## Algebra 1

__2.1 – Brackets and Simplifying__

__2.1 – Brackets and Simplifying__

- A term outside a bracket multiplies each of the terms inside the bracket. This is the distributive law
- E.g. 3(x-2y) = 3x – 6y

- ‘Like’ terms can be added:
- x’s cab be added to x’s
- y’s can be added to y’s
- x²’s can be added to x²’s
- E.g. 4x + 3y + 3x² + 4y – x = 3x +7y +3x²

__2.2 – Linear Equations__

__2.2 – Linear Equations__

- If the
*x*term is negative, add an*x*term with a positive coefficient to both sides of the equation - E.g. 8 – 6x = 4

8 = 4 + 6x

8 – 4 = 6x

4 = 6x

4/6 = x

2/3 = x

- If there are
*x*terms on both sides, collect them on one side - E.g. 4x – 8 = 6 – 6x

4x + 6x = 6+8

10x = 14

x = 14/10

x = 1.4

- If there is a fraction in the
*x*term, multiply out to simplify the equation - E.g. 4x/6 = 20

4x = 20 × 6

4x = 120

x = 120/4

x = 30

__2.3 – Simultaneous Equations__

__2.3 – Simultaneous Equations__

- Substitution Method: Used when one equation contains a unit quantity of one of the unknowns
- Obtain an equation in one unknown and solve this equation
- Substitute the results from ‘1.’ into the linear equation to find the other unknown
- E.g. 3x – 2y = 0 …[1]

2x + y = 7 …[2]

2x + y = 7 …[2]

y = 7 – 2x

Substituting into [1]

3x – 2 (7 – 2x) = 0

3x – 14 + 4x = 0

7x = 14

x = 14/7

x = 2

Substituting into [2]

2 × 2 + y = 7

4 + y = 7

y = 7 – 4

y = 3

The solutions are *x* = 2, *y* = 3

- Elimination Method:
- Choose an unknown in one of the equations and multiply the equations by a factor or factors so that this unknown has the same coefficient in both equation

- Eliminate this unknown from the two equations by adding or subtracting them, then solve for the remaining unknown

- Substitute into the first equation and solve for the eliminated unknown

- E.g. x + 2y = 8 …[1]

2x + 3y = 14 …[2]

x + 2y = 8 (×2) …[1]

2x + 4y = 16 …[3]

2x + 3y = 14 …[2]

Subtract [2] from [3]

y = 2

Substituting into [1]

x + 2 × 2 = 8

x + 4 = 8

x = 8 – 4

x = 4

The solutions are x = 4, y = 2

__2.4 – Factorising__

__2.4 – Factorising__

- Common Factors:
- E.g. 6x² + 12x

6x (x + 2)

- Difference of Two Squares:
- E.g. 49 – x²

(7 + x) (7 – x)

- Group Factorization:
- E.g. 2b + ac + ab + 2c

2 (b + c) + a (c + b)

(2 + a)(c + b)

- Quadratic Expressions:
- E.g. x² + 14x + 24
- Find two numbers which multiply to give 24 and add up to 14
- Numbers are 12 and 2
- Put these numbers into brackets

x² + 12x + 2x +24

x (x + 12) + 2 (x + 12)

**(x + 2)(x + 12)**

- So x² + 14x + 24 = (x + 2)(x + 12)

__2.5 – Quadratic Equations__

__2.5 – Quadratic Equations__

- Solution by Factors:
- E.g. x² + 14x + 24 = 0
- Find two numbers which multiply to give 24 and add up to 14
- Numbers are 12 and 2
- Put these numbers into brackets

x² + 12x + 2x +24 = 0

x (x + 12) + 2 (x + 12) = 0

(x + 2)(x + 12) = 0

- Either x + 2 = 0 or x + 12 = 0

x = -2 x = -12

- Solution by Formula:
- Formula: