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KS4 Mathematics

KS4 Mathematics. A5 Simultaneous equations. A5 Simultaneous equations. Contents. A. A5.2 The elimination method. A. A5.1 Solving simultaneous equations graphically. A5.3 The substitution method. A. A5.4 Simultaneous linear and quadratic equations. A.

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KS4 Mathematics

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  1. KS4 Mathematics A5 Simultaneous equations

  2. A5 Simultaneous equations Contents • A A5.2 The elimination method • A A5.1 Solving simultaneous equations graphically A5.3 The substitution method • A A5.4 Simultaneous linear and quadratic equations • A A5.5 Problems leading to simultaneous equations • A

  3. Simultaneous equations y 3 x + y = 3 0 x 3 Equations in two unknowns have an infinite number of solution pairs. For example, x + y = 3 is true when x = 1 and y = 2 x = 3 and y = 0 x = –2 and y = 5 and so on … We can represent the set of solutions on a graph:

  4. Simultaneous equations y 3 y–x = 1 0 x 3 Another equation in two unknowns will also have an infinite number of solution pairs. For example, y – x = 1 is true when x = 1 and y = 2 x = 3 and y = 4 x = –2 and y = –1 and so on … This set of solutions can also be represented in a graph:

  5. Simultaneous equations y–x = 1 y 3 x + y = 3 0 x 3 There is one pair of values that solves both these equations: x + y = 3 y – x = 1 We can find the pair of values by drawing the lines x + y = 3 and y – x = 1 on the same graph. The point where the two lines intersect gives us the solution to both equations. This is the point (1, 2). At this point x = 1 and y = 2.

  6. Simultaneous equations x + y = 3 y – x = 1 are called a pair of simultaneous equations. The values of x and y that solve both equations are x = 1 and y = 2, as we found by drawing graphs. We can check this solution by substituting these values into the original equations. 1 + 2 = 3 2 – 1 = 1 Both the equations are satisfied and so the solution is correct.

  7. Solving simultaneous equations graphically

  8. Simultaneous equations with no solutions Sometimes pairs of simultaneous equations produce graphs that are parallel. Parallel lines never meet, and so there is no point of intersection. When two simultaneous equations produce graphs which are parallel there are no solutions. How can we tell whether the graphs of two lines are parallel without drawing them? Two lines are parallel if they have the same gradient.

  9. Simultaneous equations with no solutions We can find the gradient of the line given by a linear equation by rewriting it in the form y = mx + c. The value of the gradient is given by the value of m. Show that the simultaneous equations y – 2x = 3 2y = 4x + 1 have no solutions. Rearranging these equations in the form y = mx + c gives, y = 2x + 3 y = 2x + ½ The gradient m is 2 for both equations and so there are no solutions.

  10. Simultaneous equations with infinite solutions Sometimes pairs of simultaneous equations are represented by the same graph. For example, 2x + y = 3 6x + 3y = 9 Notice that each term in the second equation is 3 times the value of the corresponding term in the first equation. Both equations can be rearranged to give y = –2x + 3 When two simultaneous equations can be rearranged to give the same equation they have an infinite number of solutions.

  11. Special solutions

  12. A5 Simultaneous equations Contents A5.1 Solving simultaneous equations graphically • A • A A5.2 The elimination method A5.3 The substitution method • A A5.4 Simultaneous linear and quadratic equations • A A5.5 Problems leading to simultaneous equations • A

  13. The elimination method 3x + y = 9 5x – y = 7 + 5x – y = 7 If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example, suppose Adding these equations: The y terms have been eliminated. 3x + y = 9 = 16 8x x = 2 divide both sides by 8:

  14. The elimination method To find the value of y when x = 2 substitute this value into one of the equations. 3x + y = 9 5x – y = 7 Adding the two equations eliminated the y terms and gave us a single equation in x. Solving this equation gave us the solution x = 2. Substituting x = 2 into the first equation gives us: 3 × 2 + y = 9 6 + y = 9 y = 3 subtract 6 from both sides:

  15. The elimination method This is true, so we have confirmed that x = 2 y = 3 solves both equations. We can check whether x = 2 and y = 3 solves both: 3x + y = 9 5x – y = 7 by substituting them into the second equation. 5 × 2 – 3 = 7 10 – 3 = 7

  16. The elimination method y = 4 divide both sides by 3: – 3x + 4y = 10 3x = –6 subtract 28 from both sides: x = –2 divide both sides by 3: Solve these equations: 3x + 7y = 22 3x + 4y = 10 Subtracting gives: The x terms have been eliminated. 3x + 7y = 22 3y = 12 Substituting y = 4 into the first equation gives us, 3x + 7 × 4 = 22 3x + 28 = 22

