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© Boardworks Ltd 2005 1 of 43 A5 Simultaneous equations KS4 Mathematics.

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Presentation on theme: "© Boardworks Ltd 2005 1 of 43 A5 Simultaneous equations KS4 Mathematics."— Presentation transcript:

1 © Boardworks Ltd of 43 A5 Simultaneous equations KS4 Mathematics

2 © Boardworks Ltd of 43 A A A A A A5.1 Solving simultaneous equations graphically A5 Simultaneous equations Contents A5.2 The elimination method A5.3 The substitution method A5.5 Problems leading to simultaneous equations A5.4 Simultaneous linear and quadratic equations

3 © Boardworks Ltd of 43 Simultaneous equations 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: x y x + y = 3

4 © Boardworks Ltd of 43 Simultaneous equations 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: 0 x y 3 3 y – x = 1

5 © Boardworks Ltd of 43 Simultaneous equations There is one pair of values that solves both these equations: 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. 0 x y 3 3 y – x = 1 x + y = 3 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 © Boardworks Ltd of 43 Simultaneous equations y – x = 1 are called a pair of simultaneous equations. x + y = 3 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 = 3 2 – 1 = 1 Both the equations are satisfied and so the solution is correct.

7 © Boardworks Ltd of 43 Solving simultaneous equations graphically

8 © Boardworks Ltd of 43 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 © Boardworks Ltd of 43 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 – 2 x = 3 2 y = 4 x + 1 have no solutions. Rearranging these equations in the form y = mx + c gives, y = 2 x + 3 y = 2 x + ½ The gradient m is 2 for both equations and so there are no solutions.

10 © Boardworks Ltd of 43 Simultaneous equations with infinite solutions Sometimes pairs of simultaneous equations are represented by the same graph. For example, Notice that each term in the second equation is 3 times the value of the corresponding term in the first equation. 2 x + y = 3 6 x + 3 y = 9 Both equations can be rearranged to give y = –2 x + 3 When two simultaneous equations can be rearranged to give the same equation they have an infinite number of solutions.

11 © Boardworks Ltd of 43 Special solutions

12 © Boardworks Ltd of 43 A A A A A A5.2 The elimination method Contents A5 Simultaneous equations A5.3 The substitution method A5.5 Problems leading to simultaneous equations A5.4 Simultaneous linear and quadratic equations A5.1 Solving simultaneous equations graphically

13 © Boardworks Ltd of 43 The elimination method 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 3 x + y = 9 5 x – y = 7 Adding these equations: 3 x + y = 9 5 x – y = 7+ 8x8x = 16 The y terms have been eliminated. divide both sides by 8: x = 2

14 © Boardworks Ltd of 43 The elimination method Adding the two equations eliminated the y terms and gave us a single equation in x. 3 x + y = 9 5 x – y = 7 To find the value of y when x = 2 substitute this value into one of the equations. Solving this equation gave us the solution x = 2. Substituting x = 2 into the first equation gives us: 3 × 2 + y = y = 9 y = 3 subtract 6 from both sides:

15 © Boardworks Ltd of 43 The elimination method We can check whether x = 2 and y = 3 solves both: 3 x + y = 9 5 x – y = 7 by substituting them into the second equation. 5 × 2 – 3 = 7 10 – 3 = 7 This is true, so we have confirmed that x = 2 y = 3 solves both equations.

16 © Boardworks Ltd of 43 The elimination method Solve these equations:3 x + 7 y = 22 3 x + 4 y = 10 Subtracting gives: 3 x + 7 y = 22 3 x + 4 y = 10– 3y3y = 12 The x terms have been eliminated. divide both sides by 3: y = 4 Substituting y = 4 into the first equation gives us, 3 x + 7 × 4 = 22 3 x + 28 = 22 x = –2 divide both sides by 3:subtract 28 from both sides: 3 x = –6

17 © Boardworks Ltd of 43 The elimination method We can check whether x = –2 and y = 4 solves both, 3 x + 7 y = 22 3 x + 4 y = 10 by substituting them into the second equation. 3 × –2 + 7 × 4 = 22 – = 22 This is true and so, x = –2 y = 4 solves both equations.

