n2

means n divided by 2 means multiplied by itself ( normally said as ‘n squared’)

Expression: this is a combination of letters and signs, often with numbers. e.g. 8 – n means subtract n from 8 n – 3 means subtract 3 from n 2n – 7 means n multiplied by 2 with 7 added on

Basic algebra Sum and Subtraction

Let’s express this in numbers 4–1=3 This is real simple calculation. To this calculation to be algebra,

4–□=3 And let’s say □ = x 4–x=3 x=1

Multiplication and Division I am going to start to use ‘x’, not using □ anymore. 2x = 4 x=

4 2

=2

Exercise 1. Find the value of ‘x’ 2x + 5 = 1 x = -2 2. Find the value of ‘x’ 26 – 4.5x + 0.5x = 2 x=6

Rules of algebra The rules of algebra are the same rules that are used in arithmetic 2+3=3+2 a+b=b+a 3x2=2x3 a x b = b x a or ab = ba But remember, that: 4–1 ≠ 1–4 4/6≠6/4 The other rule of algebra is the same letter can be calculated e.g. x + x = 2x 3x – 2x = x

Square and Cube 2x2=4 It can also be expressed by 22 = 4 it is called ‘square’ Therefore, if there is a equation 33 means 3 x 3 x 3 so the answer is 27. It is called ‘cube’ In algebra, square and cube are commonly used. Such as x2 = 64 Since it means x times x, the answer would be 8. (8 x 8 = 64)

Exercise 1. Find the value of ‘x’ x3 – 5 = 59 x=4 2. Find the value of ‘x’ 2x3 + x3 = 81 x=3

Expanding There are many situations in algebra where there is a need to use brackets in expressions. It is possible to EXPAND brackets. e.g. 3(2 + 3x) = 6 + 9x x(x + 2) = x2 + 2x If there are many brackets, it is necessary to simplify the equations 3(5x + 2) – 2(4x + 1) = 15x + 6 – 8x – 2 = 7x + 4

Exercise 1. Simplify in term of ‘x’ x(x2 + 2) + x2(3 – x) answer: 2x + 3x2 2. Simplify in term of ‘x’ x(x + x2) – x2(

1 x2

+ x) answer: x2 – 1

Simultaneous equations If there are two algebra equations with two unknown letters. e.g. Find values of the unknowns, x and y 2x + 2y = 1 2x + y = 5 There are steps to solve the simultaneous equation

1. Get the same number of y terms/x terms in both equations by multiplying. 2. Eliminate the y terms/x terms by adding or subtracting. 3. Solve to find x/y. 4. Substitute the value for x/y into one of the original equations. 5. Solve to find y/x Use previous example, Since both equations have the same x terms, it is easy to remove ‘x’. Therefore, 2x + 2y = 1 – 2x + y = 5 y = –4 As the value of y is -4, substitute y into one of the equations 2x + 2y = 1 2x + 2(-4) = 1 2x = 9 x=

9 2

There is no solve the simultaneous equation easier way. So you MUST practice

it to be able to solve the questions rapidly.

Exercise 1. x + 7y = 5 x - 7y = - 9 x = -2, y = 1 2. 2a + 5b = 16 10a -3b = -4b a = 0.5, b = 3 3. -2x + 7y = 4 -3x + 5y = - 5 x = 5, y = 2 4. 3x + 2y = 10 4x+ 3y = 13 x = 4, y = -1 5. 3(m - n)- 8(m + n) = 7 2(m + n) +5(m - n) = -65 m = -8, n = 3

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