1. Using and applying mathematics
Sequences & Formulae
Year 10

2. The following series of lessons will equip you with the necessary skills to complete a complex investigation at the end of the unit.
The initial lessons may seem tedious, but bear with us…

3. At the end of the unit you will be presented with this problem:x
What size must the cut-out corners be to give the maximum volume for the open box?

4. To attempt this problem you need to be able to:
LESSON 1
Simplify algebraic expressions x
LESSON 2
Solve algebraic equations
LESSON 3
Formulate your own
expressions & equations
Solve equations by
trial & improvement
LESSON 4
Accurately rearrange formulae
LESSON 5

5. Objective: Monday 21st February
To be able to simplify algebraic expressions
Level 5/6
Lesson 1

6. About 500 meteorites strike Earth each year.
A meteorite is equally likely to hit anywhere on earth.
The probability that a meteorite lands in the
Torrid Zone is
Area of Torrid Zone
Total surface area of earth
Let r represent the earth’s radius. Write an expression to estimate the area of the torrid zone.
You can think of the distance b/w the tropics as the height of a cylindrical belt around the earth at the equator.
The length of the belt is the earth’s circumference 2пr.
The surface area of a sphere is 4пr2.Write and simplify expressions for the area of,and probability that a meteor lands in, the torrid zone.
Use 3963 miles for the earth’s radius. Find the probability.

7. Objective: Wednesday 23rd February
To be able to solve algebraic equations.
Level 6/7
Lesson 2

8. Take the number of the month of your birthday…
Multiply it by 5
Add 7
Multiply by 4
Add 13
Multiply by 5
Add the day of your birth
Subtract 205

9. What have you got? Why does this work? Homework:
Write an algebraic expression and simplify it to prove why this works.

10. Objective: Thursday 24th February
To be able to formulate expressions and formulae.
Lesson 3
Level 6/7

11. The length of a rectangular field is a metres.
The width is 15m shorter than the length.
The length is 3 times the width. a
a - 15
Write down an equation in a and solve it to find the length and width of the field.
Length = 22.5m Width = 7.5m

12. Imagine a triangle
Choose a length for its base.
Call it ‘z’
Make the vertical height 3 units longer than the base
Work out the area of your triangle
Write down an equation in z that satisfies your conditions
Give it to your partner to solve for the base length of your triangle (z).

13. Problems involving quadratic equations
A rectangle has a length of ( x + 4) centimetres and a width of ( 2x – 7) centimetres.
If the perimeter is 36cm, what is the value of x?
X = 7
x + 4
2x - 7
If the area of a similar rectangle is 63cm2 show that 2x2 + x – 91 = 0 and calculate the value of x
X = 6.5

14. Monday 28th February
Objective:
Formulate equations and solve by trial and improvement.
Level 6 / 7

15. The length of a rectangular field is a metres.
The width is 15m shorter than the length.
The length is 3 times the width. a
a - 15
Write down an equation in a and solve it to find the length and width of the field.
a = 3 (a – 15)
Length = 22.5m Width = 7.5m

16. Formulating quadratic equations
Joan is x years old and her mother is 25 years older.
The product of their ages is 306.
a) Write down a quadratic equation in x
b) Solve the equation to find Joan's age.
x ( x + 25 ) = 306
x2 + 25x = 306
x x – 306 = 0
How can we solve this?
Factorisation?
Graphically
Formula

17. A more accurate method is
x x – 306 = 0
This example factorises:
( x + 34 )( x – 9 ) = 0
Either x + 34 = 0
Or, x – 9 = 0
x = -34
x = 9
Since Joan cannot be –34 years old, she must be 9.
Some quadratic equations do not factorise exactly.
Solving some equations (i.e. cubic ) by a graphical method is not very accurate.
A more accurate method is
trial and improvement

18. Solving equations by trial and improvement.
E.g. 1
A triangle has vertical height 3 cm longer than its base.
It’s area is 41 cm2. What is the length of its base to 1 d.p? 41
x ( x + 3) = 41 2
x +3
x ( x + 3) = 41 x 2 = 82
x2 + 3x – 82 = 0 x
Too small
Try x = (3 x 7) – 82 = -12
Too big
Try x = (3 x 8) – 82 = 6
Too small
Try x = ( 3 x 7.6) – 82 =
Try x = ( 3 x 7.7) – 82 = 0.39
Too big
Too small
Try x = ( 3 x 7.65) – 82 =
x = 7.7 to 1 d.p
Base = 7.7cm to 1 d.p

19. 5x2 – 12x + 5 = 0 For x > 1 x2 – 5x – 1 = 0 For x > 0
Race groups with different methods of solving I.e. formula, trial & imp, factorising…
1.9
5.2
1.8
0.5
To 1 d.p

20. Transposition of formula
Wednesday 2nd March
Transposition of formula
Objective:
To be able to accurately rearrange formula for a given subject.

21. Here are some questions and answers (by students A and B) on rearranging formulae.
Decide which answers to tick (correct) and which to trash (incorrect).
You must give reasons for your decision.
Question 2. Daniel buys n books at £4 each. He pays for them with a £20 note. He receives C pounds in change.
Write down a formula for C in terms of n.
Question 1.
Make x the subject of the following:
Y = x2 + 4 5
Student A answer
Y = x2 + 4 5
5y = x2 + 4
5y – 4 = x2
x2 = 5y – 4
x = 5y - 4 x
Student B answer
Y = x2 + 4 5
5y = x2 + 4
x2 + 4 = 5y
x2 = 5y – 4
x =
5y – 4
Trash
Trash
Books cost £4n
Change C = £4n - 20
Change = £20 – cost of books
C = 20 – 4n

22. Rearrangement of formulae
When doing these sort of problems, remember these things:
a) Whatever you do to one side of the formula, you must also do the same to the other side:
To rearrange the following formula making x the subject
Add y to both sides of the formula giving:
As (–y + y = 0) and (2y + y = 3y) we can say:
Now subtract x from both sides leaving:
So to get x we can now divide both sides by 2:

23. b) When you are dealing with more complicated formulae, try to strip off the outer layers first.
First get rid of the square root, by squaring both sides
Now get rid of the division bar, by multiplying both sides by x
To leave you with x on one side, divide both sides by g2
c) When you want to get rid of something in a formula, remember to do the opposite (inverse) to it.

24. One last example…
Make u the subject of the following:
Multiply both sides by (u + v)
Expand the bracket
Collect the u terms on one side
Factorise the LHS to isolate u
Divide both sides by (f – v)