Algebra 1

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

• 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

• 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

• 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)

• 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)

• 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:

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