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Mathematics Intermediate Tier

Mathematics Intermediate Tier. Algebra. GCSE Revision. Intermediate Tier - Algebra revision. Contents : Collecting like terms Multiplying terms together Indices Expanding single brackets Expanding double brackets Substitution Solving equations Finding nth term of a sequence

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Mathematics Intermediate Tier

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  1. Mathematics Intermediate Tier Algebra GCSE Revision

  2. Intermediate Tier - Algebra revision Contents : Collecting like terms Multiplying terms together Indices Expanding single brackets Expanding double brackets Substitution Solving equations Finding nth term of a sequence Simultaneous equations Inequalities Factorising – common factors Factorising – quadratics Other algebraic manipulations Solving quadratic equations Rearranging formulae Curved graphs Graphs of y = mx + c Graphing inequalities Graphing simultaneous equations Solving other equations graphically

  3. 4. 6. 1. 5. 2. 3. ab – 5a + 2b + b2 = x2 + 3x – 5x + 2 = – 9x2 + 5x – 4x2 – 9x = 4a2 – 7a + 6 + 4a = 4ab + 3ba – 6ba = a + a + a + a + b + b + a = Collecting like terms You can only add or subtract terms in an expression which have the same combination of letters in them e.g. 1. 3a + 4c + 5a + 2ac = 8a + 4c + 2ac e.g. 2. 3xy + 5yx – 2xy + yx – 3x2 = 7xy – 3x2 Simplify each of these expressions: 5a + 2b x2 – 2x + 2 cannot be simplified – 13x2 – 4x 4a2 – 3a + 6 ab

  4. 6. 11. 10. 9. 2. 8. 1. 7. 4. 12. 5. 3. 3s x 4s = (7a)2 = -6 x a = 4 x -3d = a x 3a = -5a x -6c = 4b2 x -9 = 5y x 7 = 6a x -a = -2t x -2s = -7f x 8a = (-9k)2 = Multiplying terms together Remember your negative numbers rules for multiplying and dividing Signs same  +ve answer Signs different  -ve answer e.g. 1. 2 x 4a = 8a e.g. 2. - 3b x 5c = - 15bc e.g. 3. (5p)2 = 5p x 5p = 25p2 Simplify each of these expressions: -6a -56af -12d -36b2 30ac 4st 12s2 35y 3a2 -6a2 49a2 81k2

  5. (F2)4 a4 Indices c0 a2 xa3 t2  t2 2e7 x3ef2 4xy3  2xy x7  x4 b1 5p5qrx6p2q6r

  6. x 8a + 12 5. 4. 8. 11. 10. 9. 2. 6. 3. 7. 1. 2(2a + 4) + 4(3a + 6) = a(a - 6) = c(d + 4) = 2p(3p + 2) - 5(2p - 1) = 4r(2r + 3) = 3(a - 4) = 6(2c + 5) = -5(2a - 3) = - (4a + 2) = 8 - 2(t + 5) = -2(d + g) = x Expanding single brackets Remember to multiply all the terms inside the bracket by the term immediately in front of the bracket e.g. 4(2a + 3) = If there is no term in front of the bracket, multiply by 1 or -1 Expand these brackets and simplify wherever possible: 3a - 12 8r2 + 12r 12c + 30 -4a - 2 -2d - 2g -2t - 2 cd + 4c 16a + 32 -10a + 15 a2 - 6a 6p2 - 6p + 5

  7. 3. 6. 5. 4. 1. 2. (4g – 1)2 = (c + 7)2 = (p + 2)(7p – 3) = (3a – 4)(5a + 7) = (2a + 1)(3a – 4) = (c + 2)(c + 6) = Expanding double brackets Split the double brackets into 2 single brackets and then expand each bracket and simplify (3a + 4)(2a – 5) “3a lots of 2a – 5 and 4 lots of 2a – 5” = 3a(2a – 5) + 4(2a – 5) = 6a2 – 15a + 8a – 20 If a single bracket is squared (a + 5)2 change it into double brackets (a + 5)(a + 5) = 6a2 – 7a – 20 Expand these brackets and simplify : c2 + 8c + 12 c2 + 14c + 49 6a2 – 5a – 4 16g2 – 8g + 1 15a2 + a – 28 7p2 + 11p – 6

  8. Substitution If a = 5 , b = 6 and c = 2 find the value of : c2 3a ac 4b2 144 15 4 10 (3a)2 ab – 2c c(b – a) 2 225 26 a2 –3b 4bc a (5b3 – ac)2 9.6 1144900 7 Now find the value of each of these expressions if a = - 8 , b = 3.7 and c = 2/3

