1. Algebra External Achievement Standard (4 Credits)

2. Contents Linear Equations Indices and Surds Rational Expressions Log & Exponential Equations Quadratic Equations Simultaneous Equations Log Laws Expanding Discriminant Complete the SquareRearranging Equations Factorising Inequations Rational Equations Quadratic Formula

3. Expanding Brackets  Hasn’t changed from Year 11 but for an extra challenge there can be up to 3 brackets (a cubic). Eg)(x + 4)(x – 6)(3x – 1)Just pick 2 brackets to start with (x 2 – 2x - 24)(3x – 1)Add in the 3 rd bracket Expand and simplify that pair Simplify like terms 3x 3 – 7x 2 – 70x +24 3x 3 – 6x 2 – 72x– x 2 + 2x+ 24

4. Expanding Brackets  Key points:  Do a common factor last Ie)3(x + 4)(x – 5)=  Remember to write out square brackets Ie)(x + 2) 2 = (x + 2)(x + 2)  Same applies in a cubic Ie)(2x + 5)(x – 2) 2 = (2x + 5)(x – 2)(x – 2) 3(x 2 + 4x – 5x – 20) 3(x 2 – 1x – 20) 3x 2 – 3x – 60

5. Expanding Brackets  Practice Theta –Square Bracket Expansions Difference of 2 Squares  Page 3 Ex 1.3  Page 3 Ex 1.4 –Expand and simplify  Page 4 Ex 1.5 –3 Bracket expansions  Page 5 Ex 1.6 –Homework  Page 1 and 2 ex 1.01 to 1.06

6. Factorising  This is the process of putting an expression into brackets.  Three Key Checks 1)Common factor 2)Co-efficient = 1 3)Co-efficient ≠ 1

7. Common Factor  Simplest of the checks but often overlooked, ask yourself; Is there a common factor? 1)5x 2 + 10x2)4x 2 – 15 3)3x 2 – 27x + 604)x 3 – 2x 2 + x

8. Common Factor  Practice –Page 20 ex 3.1  particularly 23 onwards –Page 21 ex 3.2  2 Step factorising

9. Coefficient = 1  This is what you are most used to from year 11. Factorising into 2 brackets and the number in front of the x 2 = 1 1)x 2 + 13x + 422)x 2 – 25 3)3x 2 – 27x + 60

10. Coefficient = 1  Practice –Page 22 ex 3.3 –Page 23 ex 3.4  Must take out a common factor first –Homework P13, 3.01 to 3.04

11. Coefficient ≠ 1  This is when the number in front of x 2 is not a common factor and is bigger than 1 Two different approaches, that give the same result: Guess and Check Cambridge Method

12. Coefficient ≠ 1 Eg)5x 2 + 2x – 7 Guess and Check Method Factors of 5x 2 factors of -7 5x & x-7 & 1 Cambridge Method 5 x -7 = -35 Factors of -35 that add to +2 +7, -5 works so 5x 2 + 7x – 5x – 7 Take a common factor out of the first pair, then the second pair x(5x + 7) – 1(5x + 7) (x – 1)(5x + 7) 5x -7 x 1 -35x + 1x = -34x -7x + 5x = -2x So correct combination is (5x + 7)(x – 1) Right number wrong sign so use 7 and -1 Pick and Master one method, there is no better one.

13. Coefficient ≠ 1 Eg)3x 2 – 7x + 2 Guess and Check Method Factors of 3x 2 factors of +2 3x & x2 & 1 Cambridge Method 3 x 2 = 6 Factors of 6 that add to -7 -6, -1 works so 3x 2 – 6x -1x + 2 Take a common factor out of the first pair, then the second pair 3x(x – 2) – 1(x – 2) (3x – 1)(x – 2) 3x 2 x 1 6x + 1x = 7x So correct combination is (3x – 1)(x – 2) Right number wrong sign so use -2 and -1

14. Coefficient ≠ 1 Eg)2x 2 + 7x + 3 Guess and Check Method Factors of 2x 2 factors of 3 2x & x3 & 1 Cambridge Method 2 x 3 = 6 Factors of 6 that add to +7 6, 1 works so 2x 2 + 6x + 1x + 3 Take a common factor out of the first pair, then the second pair 2x(x + 3) + 1(x + 3) (2x + 1)(x + 3) 2x 3 x 1 6x + 1x = 7x So correct combination is (2x + 1)(x + 3)

