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1.1 Algebra

1.1 Algebra. Note 1 : Adding and Subtracting Terms. Algebraic terms have numbers and variables that are sometimes raised to an exponent. Exponent is 3. To add/subtract terms they must have the same variable and the same exponent. 8 x 3. Coefficient is 8. Variable is x.

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1.1 Algebra

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  1. 1.1 Algebra

  2. Note 1: Adding and Subtracting Terms Algebraic terms have numbers and variables that are sometimes raised to an exponent Exponent is 3 To add/subtract terms they must have the same variable and the same exponent 8x3 Coefficient is 8 Variable is x

  3. Note 1: Adding and Subtracting Terms Terms with the same variable and same exponent are called Like Terms Exponent is 3 To add/subtract terms they must have the same variable and the same exponent 8x3 Coefficient is 8 Variable is x

  4. Note 1: Adding and Subtracting Terms Examples of LIKE TERMS: 2x, 9x, 0.74x, ¾ x, 300x 2x2y, -6x2y, 15x2y, 0.9x2y 3b, -2b, ½ b, 100b

  5. Note 1: Adding and Subtracting Terms Variables that have different exponents are not the same variable therefore variables can only be added if the variables are the same. e.g. 3r + r2 = 3r + r2 e.g. 4r + r2 + r + 2r2 = 5r + 3r2 e.g. 6b + 3r2 + 4r – b + 6b2 = 6b – b + 6b2 + 3r2 + 4r = 5b + 6b2 + 3r2 + 4r Start by identifying like terms GAMMA Ex 1.03 Pg 7

  6. Note 2: Exponent Rules To multiply algebraic terms, we multiply the coefficients together and add the exponents. am x an = am+n e.g. 3  4r = 3  4  r = 12r e.g. b2 b2 = b4 e.g. 2x2y  4y2 = 8x2y3 Remember : x is the same as 1x y is the same as y1

  7. Note 2: Exponent Rules 1.) Index the number. 2.) Multiply each variable index by the index outside the brackets. 3.) If the bracket can be simplified, do this first. e.g.Simplify (2x2)3 (-4h2g6)2 = 23 x23 = 16h4g12 = (4x)2 = 8x6 = 16x2

  8. Note 2: Exponents Rules  When simplifying expressions with square roots: 1.) Square root the number. 2.) Half the power of each variable. GAMMA Ex 1.02 Pg 4-5 Ex 1.04 Pg 10 e.g.Simplify = 5 k122 = 9 k62 = 25 k22 = 5 k6 = 25 k = 9 k3

  9. Note 3: Algebraic Fractions When dividing algebraic terms: Simplify the numbers by using the fraction button on your calculator. Subtract the powers for each of the different base variables. e.g. = 2 e.g. = 2a2 e.g. = am = am-n an e.g. = = Remember : x is the same as x1

  10. Note 3: Algebraic Fractions Multiply these fractions and write in simplest form e.g. x Multiply the numerators and denominators = am = am-n an e.g. x = = Remember : x is the same as x1

  11. Note 3: Algebraic Fractions Divide these fractions and write in simplest form e.g. ÷ To divide; we multiply by the reciprocal am = am-n x = e.g. ÷ an = = x GAMMA Ex 7.01 Pg 90 Ex 7.02 Pg 91 Ex 7.03 Pg 92 = = = Remember : x is the same as x1

  12. Starter + − + = + = − = + = = − = =

  13. Note 4: Expanding Brackets Multiply the number on the outside by each of the numbers/variables on the inside of the brackets e.g.Expand 7(2x – 3) x(5x – 2) 5(2a + 3) + 3(a – 4) = 7  2x + 7  -3 = x  5x + x  -2 = 10a + 15 + 3a – 12 = 5x2 –2x = 10a + 3a + 15 – 12 = 14x - 21 = 13a + 3

  14. Note 4: Expanding Two Brackets QUADRATIC EXPANSION When we expand two brackets we use: F – first (multiply the first variable or number from each bracket) O – outside (multiply the outside variables together) I – inside (multiply the two inside variables together) L – last (multiply the last variable in each bracket together) Simplify, leaving your answer with the highest power first to the lowest power (or number) last. F O I L e.g. (x + 4) (x – 2) = x2 - 2x + 4x - 8 = x2 + 2x - 8

  15. Note 4: Expanding Two Brackets e.g. (x + 10) (x + 1) QUADRATIC EXPANSION e.g. (x + 3) (x – 5) = x2 + 10x + x + 10 = x2 + 3x - 5x - 15 = x2 + 11x + 10 = x2 - 2x - 15 e.g. (x - 4) (x + 4) e.g. (x - 3) (x – 8) = x2 - 4x + 4x - 16 = x2 - 16 = x2 - 3x - 8x + 24 = x2 - 11x + 24 • Notice the middle term cancels out • DIFFERENCE OF SQUARES