  9. Try Calculation Comment x = 10 (10 x 10)+(3 x 10) = 130 Too low x = 12 (12 x 12)+(3 x 12) = 208 Too high etc. Solving equations Solve the following equation to find the value of x : 4x + 17 = 7x – 1  Take 4x from both sides 17 = 7x – 4x – 1 17 = 3x – 1  Add 1 to both sides 17 + 1 = 3x 18 = 3x  Divide both sides by 3 18 = x 3 Now solve these: 1. 2x + 5 = 17 2. 5 – x = 2 3. 3x + 7 = x + 15 4. 4(x + 3) = 20 6 = x x = 6 Find x to 1 d.p. if: x2 + 3x = 200 Some equations cannot be solved in this way and “Trial and Improvement” methods are required

  10. 4y Find the size of each angle 2y y 1500 Find the value of v 4v + 5 2v + 39 Solving equations from angle problems Rule involved: Angles in a quad = 3600 4y + 2y + y + 150 = 360 7y + 150 = 360 7y = 360 – 150 7y = 210 y = 210/7 y = 300 Angles are: 300,600,1200,1500 4v + 5 = 2v + 39 4v - 2v + 5 = 39 2v + 5 = 39 2v = 39 - 5 2v = 34 v = 34/2 v = 170 Rule involved: “Z” angles are equal Check: (4 x 17) + 5 = 73 , (2 x 17) + 39 = 73

  11. Position number (n) 1 2 3 4 5 6 Finding nth term of a sequence This sequence is the 2 times table shifted a little 2 4 6 8 10 12 5 , 7 , 9 , 11 , 13 , 15 ,…….…… Each term is found by the position number times 2 then add another 3. So the rule for the sequence is nth term = 2n + 3 100th term = 2 x 100 + 3 = 203 Find the rules of these sequences And these sequences 2n – 1 2n + 4 5n – 2 6n + 14 7n n2 n2 + 2 -2n + 22 -3n + 43 20n - 14 • 1, 3, 5, 7, 9,… • 6, 8, 10, 12,……. • 3, 8, 13, 18,…… • 20,26,32,38,……… • 7, 14, 21,28,…… • 1, 4, 9, 16, 25,… • 3, 6,11,18,27……. • 20, 18, 16, 14,… • 40,37,34,31,……… • 6, 26,46,66,……

  12. 1 Multiply the equations up until the second unknowns have the same sized number in front of them x 2 Eliminate the second unknown by combining the 2 equations using either SSS or SDA 2 x 3 3 Find the second unknown by substituting back into one of the equations Put a = 2 into: 4a + 3b = 17 Now solve: 5p + 4q = 24 2p + 5q = 13 Simultaneous equations 4a + 3b = 17 6a  2b = 6 8a + 6b = 34 18a  6b = 18 + 26a = 52 a = 52 26 a = 2 8 + 3b = 17 3b = 17 - 8 So the solutions are: a = 2 and b = 3 3b = 9 b = 3

  13. Inequalities can be solved in exactly the same way as equations -6  2x + 2 < 10 5x – 3  12 2. 4. 3. 1. 4x + 7 < x + 13 3x + 1 > 4 8 9 10 11 12 13 14 X > 1 or X = 2, 3, 4, 5, 6 …… X  3 or X = 3, 2, 1, 0, -1 …… X < 2 or X = 1, 0, -1, -2, …… -2  X < 4 or X = -2, -1, 0, 1, 2, 3 Inequalities 14  2x – 8 • Add 8 to both sides The difference is that inequalities can be given as a range of results Or on a scale: 14 + 8  2x 22  2x • Divide both sides by 2 22 x 2 Here x can be equal to : 11, 12, 13, 14, 15, …… 11  x • Remember to turn the sign round as well x  11 Find the range of solutions for these inequalities :

  14. Factorising Expanding Factorising – common factors Factorising is basically the reverse of expanding brackets. Instead of removing brackets you are putting them in and placing all the common factors in front. 5x2 + 10xy = 5x(x + 2y) Factorise the following (and check by expanding): 3(5 – x) 2(a + 5) a(b – 5) a(a + 6) 4x(2x – 1) • 15 – 3x = • 2a + 10 = • ab – 5a = • a2 + 6a = • 8x2 – 4x = • 10pq + 2p = • 20xy – 16x = • 24ab + 16a2 = • r2+ 2 r = • 3a2 – 9a3 = 2p(5q + 1) 4x(5y - 4) 8a(3b + 2a) r(r + 2) 3a2(1 – 3a)

  15. Factorising Expanding x x x2 -22 Factorising – quadratics Here the factorising is the reverse of expanding double brackets Factorise x2 – 9x - 22 x2 + 4x – 21 = (x + 7)(x – 3) To help use a 2 x 2 box x • Factor • pairs • of - 22: • 1, 22 • 22, 1 • 2, 11 • 11, 2 x x2 - 22 Factorise the following: (x + 3)(x + 1) (x – 2)(x – 1) (x + 10)(x – 3) (x + 2)(x – 6) (x + 2)(x + 5) • x2 + 4x + 3 = • x2 - 3x + 2 = • x2 + 7x - 30 = • x2 - 4x - 12 = • x2 + 7x + 10 = Find the pair which give - 9 -11 -11x 2 2x Answer = (x + 2)(x – 11)

