Developing Algebra Skills - PowerPoint PPT Presentation

Developing Algebra Skills

Session Four – 12 March 2014

• Multiply out the brackets in (x + 5)(x – 3)
• Grid

= x2 + 5x - 3x – 15

= x2 + 2x – 15

x2

+5x

-3x

-15

• Multiply out the brackets in (x + 5)(x – 3)
• FOIL
• (x + 5)(x – 3)

= x2 - 3x + 5x – 15

= x2 + 2x – 15

F First

O Outside

I Inside

L Last

• Using either method calculate (b + 3)(b - 6)

= b2 + 3b - 6b -18

= b2 - 3b - 18

b2

+3b

- 6b

-18

• What is the calculation to find the area of this rectangle
• (x - 3)(x - 5) = x2 – 5x – 3x + 15
• = x2 – 8x + 15

X

5

X

3

• Putting the brackets back in
• Eg a2 – 9a + 20 = (a )(a )
• Find two numbers that multiply to give 20 and add to give – 9
• = (a – 5)(a – 4)

This came from

This times by this

• Factorise y2 – 7 + 12
• Two numbers which multiply to give 12 and add to give -7
• So y2 – 7 + 12 = (y – 3)(y – 4)

• y2 – 7 + 12 = 0
• (y – 3)(y – 4) = 0
• Therefore y – 3 = 0 or y – 4 = 0
• So y = 3 or y = 4

For this to be true, one of the brackets must be 0

• What is a linear equation ?
• How do we solve …
• 3 + x = 7
• We can solve this only if there is one unknown value

• If we have two unknowns we must have two equations ….
• a + 2b = 12
• a – b = -3
• There are 2 methods
• Substitution
• Addition / subtraction

• Method 1 - Substitution
• a + 2b = 12
• a – b = -3

Rearrange one equation to be either

a = something or b = something

Lets rearrange to be a = b – 3

1

2

2

1

• We can now substitute a = b – 3 into
• a + 2b = 12
• So b – 3 + 2b = 12
• Rearrange and solve for b, as usual
• i.e. 3b = 15
• b = 5 Substitute back in
• a = 5 – 3
• a = 2

a = 2, b = 5 is the ONLY solution which satisfies both equations

• Method 2 – Addition or subtraction
• 2a + b = 10
• a - b = -1
• Add the equations to remove the b terms …
• 2a + b = 10
• a - b = -1
• 3a = 9 so a = 3
• Substitute back in to find b
• a – b = -1 so b = 4

We need to add or subtract to remove either the a or the b terms

• Solve the equations a + b = 9 and 4a – b =1
• By substitution …
• a = 9 – b
• So 4(9 – b) – b = 1
• Expand brackets and collect like terms
• 36 – 4b – b = 1
• So 35 = 5b b = 7 and
• a = 9 – b a = 2

• Solve the equations a + b = 9 and 4a – b =1
• By addition …
• a + b = 9
• 4a – b = 1
• 5a = 10 So a = 2
• Substitute back to give
• 2 + b = 9 So b = 7

• Try Exercise …

• e.g Formulae to find perimeter of rectangle
• P = 2l + 2w
• If the length is 23cm and the width is 14cm, what is the perimeter ?
• Substitute into formulae
• P = 2(23) + 2(14)
• P = 46 + 28
• P = 74 cm

• The area of a trapezium is given by …
• A = ½ (a + b)h
• Where a and b are the lengths of the parallel sides and h is the distance between them
• Find A when a = 2.7, b = 6.9 and h = 4.2
• NB – Use your calculator

• A = ½ (a + b)h
• A = ½ (2.7 + 6.9)4.2
• A = ½ (9.6)4.2
• A = ½ (40.32)
• A = 20.16 cm

• The connection between temperature in °C and °F is given by the formula …
• C = (F – 32)
• Find C when F = 68
• C = (68 – 32)
• C = (36) So C = 20°

• What if we have the temperature in °C and what to convert it to °F ?
• We need to transpose the formula to make F the subject…
• C = (F – 32)
• C = F – 32 So, C + 32 = F

• Changing the subject of a formula
• e.g Formulae to find perimeter of rectangle
• P = 2l + 2w
• Rearrange the formula to make l the subject
• = l + w
• So - w = l

Remember:

Whatever we do to one side we must do to the other side to maintain the equality

• Rearrange the formula to make y the subject
• 3x + 2y = 7
• 2y = 7 – 3x
• y = 7 – 3x

2

Subtract 3x from both sides

Divide both sides by 2

• The area of a trapezium is given by A = ½ (a + b)h
• Find the height when the Area = 72m2 and the lengths of the parallel sides are 8m and 10m
• First rearrange the formula to make h the subject

• A = ½ (a + b)h
• 2A = (a + b)h
• So 2A = h

(a + b)

• h = 2 x 72

(8 + 10) So h = 8m

Multiply both sides by 2

Divide both sides by (a + b)

Now substitute in the given values

• Try Exercise …

• This is a method that involves making an intelligent guess and then getting closer to the answer by improving the guess
• Eg Find √115 to 2 d.p.
• So, first guess is 10.5 10.52 = 110.25
• Improved guess is 10.75 10.752 = 115.5625
• Improved guess is 10.7 10.72 = 114.49
• Improved guess is 10.73 10.732 = 115.1329
• Improved guess is 10.72 10.722 = 114.9184
• So √115 = 10.72 (2 d.p)

We know 102 = 100 and 112 = 121

• The equation x3 – 2x2 + 3x – 9 = 0 has a solution between x = 2 and x = 3. Find the solution correct to 2 d.p.
• First guess x = 2.5 equation = 1.625
• Improved guess x = 2.3 equation = -0.513
• Improved guess x = 2.4 equation = 0.504
• Improved guess x = 2.35 equation = -0.00171
• Improved guess x = 2.37 equation = 0.18825
• Improved guess x = 2.36 equation = 0.08506
• So x3 – 2x2 + 3x – 9 = 0 when x = 2.35 (2 d.p)

• Try Exercise …

• Simultaneous Equations
• Transposition of Formulae
• Trial and Improvement