  16. Note 4: Expanding Two Brackets e.g. (x – 5) (x + 4) QUADRATIC EXPANSION e.g. (x + 7) (x – 9) = x2 - 5x + 4x - 20 = x2 + 7x - 9x - 63 = x2 - x - 20 = x2 - 2x - 63 e.g. (x - 9) (x + 9) e.g. (x - 2) (x – 6) = x2 - 9x + 9x - 81 = x2 - 2x - 6x + 12 = x2 - 81 = x2 - 8x + 12 • Notice the middle term cancels out • DIFFERENCE OF SQUARES

  17. Note 4: Expanding Two Brackets e.g. (x + 5)2 QUADRATIC EXPANSION e.g. (2x + 3) (x – 5) (x + 5) (x + 5) = 2x2 + 3x - 10x - 15 = 2x2 - 7x - 15 = x2 + 5x + 5x + 25 = x2 + 10x + 25 e.g. 5(3x + 4) (2x – 1) GAMMA (odd) Ex 2.01 Pg 12-13 Ex 2.02 Pg 14 Ex 2.03-2.04 Pg 15 = 5[6x2 + 8x - 3x - 4] = 5[6x2 + 5x - 4] NuLake Pg 14,23-25 = 30x2 + 25x - 20

  18. Note 5: Solving Linear Equations To solve simple equations, do the opposite operation to what is happening to x e.g. Solve x + 5 = 7 5x = 42 x - 7 = 7 x = 7 - 5 x = 7 + 7 x = x = 2 x = 14 = -3 x = -3 x 2 x = -6

  19. Note 5: Solving Linear Equations To solve 2 stage simple equations, do the opposite operation to what is happening to x (aim to get x on the LHS) e.g. Solve = 7 6x + 5 = 17 - 3 = 9 6x = 17 - 5 x + 3 = 7 x 2 = 9 + 3 6x = 12 x + 3 = 14 x = 2 x = 14 - 3 = 12 x = 11 x = 12 x 5 x = 60 Check that your answer works in the ORIGINAL equation

  20. Note 5: Solving Linear Equations Solve equations with like terms. Collect x terms on the LHS and collect numbers on the RHS e.g. Solve 6x + 4 = 4x - 6 5x − 1 = 7x + 9 25 – 9x – 3 + 5x = 7x – 23 -2x 6x – 4x = -6 - 4 5x – 7x = 9 + 1 -4x + 22 = 5x − 23 -2x = 10 2x = -10 -4x – 5x = -23 – 22 x = -5 x = -5 -9x = -45 x = 5 Check that your answer works in the ORIGINAL equation

  21. Note 5: Solving Linear Equations When equations have brackets, expand the brackets first and then solve the equation e.g. Solve 6(x – 1) = 12 6x -2 + 2(x+5) = 0 8(7 – x) = 6(x + 2) 6x – 6 = 12 6x – 2 +2x + 10 = 0 56 – 8x = 6x + 12 8x + 8 = 0 6x = 18 -14x = -44 x = 3 8x = -8 x = x = -1 GAMMA - odd only Ex 4.01 Pg 44 Ex 4.02 Pg 45 Ex 4.03 Pg 46 Ex 4.04 Pg 46 x = Check that your answer works in the ORIGINAL equation

  22. Starter Simplify Solve for x + 4x + 2 – (x – 3) = 0 5x = 6x + 21 – 8x 4x + 2 – x + 3 = 0 5x = -2x + 21 = + 5x + 2x = 21 3x + 5 = 0 = 3x = -5 7x = 21 x = x = 3 Simplify…..Simplify……& Simplify!

  23. Note 6: Solving equations with fractions When solving equations involving fractions, multiply both sides of the equation by a suitable number to eliminate the fractions. * Multiply both sides by 14 = 14 14 * It is possible to skip this step by cross-multiplying = 7(x+3) = 2(2x-3) * Expand and simplify 7x+21 = 4x - 6 x = -9 3x = -27

  24. Solving Linear Equations w/ fractions e.g. Try these! Cross-Multiply! Solve = = 10 = 1 4(5x+2) = 2 x – 5 = 10 x 4 4 x 5y = 7(2y+3) 20x + 8 = 2 20y = 14y + 21 x – 5 = 40 20x = -6 20y – 14y = 21 x = 45 x = 6y = 21 y = x = Check that your answer works in the ORIGINAL equation

  25. Note 7: Solving Linear Inequations Inequations are expressions that include: < (less than), > (greater than) ≤ (less than or equal to), ≥ (greater than or equal to) To solve: we need to isolate the unknown variable just like we did for solving linear equations. One exception: if at any time we multiply or divide by a negative number, we must REVERSE the inequality sign e.g. Solve the inequality 3x + 8 ≥ 29 3x ≥ 29 - 8 3x ≥ 21 x ≥ 7