  16. Other algebraic manipulations Cancelling common factors in expressions and equations Factorising a difference of two squares These two expressions are equal: 6v2 – 9x + 27z2v2 – 3x + 9z 3 Fully factorise this expression: 4x2 – 25 Look for 2 square numbers separated by a minus. Simply Use the square root of each and a “+” and a “–” to get: (2x + 5)(2x – 5) As are these: 2(x + 3)22(x + 3) (x + 3) The first equation can be reduced to the second: 4x2 – 8x + 16 = 0 x2 – 2x + 4 = 0 • Fully factorise these: • 81x2 – 1 • ¼ – t2 • 16y2 + 64 • Simplify these: • 5x2 + 15x – 10 = 0 • 4(x + 1)(x – 2) • x + 1 • Answers: • (9x + 1)(9x – 1) • (½ + t)(½ – t) • 16(y2 + 4) Answers: (a) x2 + 3x – 2 = 0 (b) 4(x – 2)

  17. x = - 7 or x = 2 Solving quadratic equations (using factorisation) Solve this equation: x2 + 5x – 14 = 0  Factorise first (x + 7)(x – 2) = 0 • Now make each bracket equal to zero separately x + 7 = 0 or x – 2 = 0  2 solutions Solve these: (x + 2)(x + 2) (x – 5)(x – 2) (x + 7)(x + 5) (x + 1)(x – 6) (x + 3)(x – 2) x = -2orx = -2  x = 5orx = 2  x = -7orx = -5  x = -1orx = 6  x = -3orx = 2 • x2 + 4x + 4 = • x2 - 7x + 10 = • x2 + 12x + 35 = • x2 - 5x - 6 = • x2 + x - 6 =

  18. Rearrange the following formula so that a is the subject 1. P = 4a+ 5 2. t -u 3. D = g2 + c xt +u 4. B = e + h 5. A = be r E = u - 4v d a = V - u t 6. Q = 4cp - st 3. g = D – c Rearranging formulae Now rearrange these V = u + at a V a V Answers: 1. a = P – 5 4 4. h = (B – e)2 2. e = Ar b 5. u = d(E + 4v) 6. p = Q + st 4c

  19. y y y y = 5/x y = 1/x x x x y = x3 + 2 y = x2 y = x3 y = x2 3 Curved graphs There are three specific types of curved graphs that you may be asked to recognise. Any curve with a number /x is this shape Any curve starting with x2 is “U” shaped If you are asked to draw an accurate curved graph (e.g. y = x2 + 3x – 1) simply substitute x values to find y values and thus the co-ordinates Any curve starting with x3 is this shape

  20. y Y = 3x + 4 c 3 m 4 1 x Graphs of y = mx + c In the equation: y = mx + c m = the gradient (how far up for every one along) c = the intercept (where the line crosses the y axis)

  21. y x Graphs of y = mx + c Write down the equations of these lines: Answers: y = x y = x + 2 y = - x + 1 y = - 2x + 2 y = 3x + 1 y = 4 y = - 3

  22. x = -2 y y = x y > 3 y = 3 x  - 2 x y < x Graphing inequalities Find the region that is not covered by these 3 regions x  - 2 y  x y > 3

  23. Finding co-ordinates for 2y + 6x = 12 using the “cover up” method: y = 0 2y + 6x = 12  x = 2 (2, 0) x = 0 2y + 6x = 12  y = 6 (0, 6) y 8 6 4 x 2 -4 -3 -2 -1 1 2 3 4 -2 -4 -6 -8 Graphing simultaneous equations Solve these simultaneous equations using a graphical method : 2y + 6x = 12 y = 2x + 1 2y + 6x = 12 y = 2x + 1 Finding co-ordinates for y = 2x + 1 x = 0 y = (2x0) + 1  y = 1 (0, 1) x = 1 y = (2x1) + 1  y = 3 (1, 3) x = 2 y = (2x2) + 1  y = 5 (2, 5) The co-ordinate of the point where the two graphs cross is (1, 3). Therefore, the solutions to the simultaneous equations are: x = 1 and y = 3

  24. y y = x2 + x – 6 x -3 2 Solving other equations graphically If an equation equals 0 then its solutions lie at the points where the graph of the equation crosses the x-axis. e.g. Solve this equation graphically: x2 + x – 6 = 0 All you do is plot the equation y = x2 + x – 6 and find where it crosses the x-axis (the line y=0) There are two solutions to x2 + x – 6 = 0 x = - 3 and x =2 Similarly to solve a cubic equation (e.g. x3 + 2x2 – 4 = 0) find where the curve y = x3 + 2x2 – 4 crosses the x-axis. These points are the solutions.

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