1 of 43. Simultaneous Equations

Transcription

1 1 of 43 Simultaneous Equations

2 Simultaneous Equations (Graphs) 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. 3 y y x = 1 The point where the two lines intersect gives us the solution to both equations. 0 3 x x + y = 3 This is the point (1, 2). At this point x = 1 and y = 2. 2 of 43 We now need to do algebraically.

3 SLO Achieve: Elimination Method (Achieve: same sign) 3 of 43 (Achieve: different sign)

4 4 of 43 What are simultaneous equations Bob goes to New World and buys the following for $11 + = $11 We can write this as 3B + 2S = 11 Although we can guess, it is impossible to say how much the beans or spaghetti cost. We need more information.

5 5 of 43 Bob goes to New World and buys the following for $11 + We can write this as 3B + 2S = 11 Jill also goes to New World and buys the following for $ We can write this as 3B + 3S = How much does the spaghetti cost? 1 extra can of spaghetti adds an extra $2.50 to the bill

6 6 of 43 Elimination method Algebraically this is what we were doing 3B + 3S = B + 2S = S = B 3B = 0 (The B has been eliminated) = S 2S = S

7 7 of 43 Copy into your notes Elimination method Step 1: Rewrite both equations so the = sign is the same place Step 2: Write one equation above the other Step 3: Identify the letter to eliminate. It is the letter with the same number in front (coefficient), regardless of sign. Step 4: Re coefficient: Sign Same Subtract SSS Sign Different Add SDA Step 5: Find the value of the first letter. Merit Students: leave gap for an additional step Step 6: Substitute to find the value of the second letter. Step 7: Check your answers.

8 8 of 43 Example1: Elimination method Solve the simultaneous equations 3x + y = 9 and 5x y = 7 Step 1: Rewrite both equations so the = sign is the same place Already done Step 2: Write one equation above the other 3x + y = 9 5x y = 7 Step 3: Identify the letter to eliminate. It is the letter with the same number in front (coefficient), regardless of sign. y has either -1 or 1 as the coefficient so we will eliminate y

9 9 of 43 Step 4: Re coefficient: Sign Same Subtract SSS Sign Different Add SDA As the signs are different: Add these equations: 3x + y = 9 Step 5: Find the value of the first letter. 8x = 16 x = 2 + 5x y = 7 8x = 16 divide both sides by 8: Step 6: Substitute to find the value of the second letter. Substituting x = 2 into the first equation gives us: subtract 6 from both sides: y = y = 9 y = 3

10 10 of 43 Step 7: Check your answers. We can check whether x = 2 and y = 3 solves both: 3x + y = 9 5x y = 7 by substituting them into the second equation = = 7 This is true, so we have confirmed that x = 2 y = 3

11 11 of 43 Copy into your notes Example 2: Elimination method Solve the simultaneous equations 3x + 7y = 22 and 3x + 4y = 10 Step 1: Rewrite both equations so the = sign is the same place Already done Step 2: Write one equation above the other 3x + 7y = 22 3x + 4y = 10 Step 3: Identify the letter to eliminate. It is the letter with the same number in front (coefficient), regardless of sign. Both x s have the same coefficient: eliminate x

12 12 of 43 Copy into your notes Step 4: Re coefficient: Sign Same Subtract SSS Sign Different Add SDA As the signs are the same: subtract Step 5: Find the value of the first letter. = 12 divide both sides by 3: y = 4 3x + 7y = 22 3x + 4y = 10 3y = 12 Step 6: Substitute to find the value of the second letter. Substituting y = 4 into the first equation gives us, 3x = 22 3x + 28 = 22 subtract 28 from both sides: 3x = 6 3y divide both sides by 3: x = 2

13 Copy into your notes 13 of 43 Step 7: Check your answers. We can check whether x = 2 and y = 4 solves both, 3x + 7y = 22 3x + 4y = 10 by substituting them into the second equation = = 22 This is true and so, x = 2 y = 4

14 14 of 43 Example 3: Elimination method Solve the simultaneous equations x + y = 3 and x + 5y = 11 Step 1: Rewrite both equations so the = sign is the same place Already done Step 2: Write one equation above the other x + y = 3 x + 5y = 11 Step 3: Identify the letter to eliminate. It is the letter with the same number in front (coefficient), regardless of sign. Both x s have the same coefficient so eliminate x

15 15 of 43 Step 4: Re coefficient: Sign Same Subtract SSS Sign Different Add SDA As the signs are the same: subtract Notice it is easier to subtract top from bottom Step 5: Find the value of the first letter. 4y = 8 divide both sides by 4: y = 2 x + y = 3 x + 5y = 11 4y = 8 Step 6: Substitute to find the value of the second letter. Substituting y = 2 into the first equation gives us, x + 2 = 3 x = 1