15. Coefficient ≠ 1  Practice –Theta Page 23 and 24 ex 3.5 –Homework Page 14 ex 3.05, 3.06 & 3.07

16. Indices (Positive and negative exponents) Indices Laws ax m.bx n = abx m+n ax m.bx n = abx m+n x m = x m-n x m = x m-n x n x n (ax m ) n = a n x mn Where a, b, m and n are all numbers

17. Indices (Positive and negative exponents) A negative indices flips the fraction over = x -m = x -m1 = m = m1 = 1 = 1 xmxmxmxm Or negative indices above the vinculum (fraction line) are positive below and vice versa. x as a fraction is x/1 Flip the fraction, change the sign Expand the brackets and simplify x - m eg)x -m

18. Indices (Positive and negative exponents) Some Practice 1)(¾) -2 2)x 2.x -4 3)x 3 x -2 Page 72, ex 9.3

19. Surds (Fractional exponents) Surds a m/n = n √a m eg) 4 3/4 = 4 3 = 4 √64= 2 = 4 √64= 2 Surds behave in exactly the same way as exponents, we must convert them to fractional exponents first though. / 4 √

20. Surds (Fractional exponents) Simplify and convert to surd form 1)5 ¼ 2)( 4 √x 3 ) 3 3)√x 3 √x 5 Page 73 ex 9.4 Page 74 ex 9.5 Surds behave in exactly the same way as exponents, we must convert them to fractional exponents first though.

21. Rational Expressions  These are expressions involving algebraic fractions that we generally want to simplify.  The main ones are –Multiplication –Division –Addition, Subtraction –Simplifying (Cancelling Common Factors)

22. Multiplication  Multiply the numerators, multiply the denominators 1)5x ×2x 4 7 4 7 2)5x + 2 × 2x x 3 x 3 3) 2 × 4 x - 3 (x – 3) 2 x - 3 (x – 3) 2

23. Division  Take the reciprocal of the second fraction then multiply 1)5x ÷2x 4 7 4 7 2)5x + 2 ÷ 2x x 3 x 3 3) 2 ÷ 4 x - 3 (x – 3) 2 x - 3 (x – 3) 2

24. Addition and Subtraction  Key thing to remember is to get a common denominator 1)5x +2x 4 7 4 7 2)5x + 2 – 2x x 3 x 3 3) 2 – 4 x - 3 (x – 3) 2 x - 3 (x – 3) 2 Note It helps to look for the lowest common denominator rather than cross multiplying straight away

25. Simplifying Algebraic Fractions Eg:1) (x + 2) 3(x + 2) 3(x + 2) 2)x 2 + 5x – 14 (x - 2) (x - 2) Factorise then find a common factor and cancel it out. You can only cancel out things that are multiplied.

26. Simplifying Algebraic Fractions Eg:1) (x + 2) 3(x + 2) Already Factorised, Cancel out common factor = 1 3

27. Simplifying Algebraic Fractions Eg:2) x 2 + 5x – 14 (x - 2) (x - 2) Factorise Top line (x + 7)(x – 2) (x – 2) (x – 2) Cancel out common factor = (x + 7)

28. Simplifying Algebraic Fractions To sum up Step 1Factorise top and bottom Step 2Cancel out common factors Eg:1)x 2 + 5x – 14 (x - 2) (x - 2) (x + 7)(x – 2) = x + 7 (x – 2) (x – 2) 2)x 2 – 7x – 44 x 2 + 6x + 8 (x – 11)(x + 4) (x + 2)(x + 4) (x + 2)(x + 4) = (x – 11) (x + 2)

29. Rational expressions  To make the task of simplifying the expression later on easier it helps not to expand the bottom line.  Practice –Theta Page 16 ex 2.2, 2.3 Multiplication, Division –Theta Page 16 ex 2.4Mixed problems –Theta Page 18 ex 2.5 Addition, Subtraction –Theta Page 19 ex 2.6 –Homework Pages 9 to 12

30. Logs  Logs were designed by Scott, John Napier as a way of dealing with very difficult multiplication problems using addition.  It is also useful for solving equations with an unknown exponent. –Eg 3 x = 729  Logs can have different bases but the calculator only uses 10 and e (a special number like π)