  17. The elimination method This is true and so, x = –2 y = 4 solves both equations. We can check whether x = –2 and y = 4 solves both, 3x + 7y = 22 3x + 4y = 10 by substituting them into the second equation. 3 × –2 + 7 × 4 = 22 –6 + 28 = 22

  18. The elimination method 1

  19. The elimination method Call these equations 1 and 2 . 1 3 2 2 × 1 : + 3x + 2y = 19 3 + 2 : Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example, 4x – y = 29 3x + 2y = 19 We need to have the same number in front of either the x or the y before adding or subtracting the equations. 8x – 2y = 58 = 77 11x x = 7 divide both sides by 11:

  20. The elimination method 4x – y = 29 1 Substituting x = 7 into 1 gives, 2 –y = 1 subtract 28 from both sides: y = –1 multiply both sides by –1: Check by substituting x = 7 and y = –1 into2 , To find the value of y when x = 7 substitute this value into one of the equations, 3x + 2y = 19 4 × 7 – y = 29 28 – y = 29 3 × 7 + 2 × –1 = 19 21 – 2 = 19

  21. The elimination method Solve: 2x – 5y = 25 3x + 4y = 3 Call these equations 1 and 2 . 3 1 3 × 1 6x + 8y = 6 4 2 2 × 2 – y = –3 divide both sides by –23: Substitute y = –3 in 1 , 3 – 4 , 2x = 10 subtract 15 from both sides: x = 5 divide both sides by 2: 6x – 15y = 75 – 23y = 69 2x – 5 × –3 = 25 2x + 15 = 25

  22. The elimination method 2

  23. A5 Simultaneous equations Contents A5.1 Solving simultaneous equations graphically • A A5.2 The elimination method • A A5.3 The substitution method • A A5.4 Simultaneous linear and quadratic equations • A A5.5 Problems leading to simultaneous equations • A

  24. The substitution method Call these equations 1 and 2 . 1 2 Substitute equation 1 into equation 2 . 2x + 6x – 9 = 23 expand the brackets: 8x – 9 = 23 simplify: 8x= 32 add 9 to both sides: x= 4 divide both sides by 8: Two simultaneous equations can also be solved by substituting one equation into the other. For example, y = 2x – 3 y = 2x – 3 2x + 3y = 23 2x + 3(2x – 3) = 23

  25. The substitution method y = 2x – 3 2x + 3y = 23 1 Substituting x = 4 into 1 gives 2 Check by substituting x = 4 and y = 5into2 , To find the value of y when x = 4 substitute this value into one of the equations, y = 2 × 4 – 3 y = 5 2 × 4 + 3 × 5 = 23 8 + 15 = 23 This is true and so the solutions are correct.

  26. The substitution method 1 2 Call these equations 1 and 2 . Rearrange equation 1 . 3x = 9 + y add y to both sides: 3x – 9 = y subtract 9 from both sides: How could the following pair of simultaneous equations be solved using substitution? 3x – y = 9 8x + 5y = 1 One of the equations needs to be arranged in the form x = … or y = … before it can be substituted into the other equation. 3x – y = 9 y = 3x – 9

  27. The substitution method 1 8x + 15x – 45 = 1 expand the brackets: Now substitute y = 3x – 9 into equation 2 . 2 23x – 45 = 1 simplify: 23x = 46 add 45 to both sides: x = 2 divide both sides by 23: Substitute x = 2 into equation 1 to find the value of y. 3x – y = 9 8x + 5y = 1 8x + 5(3x – 9) = 1 3 × 2 – y = 9 6 – y = 9 –y = 3 y = –3

  28. The substitution method Check the solutions x = 2and y = –3 by substituting them into equation 2 . 1 2 3x – y = 9 8x + 5y = 1 8 × 2 + 5 × –3 = 1 16 – 15 = 1 This is true and so the solutions are correct. Solve these equations using the elimination method to see if you get the same solutions for x and y.