  26. Note 7: Solving Linear Inequations Inequations are expressions that include: < (less than), > (greater than) ≤ (less than or equal to), ≥ (greater than or equal to) To solve: we need to isolate the unknown variable just like we did for solving linear equations. One exception: if at any time we multiply or divide by a negative number, we must REVERSE the inequality sign e.g. Solve the inequality 2x + 10 ≥ 4x + 6 2x – 4x ≥ 6 - 10 -2x ≥ -4 x ≤ 2

  27. Note 7: Solving Linear Inequations Try These! e.g. 5x + 8 < -12 -3(w – 6) > 9 < 7 5x < -12 - 8 -3w + 18 > 9 2 – 3x < 7 x 2 -3w > 9 - 18 5x < -20 2 – 3x < 14 5 5 -3w > -9 -3x < 12 x < - 4 -3 -3 -3 -3 w < 3 x > -4 NuLake Pg 20-21

  28. Note 8: Substitution & Rearranging Formulae Substitution means replacing a variable with a number. e.g.Calculate the value of these expressions: 7x – 1 when x = 2 = 7 2 – 1 = 14 – 1 = 13 when f = 2 and g = 6 = = 4

  29. Note 8: Substitution & Rearranging Formulae 5x2 - 3x + 2 when x = -3 = 5×(-3)2 - 3 -3 + 2 = 45 - −9 + 2 = 45 + 9 + 2 = 56

  30. Note 8: Substitution & Rearranging Formulae Rearranging formulae and equations involves manipulating them to make a certain variable ‘the subject’ of the equation (all by itself). Just like how we do when we ‘solve for x’. e.g. Make d the subject v2 = u2 + 2ad ad + bd = 5 v2 – u2 = 2ad d (a + b) = 5 2a 2a (a + b) (a + b) d = 5 v2 – u2 = d (a + b) 2a

  31. Note 8: Substitution & Rearranging Formulae e.g. Make r the subject A = πr2 A = πr2 π π √ √ √ A = r2 π A = r ± GAMMA - odd only Ex 3.01 Pg 25-28 Ex 3.04 Pg 39 π NuLake Pg 26-28

  32. Starter - Expand e.g. (x – 5) (x + 9) QUADRATIC EXPANSION e.g. (x + 7) (x + 2) = x2 - 5x + 9x - 45 = x2 + 7x + 2x + 14 = x2 + 4x - 45 = x2 + 9x + 14 e.g. (x - 8) (x + 8) e.g. (x – 2) (x – 3) = x2 - 8x + 8x - 64 = x2 – 2x – 3x + 6 = x2 - 64 = x2 – 5x + 6 • Notice the middle term cancels out • DIFFERENCE OF SQUARES

  33. Note 9: Factorising Factorising is the reverse procedure of expanding. Expanding (x + 7) (x + 2) x2 + 9x + 14 Factorising

  34. Note 9: Factorising Look for common factors first. e.g. 15ab + 10b e.g. 4x2 + 16x + 8 = 5b(3a + 2) = 4(x2 + 4x + 2) e.g. 30a2 – 15a e.g. 3x + 6y – 9z = 15a(2a – 1) = 3(x + 2y – 3z)

  35. Note 9: Factorising If there is no common factor, try splitting the expression into two groups, and look for a common factor in each pair. e.g. 2ac + 2bc + 3ad + 3bd 3d(a + b) = 2c(a + b) + (a + b) is the common factor = (a + b) (2c + 3d) e.g. 3x + 6y + 4xz + 8yz 4z(x + 2y) = 3(x + 2y) + (x + 2y) is the common factor = (x + 2y)(3 + 4z)

  36. Note 9: Factorising GAMMA - odd Ex 2.06 Pg 18 Ex 2.07 Pg 19-20 e.g.Factorise simple quadratics: a.) x2 + 6x + 8 Find 2 numbers which multiply to 8 and add to 6 (x+2) (x+4) 4 and 2 Put these number into brackets The factors are (x+2) and (x+4) b.) x2 - 5x - 24 Find 2 numbers which multiply to -24 and add to -5 (x-8) (x+3) -8 and 3 Put these number into brackets The factors are (x-8) and (x+3) x2 + 5x - 24 c.) Find 2 numbers which multiply to -24 and add to 5 (x+8) (x-3) 8 and -3 Put these number into brackets The factors are (x+8) and (x-3)

  37. Note 10: Factorising – Difference of squares Expand (x + y)(x – y) = x2 – y2 Remember this result The middle term cancels out a.) a2-16b2 (a)2-(4b)2 (a-4b) (a+4b) The factors are (a-4b) and (a+4b) 36a3b - 4ab3 b.) There is a common factor of 4ab 4ab(9a2 - b2) 4ab[(3a)2 – b2] 4ab [(3a – b)(3a+b)] Check your solution by expanding