31. Logs and Exponents  Logs have a base, and work as below x = b y x = b y log b x = y  Ie log 10 (3) = 0.4771 because 10 0.4771 = 3  Quickly work out the log statements below log 10 (10000) = alog 2 (16) = b log 9 (81) = clog 3 (81) = d Log e (32) = 5log 5 (f) = 4

32. What Logs mean  Practice –Theta Page 89 ex 11.1 Particularly Q3 onwards –Homework book Page 35

33. Log Laws  Because logs are very similar to exponents there are some laws that go with them that make them easier to simplify. log(ab) = log(a/b) = log(a n )= These laws are used when simplifying and solving log equations so although the are given, it helps to know them. log(a) + log(b) log(a) – log(b) nlog(a)

34. Log Laws  Some Examples log(3) + log(5) log(12) – log(2) 3log(2) 2log(6) – log(9) Practice Theta Page 91 ex 11.2 Homework Page 36

35. Linear Equations  The simplest type of equation it involves no exponents and if graphed would be a straight line. eg)4x – 8 = 3x + 12 Homework book Page 3

36. Rational Equations  These are equations containing fractions eg)2x – 3 = -3 5 4 5 4 Easiest way to deal with these is to multiply through by the lowest common multiple of the denominators to get rid of the fraction.

37. Rational Equations eg)2x – 3 = -3 5 4 5 4 20 × 2x – 3 = -3 × 20 5 4 5 4 4 × (2x – 3) = (-3) × 5 8x – 12 = -15 8x – 12 = -15 8x = -3 8x = -3 x = -3/8 x = -3/8 Lowest common multiple of 5 and 4 is 20 Multiply through by 20 Cancel down 20/5 = 4 20/4 = 5 Solve the linear equation Some people use cross multiplying, multiply each side by the other sides denominator, takes you straight to the third line.

38. Rational Equations 1)x = 4 2 2)3x = 8 5 3)x + 1 = x 5 3 5 3 Practice Theta page 8 ex 1.10 even no’s Homework book Page 4

39. Inequations  Very similar too linear equations but have an inequality instead of an equals sign. eg)4x – 8 < 3x + 12 The only trick is if you divide by a negative you have to reverse the sign.

40. Inequations 1)4x – 8 < 3x + 12 2)11 – 3x > 32 3) 3x ≤ 5(2x + 4) Practice –Theta Page 9 ex 1.11 –Theta Page 9 ex 1.12 (Applications) –Homework Page 5

41. Solving Quadratic Equations Eg:1)(x + 6)(2x – 7)= 0 2) x 2 + 12x + 20 = 0 3)x 2 + 7x= 18 4) 3x 2 + 5x – 8= 2x 2 – 3x + 40 Step 1:Make it equal to zero by adding and subtracting Step 2:Factorise into 1 or 2 brackets Step 3:Solve

42. Solving Quadratic Equations Eg:1)(x + 6)(2x – 7)= 0 Eitherx + 6 = 0 x = -6 x = -6 Step 1:Make it equal to zero by adding and subtracting Step 2:Factorise into 1 or 2 brackets Step 3:Solve or 2x – 7= 0 2x= 7 2x= 7 x = 3.5 x = 3.5

43. Solving Quadratic Equations Eg: 2)x 2 + 12x + 20 = 0 (x + 2)(x + 10)= 0 Eitherx + 2 = 0 x = -2 x = -2 Step 1:Make it equal to zero by adding and subtracting Step 2:Factorise into 1 or 2 brackets Step 3:Solve or x + 10= 0 x= -10 x= -10

44. Solving Quadratic Equations Eg: 3)x 2 + 7x= 18 x 2 + 7x – 18= 0 x 2 + 7x – 18= 0 (x + 9)(x – 2)= 0 (x + 9)(x – 2)= 0 Eitherx + 9 = 0 x = -9 x = -9 Step 1:Make it equal to zero by adding and subtracting Step 2:Factorise into 1 or 2 brackets Step 3:Solve or x – 2 = 0 x = 2 x = 2

45. Solving Quadratic Equations Eg: 4) 3x 2 + 5x – 8= 2x 2 – 3x + 40 x 2 + 8x – 48= 0 x 2 + 8x – 48= 0 (x + 12)(x – 4)= 0 (x + 12)(x – 4)= 0 Eitherx + 12 = 0 x = -12 x = -12 Step 1:Make it equal to zero by adding and subtracting Step 2:Factorise into 1 or 2 brackets Step 3:Solve or x – 4 = 0 x = 4 x = 4