16 16 of 43 Step 7: Check your answers. We can check whether x = 1 and y = 2 solves both, x + y = 3 x + 5y = 11 by substituting them into the second equation. This is true and so, x 2 = = 11 x = 2 y = 1

17 17 of 43 Elimination: coefficient the same

18 18 of 43 Your Turn Question 7x + 2y = 24 8x + 2y = 30 2x + 3y = 8 x 3y = 14-4x 2y = -12 4x + 8y = -24 x y = 11 2x + y = 19 Answer 6 and -9-2 and 4-6 and 6 10 and -1

19 19 of 43 Questions to do from the books Elimination method Gamma Achieve Merit Excellence P78 Ex6.01 Q2 16 P78 Ex6.02 CAT 1.2 P45 Q Merit students: do a couple questions only as we need to move on to harder work.

20 SLO Merit: Elimination Method 20 of 43 (Merit: Elimination method (multiply first))

21 21 of 43 Copy into your notes Elimination method Step 1: Rewrite both equations so the = sign is the same place Step 2: Write one equation above the other Step 3: Identify the letter to eliminate. It is the letter with the same number in front (coefficient), regardless of sign. Step 3a: If letters are not the same, multiply one or both equations. Step 4: Re coefficient: Sign Same Subtract SSS Sign Different Add SDA Step 5: Find the value of the first letter. Step 6: Substitute to find the value of the second letter. Step 7: Check your answers.

22 No common coefficients Bob goes to New World and buys the following for $11 + = 11 If Jill buys double what Bob did of 43 How much did Jill spend? $22

23 Copy into your notes 23 of 43 Not all simultaneous can be eliminated straight away e.g. 4x y = 29 3x + 2y = To keep track of equations, number them If we want to eliminate the y s, multiply 8x 2y = 58 3x + 2y = 19 1 by 2 to give What do we need to do to eliminate the x s (from top 2 equations)? Multiply by 3 and multiply by x 3y = 87 12x + 8y = 76

24 24 of 43 Elimination method example 1 Solve 4x y = 29 3x + 2y = We need to have the same number in front of either the x or the y before adding or subtracting the equations. 2 1 : 8x 2y = x + 2y = : 11x = 77 x = 7 divide both sides by 11:

25 25 of 43 To find the value of y when x = 7 substitute this value into one of the equations, 4x y = 29 1 Substituting x = 7 into 1 gives, 3x + 2y = y = y = 29 subtract 28 from both sides: y = 1 multiply both sides by 1: y = 1 Check by substituting x = 7 and y = 1 into 2, = = 19

26 26 of 43 Copy into your notes Merit: Elimination example 2 Solve 5x + 4y = -14 3x + 6y = 6 To have the same number in front of x, multiply both equations. 3 1 : 15x + 12y = : 15x + 30y = : 18y = 72 y = 4 Equation 4 equation 3 avoids negative numbers 1 2 Take care with double negative: = = 72 divide both sides by 18:

27 Copy into your notes To find the value of y when y = 4 substitute this value into any of the equations, Substituting y = 4 into 2 gives, 3x + 6y = 6 2 3x + 6 x 4 = 6 3x + 24 = 6 subtract 24 from both sides: 3x = -18 divide both sides by 3: 27 of 43 x = -6 Check by substituting x = -6 and y = 4 into 2, 5 x x 4 = = -14 5x + 4y = -14

28 28 of 43 Elimination method example 3 Solve: 2x 5y = 25 3x + 4y = 3 Call these equations 1 and x 15y = x + 8y = 6 3 4, 23y = 69 y = 3 divide both sides by 23: Substitute y = 3 in 1, 2x 5 3 = 25 2x + 15 = 25 subtract 15 from both sides: 2x = 10 x = 5 divide both sides by 2:

29 The elimination: different coefficients 29 of 43

30 30 of 43 Your Turn Question 2x + 3y = 10 x + 5y = 26 4a + 3b = 17 6a 2b = 6 5p + 4q = 24 2p + 5q = x y = 0-3 7y = 10x Answer -4 and 6 2 and 3 4 and 1-1 and 1

31 31 of 43 Questions to do from the books Elimination method Gamma Achieve Merit Excellence P78 Ex6.01 Q2 16 P78 Ex6.02 P80 Ex6.03 CAT 1.2 P45 Q P45 Q

32 32 of 43 SLO Achieve: substitution method (no rearrangement first)

33 33 of 43 Substitution method Substitution method involves taking one of the equations and substituting into the other. y = x - 1 2x - 3 y = -1 ( ) We need to be left with one equation with one unknown. Expand and solve. 2x 3x + 3 = -1 -x = -4 x = 4 Substitute x = 4 into y = x 1 and solve for y y = 4 1 Therefore y = 3 Check in both equations that solutions work.