18 © Boardworks Ltd of 43 The elimination method 1

19 © Boardworks Ltd of 43 The elimination method Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example, 4 x – y = 29 3 x + 2 y = 19 We need to have the same number in front of either the x or the y before adding or subtracting the equations. 8 x – 2 y = x = 77 divide both sides by 11: x = 7 Call these equations 1 and × 1 : 3 3 x + 2 y = :

20 © Boardworks Ltd of 43 The elimination method To find the value of y when x = 7 substitute this value into one of the equations, 4 x – y = x + 2 y = × 7 – y = – y = 29 Substituting x = 7 into 1 gives, subtract 28 from both sides: – y = 1 y = –1 multiply both sides by –1: Check by substituting x = 7 and y = –1 into 2, 3 × × –1 = 9 21 – 2 = 19

21 © Boardworks Ltd of 43 The elimination method 6 x – 15 y = 75 Call these equations 1 and 2. Solve:2 x – 5 y = 25 3 x + 4 y = × 1 – 6 x + 8 y = 6 2 × 2 – 23 y = – 4, y = –3 divide both sides by –23: Substitute y = –3 in 1,2 x – 5 × –3 = 25 2 x + 15 = 25 2 x = 10 subtract 15 from both sides: x = 5 divide both sides by 2: 69

22 © Boardworks Ltd of 43 The elimination method 2

23 © Boardworks Ltd of 43 Contents A A A A A A5.3 The substitution method A5 Simultaneous equations A5.2 The elimination method A5.5 Problems leading to simultaneous equations A5.4 Simultaneous linear and quadratic equations A5.1 Solving simultaneous equations graphically

24 © Boardworks Ltd of 43 The substitution method Two simultaneous equations can also be solved by substituting one equation into the other. For example, Call these equations 1 and 2. y = 2 x – 3 2 x + 3 y = Substitute equation 1 into equation 2. y = 2 x – 3 2 x + 3(2 x – 3) = 23 expand the brackets: 2 x + 6 x – 9 = 23 simplify: 8 x – 9 = 23 add 9 to both sides: 8 x = 32 x = 4 divide both sides by 8:

25 © Boardworks Ltd of 43 The substitution method To find the value of y when x = 4 substitute this value into one of the equations, y = 2 x – x + 3 y = 23 2 y = 2 × 4 – 3 y = 5 Substituting x = 4 into 1 gives Check by substituting x = 4 and y = 5 into 2, 2 × × 5 = = 23 This is true and so the solutions are correct.

26 © Boardworks Ltd of 43 The substitution method How could the following pair of simultaneous equations be solved using substitution? Call these equations 1 and 2. 3 x – y = 9 8 x + 5 y = One of the equations needs to be arranged in the form x = … or y = … before it can be substituted into the other equation. Rearrange equation 1. 3 x – y = 9 add y to both sides: 3 x = 9 + y subtract 9 from both sides: 3 x – 9 = y y = 3 x – 9

27 © Boardworks Ltd of 43 The substitution method 3 x – y = 9 8 x + 5 y = Now substitute y = 3 x – 9 into equation 2. 8 x + 5(3 x – 9) = 1 expand the brackets: 8 x + 15 x – 45 = 1 simplify: 23 x – 45 = 1 add 45 to both sides: 23 x = 46 divide both sides by 23: x = 2 Substitute x = 2 into equation 1 to find the value of y. 3 × 2 – y = 9 6 – y = 9 – y = 3 y = –3

28 © Boardworks Ltd of 43 The substitution method 3 x – y = 9 8 x + 5 y = Check the solutions x = 2 and y = –3 by substituting them into equation 2. 8 × × –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 © Boardworks Ltd of 43 Contents A A A A A A5.4 Simultaneous linear and quadratic equations A5 Simultaneous equations A5.3 The substitution method A5.2 The elimination method A5.5 Problems leading to simultaneous equations A5.1 Solving simultaneous equations graphically

