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WELCOME Mathematics Department. PERIOD STARTER 1)Calculate a)123 + 39b) 236 + 149 c) 321 - 56d) 453 - 164.

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Presentation on theme: "WELCOME Mathematics Department. PERIOD STARTER 1)Calculate a)123 + 39b) 236 + 149 c) 321 - 56d) 453 - 164."— Presentation transcript:

1 WELCOME Mathematics Department

2 PERIOD STARTER 1)Calculate a)123 + 39b) 236 + 149 c) 321 - 56d) 453 - 164

3 PERCENTAGES To remember 50% = ½25% = ¼75% = ¾ 10% = 20% =33 % = 5% =

4 Multiples of 10 Divide by 10 Divide by 5? 2 lots of 10% c) You try……..30% of 32 a)10% of 80 = 8 b)20% of 34 10% of 34 = 3·4 20% of 34 2  3·4 = 6·8

5 30% of 32 10% of 32 = 3·2 30% of 32 3·2×3 = 9·6

6 Fractions Divide by 4? b) You try…… 20% of 155 a)25% of 84 ¼ of 84 = 21

7 20% 0f 155 1/5 of 155 = 31

8 Tricky 5% is half of 10% 1) 15% of 260 10% of 260 = 26 5 % of 260 = 13 15% = 10% + 5% = 26 + 13 = 39

9 2a) 17·5% of 480 So, 10% of 480 = 48 5% of 480 = 24 2·5% of 480 = 12 Therefore: 17·5% = 10% + 5% + 2·5% = 48 + 24 + 12 = 84 b) You try 35% of 86

10 3. 35% of 120 10% of 120 = 12 30% of 120 3×12 = 36 5% of 120 = 6 36 + 6 = 42

11 ALGEBRA + 4 x 20  5 x 15 What is x?......

12 Traditional Methods ‘Change side, Change sign’ Works for equations such as: X+7=18 However, they can cause confusion in equations like -5x = 25 Does multiply by -5 become divide by +5?

13 ‘Cover up’ method Works for equations like: 2x+1=11 As soon as the unknown appears on both sides of the equation, this method becomes unsuitable. 4x+1=x+4

14 x x x x x Keep the balance at all times!! The Golden Rule

15 Solving Equations a)X + 3 = 5  3 from both sides Balance! How would you get x on its own? X = 2

16 b) w  2 = 9 Add 2 to both sides Subtract 4 from both sides c) You try r + 4 = 6 w = 11 r = 2

17 X on its own…. Balance! Divide both sides by 3 Divide both sides by 4 Divide both Sides by 5 a) 3x = 9 x = 3 b) 4y = 16 y = 4 You try.. c) 5h = 25 h = 5

18 Subtract 1 from both sides Divide both sides by 2 Add 1 to both sides Divide both sides by 4 a) 2x + 1 = 9 2x = 8 x = 4 b) 4t  1 = 15 4t = 16 t = 4 c) You try…… 3g + 2 = 11

19 3g = 9 g = 3 Subtract 2 from both sides Divide both sides by 3

20 Tricky Algebra….. Variables on both sides Add/subtract smallest variable Subtract p from both sides Add 6 to both sides Subtract 2g from both sides Add 4 to both sides a) 2p  6 = p p  6 = 0 p = 6 b) 3g  4 = 2g g  4 = 0 g = 4 c) You try…… 2m  1 = m

21 By adding or subtracting the smallest letter first, we avoid being left with negative letters y+3=2y

22 Subtract m from both sides Add 1 to both sides Subtract 3r from both sides Add 10 to both sides Divide both sides by 2 2m  1 = m m  1 = 0 m = 1 You try…. 5r  10 = 3r 2r  10 = 0 2r = 10 r = 5

23 LETTERS AND NUMBERS ON BOTH SIDES Subtract x from both sides Subtract 5 from both sides Subtract x from both sides Add 1 to both sides Divide both sides by 2 Look for smallest variable a)2x + 5 = x + 10 x + 5 = 10 x = 5 b)3x  1 = x + 9 2x  1 = 9 2x = 10 x = 5

24 Pupils are encouraged to deal with the letters first Get all the letters on one side Then work towards obtaining the value of one letter

25 Subtract 2p from both sides Subtract 2 from both sides Divide both sides by 3 Subtract 2m from both sides Add one to both sides Divide both sides by 2 c) 5p + 2 = 2p + 8 3p + 2 = 8 3p = 6 p = 2 d) You try…… 4m  1 = 2m + 7 2m  1 = 7 2m = 8 m = 4

26 For further information, please visit the school website’s Maths Department http://www.ea.e- renfrew.sch.uk/stninians Departments, Maths, About us, Common Methodology


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