46. Solving Quadratic Equations Eg:1)(x + 6)(2x – 7)= 0 Eitherx + 6 = 0 x = -6 x = -6 or 2x – 7= 0 2x= 7 2x= 7 x = 3.5 x = 3.5 Step 1:Make it equal to zero by adding and subtracting Step 2:Factorise into 1 or 2 brackets Step 3:Solve Eg: 4) 3x 2 + 5x – 8= 2x 2 – 3x + 40 x 2 + 8x – 48= 0 x 2 + 8x – 48= 0 (x + 12)(x – 4)= 0 (x + 12)(x – 4)= 0 Eitherx + 12 = 0 x = -12 x = -12 or x – 4= 0 x= 4 x= 4

47. Solving Quadratic Equations Practice Theta Page 56 ex 8.1 & 8.2 Theta Page 57 ex 8.3 Theta Page 58 ex 8.4 (Applications) Homework book Page 23 and 24

48. Completing the Square This uses the square brackets to help us solve complicated quadratic equations reasonably simply. The equation (x – 2) 2 – 8 = 0 is relatively easy to solve, the answers however are difficult decimals meaning the equation won’t factorise. The trick is fitting a quadratic equation into square brackets to start with.

49. Completing the Square Eg:x 2 + 8x + 12= 0 x 2 + 8x + 16 + 12= 16 (x + 4) 2 + 12= 16 (x + 4) 2 + 12= 16 (x + 4) 2 = 4 x + 4= √4 x + 4= √4 so x= -2 Why is there only 1 solution? There used to be 2 what went wrong? Answer √4 = 2 or -2 x= 2 – 4 or -2 – 4 x= 2 – 4 or -2 – 4 = -2 -6

50. Completing the Square A step by step guide: Eg:2)x 2 + 12x + 14= 0 x 2 + 12x + 36 + 14= 36 x 2 + 12x + 36 + 14= 36 (x + 6) 2 + 14= 36 (x + 6) 2 + 14= 36 (x + 6) 2 = 22 x + 6= ±√22 x + 6= ±√22 so x= -6 ± √22 Step 1:Make it equal to zero by adding and subtracting Step 2:Make the number in front of x 2 a 1 by dividing Step 3:Pick your square bracket Step 4:Solve

51. Completing the Square A step by step guide: Eg:3)5x 2 + 30x + 24= 0 x 2 + 6x + 4.8= 0 x 2 + 6x + 4.8= 0 x 2 + 6x + 9 + 4.8= 9 (x + 3) 2 + 4.8= 9 (x + 3) 2 + 4.8= 9 (x + 3) 2 = 4.2 x + 3= ±√4.2 x + 3= ±√4.2 so x= -3 ± √4.2 Step 1:Make it equal to zero by adding and subtracting Step 2:Make the number in front of x 2 a 1 by dividing Step 3:Pick your square bracket Step 4:Solve

52. Completing the Square Eg:3)5x 2 + 30x + 24= 0divide it all by 5 x 2 + 6x + 4.8= 0 half 6 = 3, 3 2 = 9, add 9 both sides x 2 + 6x + 4.8= 0 half 6 = 3, 3 2 = 9, add 9 both sides x 2 + 6x + 9 + 4.8= 9factorise the square bracket (x + 3) 2 + 4.8= 9- 4.8 both sides (x + 3) 2 + 4.8= 9- 4.8 both sides (x + 3) 2 = 4.2Square root both sides x + 3= ±√4.2-3 both sides x + 3= ±√4.2-3 both sides so x= -3 ± √4.2Calculator Overview Step 1:Make it equal to zero by adding and subtracting Step 2:Make the number in front of x 2 a 1, by dividing Step 3:Pick your square bracket Step 4:Solve

53. Completing the Square Practice Theta Page 61 ex 8.5

54. Quadratic Formula If ax 2 + bx + c= 0 Then x= -b ± √b 2 - 4ac 2a Eg:if 3x 2 + 5x – 9= 0 a = 3, b = 5, c = -9 then x = -(5) ± √(5) 2 – 4(3)(-9) 2(3) 2(3) x = -5 ± √25 – -108 x = -5 ± √25 – -108 6 x = (-5 + √132) x = (-5 + √132) 6 or x= (-5 – √132) 6 TAKE CARE and TAKE YOUR TIME

