Intermediate Tier - Algebra revision

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Grant Dennis
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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 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 3x 2 = 7xy 3x 2 Simplify each of these expressions: 1. a + a + a + a + b + b + a = 5a + 2b 2. x 2 + 3x 5x + 2 = x 2 2x ab 5a + 2b + b 2 = cannot be simplified 4. 9x 2 + 5x 4x 2 9x = 13x 2 4x 5. 4a 2 7a a = 6. 4ab + 3ba 6ba = 4a 2 3a + 6 ab

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

5 Indices (F 2 ) 4 a 2 x a 3 c 0 a 4 2e 7 x 3ef 2 t 2 t 2 x 7 x 4 4xy 3 2xy b 1 5p 5 qr x 6p 2 q 6 r

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

7 Expanding double brackets Split the double brackets into 2 single brackets and then expand each bracket and simplify (3a + 4)(2a 5) = 3a(2a 5) + 4(2a 5) = 6a 2 15a + 8a 20 = 6a 2 7a 20 3a lots of 2a 5 and 4 lots of 2a 5 If a single bracket is squared (a + 5) 2 change it into double brackets (a + 5)(a + 5) Expand these brackets and simplify : 1. (c + 2)(c + 6) = c 2 + 8c (c + 7) 2 = c c (2a + 1)(3a 4) = 6a 2 5a 4 6. (4g 1) 2 = 16g 2 8g (3a 4)(5a + 7) = 15a 2 + a (p + 2)(7p 3) = 7p p 6

8 Substitution If a = 5, b = 6 and c = 2 find the value of : 3a c 2 4b 2 ac ab 2c c(b a) a 2 3b 4bc a (3a) (5b 3 ac) Now find the value of each of these expressions if a = - 8, b = 3.7 and c = 2 / 3

9 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 = 3x 18 = 3x 18 = x 3 6 = x x = 6 Divide both sides by 3 Now solve these: 1. 2x + 5 = x = x + 7 = x (x + 3) = 20 Some equations cannot be solved in this way and Trial and Improvement methods are required Find x to 1 d.p. if: x 2 + 3x = 200 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.

10 Solving equations from angle problems Find the size of each angle 2y 4y y Rule involved: Angles in a quad = y + 2y + y = 360 7y = 360 7y = y = 210 y = 210/7 y = 30 0 Angles are: 30 0,60 0,120 0, v + 5 Find the value of v 4v + 5 = 2v v - 2v + 5 = 39 2v + 39 Rule involved: Z angles are equal 2v + 5 = 39 2v = v = 34 v = 34/2 v = 17 0 Check: (4 x 17) + 5 = 73, (2 x 17) + 39 = 73

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

12 Simultaneous equations 1 Multiply the equations up until the second unknowns have the same sized number in front of them x 2 4a + 3b = 17 6a 2b = 6 x 3 2 Eliminate the second unknown by combining the 2 equations using either SSS or SDA 8a + 6b = 34 18a 6b = 18 26a = 52 a = a = 2 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 = b = 17 3b = b = 9 b = 3 + So the solutions are: a = 2 and b = 3

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

14 Factorising common factors Factorising 5x xy = 5x(x + 2y) Expanding 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. Factorise the following (and check by expanding): 15 3x = 2a + 10 = ab 5a = a 2 + 6a = 8x 2 4x = 3(5 x) 2(a + 5) a(b 5) a(a + 6) 4x(2x 1) 10pq + 2p = 20xy 16x = 24ab + 16a 2 = r r = 3a 2 9a 3 = 2p(5q + 1) 4x(5y - 4) 8a(3b + 2a) r(r + 2) 3a 2 (1 3a)

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

16 Other algebraic manipulations Factorising a difference of two squares Fully factorise this expression: 4x 2 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) Fully factorise these: (a) 81x 2 1 (b) ¼ t 2 (c) 16y Answers: (a) (9x + 1)(9x 1) (b) (½ + t)(½ t) (c) 16(y 2 + 4) Cancelling common factors in expressions and equations These two expressions are equal: 6v 2 9x + 27z 2v 2 3x + 9z 3 As are these: 2(x + 3) 2 2(x + 3) (x + 3) The first equation can be reduced to the second: 4x 2 8x + 16 = 0 x 2 2x + 4 = 0 Simplify these: (a) 5x x 10 = 0 (b) 4(x + 1)(x 2) x + 1 Answers: (a) x 2 + 3x 2 = 0 (b) 4(x 2)

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

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

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

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

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

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

23 Graphing simultaneous equations 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) 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 Solve these simultaneous equations using a graphical method : 2y + 6x = 12 y = 2x + 1 y 2y + 6x = 12 y = 2x x

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

Higher Tier - Algebra revision

Higher Tier - Algebra revision Contents: Indices Epanding single brackets Epanding double brackets Substitution Solving equations Solving equations from angle probs Finding nth term of a sequence Simultaneous