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Developing Algebra Skills

Developing Algebra Skills. Session Four – 12 March 2014. A Quick Recap …. Multiply out the brackets in (x + 5)(x – 3) Grid = x 2 + 5 x - 3 x – 15 = x 2 + 2 x – 15. x 2. +5 x. -3 x. -15. A Quick Recap …. Multiply out the brackets in (x + 5)(x – 3) FOIL ( x + 5)( x – 3)

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Developing Algebra Skills

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  1. Developing Algebra Skills Session Four – 12 March 2014

  2. A Quick Recap … • Multiply out the brackets in (x + 5)(x – 3) • Grid = x2 + 5x - 3x – 15 = x2 + 2x – 15 x2 +5x -3x -15

  3. A Quick Recap … • 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

  4. A Quick Recap … • Using either method calculate (b + 3)(b - 6) = b2 + 3b - 6b -18 = b2 - 3b - 18 b2 +3b - 6b -18

  5. Multiplying Expressions • 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

  6. A Quick Recap … • 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

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

  8. Solving Quadratics by Factorisation • 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

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

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

  11. Simultaneous Linear Equations • 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

  12. Simultaneous Linear Equations 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

  13. Simultaneous Linear 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

  14. Simultaneous Linear Equations • 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

  15. Simultaneous Linear Equations • 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

  16. Simultaneous Linear Equations • Try Exercise …

  17. Formulae • 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

  18. Formulae • 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

  19. Formulae • 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

  20. Formulae • 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°

  21. Formulae • 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

  22. Transposition of Formulae • 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

  23. Transposition of Formulae • 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

  24. Transposition of Formulae • 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

  25. Transposition of Formulae • 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

  26. Transposition of Formulae • Try Exercise …

  27. Trial and Improvement • 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

  28. Trial and Improvement • 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)

  29. Trial and Improvement • Try Exercise …

  30. Session Summary • Simultaneous Equations • Transposition of Formulae • Trial and Improvement

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