55. Quadratic Formula On a graphics calculator Eg:if 4x 2 + 8x – 7= x 2 – 3x Make one side = 0 by adding and subtracting 3x 2 + 11x – 7= 0 3x 2 + 11x – 7= 0 Menu : Equation F2 : Polynomial F1 : Degree 2 (x 2 ) Degree 3 (x 3 ) a = 3, b = 11, c = -7 F1 : Solve x 1 = 0.5529 x 2 = -4.219 EQUA Equation Select Type F1:Simultaneous F2:Polynomial F3:Solver Polynomial Degree? aX 2 +bX+c=0 a b c. [ 0 0 0 ] SIMLPOLYSOLV23CLRSOLVDEL aX 2 +bX+c=0 a b c. [ 3 0 0 ] aX 2 +bX+c=0 a b c. [ 3 11 0 ] aX 2 +bX+c=0 a b c. [ 3 11 -7 ] 3_1_11_-_-7 aX 2 +bX+c=0 X. 1 0.5529 2 -4.219 REPT

56. Quadratic Formula  A good challenge is to try and prove the quadratic formula by completing the square on ax 2 + bx + c = 0 ax 2 + bx + c = 0Practice Theta Page 61 ex 8.6 Page 62 ex 8.8 Applications HwkbkPage 24 ex 6.06 Page 25 and 26

57. Discriminant  Not every quadratic equation can be solved to give ‘real’ solutions.  At this stage in your Math careers, you cannot square root a negative number so any time b 2 – 4ac is less than zero (negative) we know there are no real solutions.  b 2 – 4ac is called the discriminant because it tells us how many solutions there will be;

58. Discriminant b 2 – 4ac > 0 (a positive number) means there will be 2 ‘real’ solutions b 2 – 4ac = 0 means there will be 1 ‘real’ solutions b 2 – 4ac < 0 (a negative number) means there will be no ‘real’ solutions

59. Discriminant Practice examples; 1)Show that the equation 7x 2 + 13x + 73 = 0 has no real solutions 2) For what value of k does the equation 3x 2 + 24x + k = 0 have only one solution?

60. Discriminant 1)7x 2 + 13x + 73 = 0 a = 7,b = 13,c = 73 b 2 – 4ac is the discriminant b 2 – 4ac is the discriminant 13 2 – 4×7×73 = 169 – 2044 = -1875 this is less than zero so there are no solutions

61. Discriminant 2)3x 2 + 24x + k = 0 One solution when the discriminant = 0 ie) b 2 – 4ac = 0 24 2 – 4×3×k = 0 24 2 – 4×3×k = 0 576 – 12k= 0 576 – 12k= 0 576= 12k 576= 12k k= 48 So 3x 2 + 24x + k = 0 has only one solution when k = 48

62. Discriminant Practice Textbook Page 66 ex 8.9 page 67 ex 8.10

63. Exponential Equations  These are equations containing an unknown exponent  Eg)3 x = 243  Some can be solved by inspection (or trial and error) but for more difficult ones we need a more reliable method.  We can use Logs to help us out here as they can change a power into a multiplication

64. Exponential Equations  Using the third log law  n loga = loga n  Now apply this in reverse to our problem 3 x = 243 3 x = 243 log(3 x )= log(243) x log(3)= log(243) x log(3)= log(243) x= log(243) / log(3) x= log(243) / log(3) Calculate the answer on a calculator x= 5 x= 5 Log Both Sides Move the exponent out the front Divide by log(3)

65. Exponential Equations  Practice 5 x = 3125 12 (x+4) = 20736 log(5 x ) = log(3125) log(12 (x+4) ) = log(20736) x = log(3125) / log(5) x + 4 = log(20736)/log12 x log(5) = log(3125) (x + 4)log(12) = log(20736) x + 4 = 4 x = 0 x = 5

66. Rearranging Equations  I approach these in exactly the same way as I would if I had to solve an equation.  The first thing I do is decide which letter I have to get alone.  Then I ask myself what I would do if everything else were numbers, –Get Rid of Fractions –Group like terms (get my letter on one side and everything else on the other) –Simplify, (factorise) –Divide through by whatever I have in front of my letter (the coefficient)

67. Rearranging Equations  The examples we are working through are 1.Make ‘x’ the subject of Y = m x + c Y = m x + c 2.Make ‘a’ the subject of S = v t + 0.5 a t 2 S = v t + 0.5 a t 2 3. Make ‘x’ the subject of 12 – y x = 3y + 4 x 12 – y x = 3y + 4 x