  38. Starter Factorise 4x2 – 49y2 9x2 – 36y2 144x2 – 9y8 ( ) ( ) 9( ) ( ) ( )( ) x - 2y 2x – 7y 2x + 7y x + 2y 12x – 3y4 12x + 3y4 x2 + 6x + 9 x2 – 12x + 36 ( ) ( ) ( ) ( ) x + 3 x + 3 x – 6 x – 6 ( )2 x + 3 ( )2 x – 6

  39. Note 10: Factorising – Difference of Squares Evaluate GAMMA - odd Ex 2.07 Pg 19-20 Ex 2.08 Pg 21 Ex 2.09 Pg 21 Factorise a.) 812 – 802 = (81-80) (81+80) = (1) (161) = 161 12342-12352 b.) = (1234-1235) (1234+1235) = (-1) (2469) = -2469 Check your solution

  40. Factorisation of quadratic expressions with a ≠ 1 Mulitply the coefficient of x2 and the constant. 3 x 8 Find 2 numbers that multiply to give this value and add to give the coefficient of x Write the quadratic with the x-term split into two x-terms using these numbers Factorise the pairs of terms Factorise again, taking the bracket as the common factor Check by expanding

  41. Factorisation of quadratic expressions with a ≠ 1 Mulitply the coefficient of x2 and the constant. 1x 6 Find 2 numbers that multiply to give this value and add to give the coefficient of x Write the quadratic with the x-term split into two x-terms using these numbers Factorise the pairs of terms Factorise again, taking the bracket as the common factor Check by expanding

  42. Factorisation of quadratic expressions with a ≠ 1 Mulitply the coefficient of x2 and the constant. 5 x -8 Find 2 numbers that multiply to give this value and add to give the coefficient of x Write the quadratic with the x-term split into two x-terms using these numbers Factorise the pairs of terms Factorise again, taking the bracket as the common factor Check by expanding

  43. Factorisation of quadratic expressions with a ≠ 1 GAMMA - odd Ex2.09 Pg21 Ex2.10 Pg22 Ex2.11 Pg23 4x 4 Check by expanding

  44. Note 11: Quadratic Equations Quadratic equations always have an x2 term and generally have 2 different solutions If a x b = 0, this means that eithera or b (or both) must be zero. e.g. Solve the equation x2 + x – 12 = 0 (x-3)(x+4) = 0 factorise first! either x - 3 = 0 or x + 4 = 0 x = 3 x = -4 2 solutions

  45. Note 11: Quadratic Equations Quadratic equations always have an x2 term and generally have 2 different solutions 12 x -14 = -168 24, -7 e.g. Solve the equation 12x2 + 17x – 14 = 0 12x2 + 24x – 7x – 14 = 0 split the x term 12x (x+2) -7 (x+2) = 0 factorise pairs (12x-7) (x+2) = 0 either 12x - 7 = 0 or x + 2 = 0 x = -2 x =

  46. Note 11: Quadratic Equations SOLVE THE EQUATIONS e.g. x2 + 9x = 0 9x2 -16 = 0 e.g. x (x+9) = 0 factorise 9x2 = 16 either x = 0 or x + 9 = 0 x2 = x = -9 x = GAMMA - odd Ex5.01Pg65 Ex5.02Pg 66 Ex5.03 Pg68 x = ±

  47. Use the quadratic formula to solve 2x2 + 7x + 5 = 0 5 a = 2 b = 7 c = x = -7 ± √9 4 7 72 (2)(5) x = -7 ± 3 (2) 4 IGCSE Ex 28 Pg 73 odd Ex 30 Pg 76 odd x = -5 Two Solutions or x = -1 2

  48. Starter A wallet containing $40 has three times as many $1 notes as $5 notes. Find the number of each kind. Let x be the number of $1 notes and y be the number of $5 notes (1) 1x + 5y = 40 x + 5y = 40 (2) 3y = x x – 3y = 0 8y = 40 (1) – (2) y = 5 3(5) = x There are 15 $1 notes and 5 $5 notes x = 15

  49. Note 12: Simultaneous Equations In order to solve for the value of two unknowns in a problem, you must have two different equations that relate to the unknowns Substitution Substitution is often used when one of the equations contains a single unit quantity of an unknown. e.g. 2x – 3y = 34 (1) Label the equations 5x + y = 0 (2) y = -5x Rearrange for the single unit quantity (y here) 2x – 3(-5x) = 34 Substitute this expression for y into (1) & solve for x 17x = 34 x = 2 Solve for y by substituting x = 2 into (2) 5(2) + y = 0 y = -10

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