30 © Boardworks Ltd of 43 Simultaneous linear and quadratic equations 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 = x + 3 y = x x = x + 3 We have to collect all the terms onto the left-hand side to give a quadratic equation of the form ax 2 + bx + c = 0. x 2 – x – 2 = 0 factorize: ( x + 1)( x – 2) = 0 x = –1 or x = 2 Substituting equation 1 into equation 2,

31 © Boardworks Ltd of 43 Simultaneous linear and quadratic equations We can substitute these values of x into one of the equations y = x + 3 y = x When x = –1 we have, It is easiest to substitute into equation 2 because it is linear. to find the corresponding values of y. y = –1 + 3 y = 2 When x = 2 we have, y = y = 5 The solutions are x = –1, y = 2 and x = 2, y = 5.

32 © Boardworks Ltd of 43 Using graphs to solve equations –1–2–3– – We can also show the solutions to using a graph. y = x y = x + 3 y = x y = x + 3 (–1,2) (2, 5) The points where the two graphs intersect give the solution to the pair of simultaneous equations. It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs, particularly when the solutions are not exact.

33 © Boardworks Ltd of 43 Linear and quadratic graphs

34 © Boardworks Ltd of 43 Simultaneous linear and quadratic equations Look at this pair of simultaneous equations: x 2 + y 2 = 13 y = x The graph of x 2 + y 2 = 13 is a circular graph with its centre at the origin and a radius of √13. What shape is the graph given by x 2 + y 2 = 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 © Boardworks Ltd of 43 Simultaneous linear and quadratic equations x 2 + y 2 = 13 y = x x 2 + ( x + 1) 2 = 13 factorize: ( x + 3)( x – 2) = 0 x = –3 or x = 2 Substituting equation 1 into equation 2, x 2 + x x + 1 = 13 expand the bracket:simplify: 2 x x + 1 = 13 2 x x – 12 = 0 subtract 13 from both sides: x 2 + x – 6 = 0 divide through by 2:

36 © Boardworks Ltd of 43 Simultaneous linear and quadratic equations We can substitute these values of x into one of the equations x 2 + y 2 = 13 y = x When x = –3 we have, It is easiest to substitute into equation 1 because it is linear. to find the corresponding values of y. y = –3 + 1 y = –2 When x = 2 we have, y = y = 3 The solutions are x = –3, y = –2 and x = 2, y = 3.

37 © Boardworks Ltd of 43 Linear and circular graphs

38 © Boardworks Ltd of 43 Contents A A A A A A5.5 Problems leading to simultaneous equations A5 Simultaneous equations A5.3 The substitution method A5.2 The elimination method A5.4 Simultaneous linear and quadratic equations A5.1 Solving simultaneous equations graphically

39 © Boardworks Ltd of 43 Solving problems The sum of two numbers is 56 and the difference between the two numbers is 22. Find the two numbers. a + b = 56 a – b = 22 Adding these equations gives: 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, 2 a = 78 a = 39

40 © Boardworks Ltd of 43 Solving problems 39 + b = 56 So the two numbers are 39 and – 17 = 22 This is true and so our solution is correct. Substituting a = 39 into the first equation gives, b = 17 subtract 39 from both sides: We can check these solutions by substituting them into the second equation, a – b = 22:

41 © Boardworks Ltd of 43 Solving problems The cost of theatre tickets for 4 adults and 3 children is £ 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, 4 a + 3 c = a + 6 c = 44 2 Dividing equation 2 by 2 gives, a + 3 c = 22 3 We can now subtract equation 3 from equation 1 to eliminate the terms containing c.

42 © Boardworks Ltd of 43 Solving problems 4 a + 3 c = a + 3 c = 22 3 The cost of an adult’s ticket is £8.50 and the cost of a child’s ticket is £4.50. – 3a3a 1 – 3, a = 8.50 divide both sides by 3: Substitute a = 8.50 in 3 : c = 22 3 c = subtract 8.50 from both sides: c = 4.50 divide both sides by 3: = 25.50

43 © Boardworks Ltd of 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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