68. Rearranging Equations  Practice –Make x the subject of Y = m x + c Y – c = m x Y – c = m x (Y – c)/m= x –Make a the subject of S = v t + 0.5 a t 2 S = v t + 0.5 a t 2 S – v t = 0.5 a t 2 S – v t = 0.5 a t 2 2(S – v t) = a t 2 2(S – v t) = a t 2 2(S – v t) / t 2 = a What letter are we trying to get alone? Get everything away from x. Divide through by the coefficient (m). What letter are we trying to get alone? Get everything away from a. Divide through by the coefficients.

69. Rearranging Equations  Practice –Make x the subject of 12 – y x = 3 y + 4 x 12 = 3 y + 4 x + y x 12 = 3 y + 4 x + y x 12 – 3 y= 4 x + y x 12 – 3y = x(4 + y) x = (12 – 3y) (4 + y) (4 + y) What letter are we trying to get alone? Get x’s to one side. Get everything away from x’s Make it only one x, (by factorising) Divide through by the coefficients.

70. Rearranging Equations  Practice  Theta textbook Page 29 exercise 4.2 (Applications)  Homework Page 15 and 16  Theta Textbook Page 31 ex (Harder)  Theta Textbook Page 35 ex (Harder applications)  Homework Page 17 and 18

71. Simultaneous Equations  These are still similar to year 11, still involve only 2 equations and 2 unknowns.  We now have to be able to solve both linear and non-linear simultaneous equations.

72. Simultaneous Equations  Linear Simultaneous Equations –The challenge this year is that any linear simultaneous equations you are asked to solve will not have equations given. They will just be word problems. So you need to be able to interpret the information given, write 2 equations and then solve the simultaneous equations. –The methods to solve them haven’t changed

73. Simultaneous Equations  Eg  Tara buys 8 cakes for her family. Five of the cakes are cream and three are plain. She spends $16.25 altogether. A cream cake costs 45 cents more than a plain cake. Calculate the price of ONE cream cake. –Step one: List the variables c = Cost of cream cake, p = Cost of plain cake –Step two: Write the Equations 5c + 3p = 16.25 c= p + 0.45

74. Simultaneous Equations –Step one: List the variables c = Cost of cream cake, p = Cost of plain cake –Step two: Write the Equations 5c + 3p = 16.25 c= p + 0.45 –Step three: Solve the Equations Substitution Method Make ‘c’ the subject of one side c = p + 0.45 Substitute ‘c’ into other equation 5(p + 0.45) + 3p = 16.25 Solve the equation Elimination Method 5c + 3p = 16.25 (c – p = 0.45) × 5 5c + 3p = 16.25 -(5c – 5p= 2.25) 0c + 8p = 14.00

75. Simultaneous Equations  Practice –Theta TextbookPage 39 ex 5.2 Page 40 ex 5.4 Page 41 ex 5.5

76. Simultaneous Equations  Non-linear Simultaneous Equations –This is really no more difficult than linear simultaneous equations. The only difference is that on or both of the equations would draw a curve if graphed. (Normally a circle or Parabola) –Some people find it easier as the equations are given. –The best method to use is Substitution, as it will always work, Elimination is much trickier.

77. Simultaneous Equations  Eg Solve x 2 + y 2 = 25 and y = x + 1 simultaneously.  Step 1: Rearrange one equation to make ‘y’ or ‘x’ the subject. Y in the second equation is easiest so use that.  Step 2: Substitute ‘y’ into the equation x 2 + (x + 1) 2 = 25  Step 3: Solve the equation to find ‘x’. x = 3 or -4  Step 4: Substitute ‘x’ into one of the original equations to find ‘y’. y = 4 or -3

78. Simultaneous Equations  Eg Find the x-coordinates of the point of intersection of y = 2x – 1 and x 2 + y 2 – 4x – 5= 0.  Step 1: Rearrange one equation to make ‘y’ or ‘x’ the subject. Y in the second equation is easiest so use that.  Step 2: Substitute ‘y’ into the equation x 2 + (2x – 1) 2 – 4x – 5 = 0  Step 3: Solve the equation to find ‘x’. x = 2 or -0.4  Step 4: Substitute ‘x’ into one of the original equations to find ‘y’. y = 3 or -1.8