  29. A5 Simultaneous equations Contents A5.1 Solving simultaneous equations graphically • A A5.2 The elimination method • A A5.4 Simultaneous linear and quadratic equations A5.3 The substitution method • A • A A5.5 Problems leading to simultaneous equations • A

  30. Simultaneous linear and quadratic equations 1 2 Substituting equation 1 into equation 2 , (x + 1)(x – 2) = 0 factorize: When one of the equations in a pair of simultaneous equations is quadratic, we often end up with two pairs of solutions. For example, y = x2 + 1 y = x + 3 x2 + 1 = x + 3 We have to collect all the terms onto the left-hand side to give a quadratic equation of the form ax2 + bx + c = 0. x2 – x – 2 = 0 x = –1 or x = 2

  31. Simultaneous linear and quadratic equations 1 2 It is easiest to substitute into equation 2 because it is linear. We can substitute these values of x into one of the equations y = x2 + 1 y = x + 3 to find the corresponding values of y. When x = –1 we have, When x = 2 we have, y = –1 + 3 y = 2 + 3 y = 2 y = 5 The solutions are x = –1, y = 2 and x = 2, y = 5.

  32. Using graphs to solve equations We can also show the solutions to using a graph. 10 8 y = x2 + 1 6 y = x + 3 4 2 –4 –3 –2 –1 0 1 2 3 4 –2 y = x2 + 1 y = x + 3 (2, 5) The points where the two graphs intersect give the solution to the pair of simultaneous equations. (–1,2) It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs.

  33. Linear and quadratic graphs

  34. Simultaneous linear and quadratic equations 1 2 Look at this pair of simultaneous equations: y = x + 1 x2 + y2 = 13 What shape is the graph given by x2 + y2 = 13? The graph ofx2 + y2 = 13 is a circular graph with its centre at the origin and a radius of √13. We can solve this pair of simultaneous equations algebraically using substitution. We can also plot the graphs of the equations and observe where they intersect.

  35. Simultaneous linear and quadratic equations 1 2 Substituting equation 1 into equation 2 , x2 + x2 + 2x + 1 = 13 expand the bracket: 2x2 + 2x – 12 = 0 2x2 + 2x + 1 = 13 simplify: subtract 13 from both sides: x2 + x – 6 = 0 divide through by 2: (x + 3)(x – 2) = 0 factorize: y = x + 1 x2 + y2 = 13 x2 + (x + 1)2 = 13 x = –3 or x = 2

  36. Simultaneous linear and quadratic equations 1 2 It is easiest to substitute into equation 1 because it is linear. We can substitute these values of x into one of the equations y = x + 1 x2 + y2 = 13 to find the corresponding values of y. When x = –3 we have, When x = 2 we have, y = –3 + 1 y = 2 + 1 y = –2 y = 3 The solutions are x = –3, y = –2 and x = 2, y = 3.

  37. Linear and circular graphs

  38. A5 Simultaneous equations Contents A5.1 Solving simultaneous equations graphically • A A5.2 The elimination method • A A5.5 Problems leading to simultaneous equations A5.3 The substitution method • A A5.4 Simultaneous linear and quadratic equations • A • A

  39. Solving problems The sum of two numbers is 56 and the difference between the two numbers is 22. Find the two numbers. Let’s call the unknown numbers a and b. We can use the given information to write a pair of simultaneous equations in terms of a and b, a + b = 56 a – b = 22 Adding these equations gives: 2a = 78 a = 39

  40. Solving problems b = 17 subtract 39 from both sides: Substituting a = 39 into the first equation gives, 39 + b = 56 So the two numbers are 39 and 17. We can check these solutions by substituting them into the second equation, a – b = 22: 39 – 17 = 22 This is true and so our solution is correct.

  41. Solving problems 1 2 3 4a + 3c = 47.50 2a + 6c = 44 Dividing equation 2 by 2 gives, a + 3c = 22 We can now subtract equation 3 from equation 1 to eliminate the terms containing c. The cost of theatre tickets for 4 adults and 3 children is £47.50. The cost for 2 adults and 6 children is £44. How much does each adult and child ticket cost? Let’s call the cost of an adult’s ticket a and the cost of a child’s ticket c. We can write,

  42. Solving problems 1 a = 8.50 divide both sides by 3: 3 Substitute a = 8.50 in 3 : – 4a + 3c = 47.50 a + 3c = 22 3c = 13.50 subtract 8.50 from both sides: c = 4.50 divide both sides by 3: 1 – 3 , 3a = 25.50 8.50 + 3c = 22 The cost of an adult’s ticket is £8.50 and the cost of a child’s ticket is £4.50.

  43. Solving problems Remember, when using simultaneous equations to solve problems: 1) Decide what letters to use to represent each of the unknown values. 2) Use the information given in the problem to write down two equations in terms of the two unknown values. 3) Solve the simultaneous equations using the most appropriate method. 4) Check the values by substituting them back into the original problem.

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