Using and applying mathematics Sequences & Formulae Year 10

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

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

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

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

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

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

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

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

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

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 = 22.5mWidth = 7.5m

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).

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 + 4 2x - 7 If the area of a similar rectangle is 63cm 2 show that 2x 2 + x – 91 = 0 and calculate the value of x X = 7 X = 6.5

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

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 = 22.5mWidth = 7.5m a = 3 (a – 15)

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 x x = 306 x x – 306 = 0 How can we solve this? Factorisation? Formula Graphically

x x – 306 = 0 This example factorises: ( x + 34 )( x – 9 ) = 0 Either x + 34 = 0Or, x – 9 = 0 x = -34x = 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

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 cm 2. What is the length of its base to 1 d.p? 41 x x +3 x ( x + 3) = 41 2 x ( x + 3) = 41 x 2 = 82 x 2 + 3x – 82 = 0 Try x = (3 x 7) – 82 = -12 Too small Try x = (3 x 8) – 82 = 6 Too big Try x = ( 3 x 7.6) – 82 = Too small Try x = ( 3 x 7.7) – 82 = 0.39 Too big Try x = ( 3 x 7.65) – 82 = Too small x = 7.7 to 1 d.p Base = 7.7cm to 1 d.p

a) 5x 2 – 12x + 5 = 0For x > 1 b) x 2 – 5x – 1 = 0For x > 0 c) 2x 2 – 2x – 3 = 0For x > 0 d) 5x 2 + 9x – 6 = 0For x > 1 To 1 d.p a)1.9 b)5.2 c)1.8 d)0.5

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

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 1. Make x the subject of the following: Y = x Student A answer Y = x y = x y – 4 = x 2 x 2 = 5y – 4 x = 5y - 4 x Student B answer Y = x y = x x = 5y x 2 = 5y – 4 x = 5y – 4 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. Books cost £4n Change C = £4n - 20 Change = £20 – cost of books C = 20 – 4n

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:

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 g 2 c) When you want to get rid of something in a formula, remember to do the opposite (inverse) to it.

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)