34 34 of 43 Substitution method example 1 Solve y = 2x 3 2x + 3y = Substitute equation 1 into equation 2. 2x + 3(2x 3) = 23 Solve for x expand the brackets: 2x + 6x 9 = 23 simplify: 8x 9 = 23 add 9 to both sides: 8x = 32 divide both sides by 8: x = 4

35 35 of 43 To find the value of y when x = 4 substitute this value into one of the equations, y = 2x 3 1 Substituting x = 4 into 1 gives 2x + 3y = 23 2 y = y = 5 Check by substituting x = 4 and y = 5 into 2, = = 23

36 36 of 43 Copy into your notes Substitution: example 2 Solve 4x 3y = 9 x = 2y Substitute equation 2 into equation 1. 4(2y + 1) 3y = 9 Solve for x expand the brackets: 8y + 4 3y = 9 simplify: 5y + 4 = 9 Subtract 4 from both sides: 5y = 5 divide both sides by 5: y = 1

37 Copy into your notes 37 of 43 To find the value of y when y = 1 substitute this value into one of the equations, 4x 3y = 9 1 Substituting y = 1 into 2 gives x = 2y x = x = 3 Check by substituting x = 3 and y = 1 into 1, = = 9

38 38 of 43 Your Turn Question 6x + 5y = -2 y = 2x + 6 y = -3x + 5 5x 4y = -3 6x + 6y = 0 Y = 2x 24 y = 5x 7-3x 2y = -12 Answer -2 and 2 1 and 2 8 and -8 2 and 3

39 39 of 43 Questions to do from the books Substitution method Gamma Achieve Merit Excellence P82 Ex6.04 P83 Ex6.05 Q1 9 CAT 1.2 P48 Q

40 SLO Merit: Substitution method (Rearrangement required first) 40 of 43 (merit: elimination method, multiplication first)

41 41 of 43 Merit Substitution method example 1 Solve 3x y = 9 8x + 5y = Unlike previous examples, 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. 3x y = 9 add y to both sides: subtract 9 from both sides: 3x = 9 + y 3x 9 = y y = 3x 9

42 3x y = 9 8x + 5y = Substitute y = 3x 9 into equation 2. 8x + 5(3x 9) = 1 Solve expand the brackets: 8x + 15x 45 = 1 simplify: 23x 45 = 1 add 45 to both sides: 23x = 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 42 of 43

43 43 of 43 3x y = 9 1 8x + 5y = 1 Check the solutions x = 2 and y = 3 by substituting them into equation 2. 8x + 5y = = = 1 This is true and so the solutions are correct. 2

44 Copy into your notes Merit Substitution example 2 Solve x + y = 3 x + 5y = 11 Rearrange one equation in the form x = or y = 1 2 Rearrange equation 1. x + y = 3 Take y from both sides: x = 3 y Substitute x = 3 y into equation 2. x + 5y = y + 5y = of 43 Solve simplify: 3 + 4y = 11 Subtract 3 from both sides: 4y = 8 divide both sides by 4: y = 2

45 45 of 43 Copy into your notes x + 5y = 11 x + 5y = Substitute y = 2 into equation 1 to find the value of x. x + y = 3 x + 2 = 9 x = 1 Check the solutions x = 1 and y = 2 by substituting them into equation 2. x + 5y = = = 11

46 Merit Substitution example 3 46 of 43 Solve -3x 3y = 3 y + 5x = -17 Rearrange one equation in the form x = or y = Rearrange equation 2. y = -5x 17 Substitute y = -5x 17 into equation 1. -3x 3(-5x 17) = 3 Solve x = -4 Substitute x = -4 into equation 3 to find the value of y y = 3 Check the solutions x = -4 and y = 3 by substituting them into equation 1. -3x 3y = 3-3(-4) 3(3) = = 3

47 47 of 43 Your Turn Question 2x + y = 20 6x 5y = 12 6x + 6y = -6 5x + y = -13-3x 4y = 2 3x + 3y = -3-2x + 6y = 6-7x + 8y = -5 Answer 7 and 6-3 and 2-2 and 1 3 and 2 Harder

48 48 of 43 Questions to do from the books Substitution method Gamma Achieve Merit Excellence P82 Ex6.04 P83 Ex6.05 Q1 9 P83 Ex6.05 Q10 12 P83 Ex6.06 Q2 6, 8 10 CAT 1.2 P48 Q P48 Q Do by substitution

49 Copy into your notes SLO Merit: Word Problems 49 of 43 (word simultaneous problems)

50 50 of 43 Solving problems example 1 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. 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

51 51 of 43 a + b = 56 a b = 22 Substituting a = 39 into the first equation gives, 39 + b = 56 subtract 39 from both sides: b = 17 So the two numbers are 39 and 17. We can check these solutions by substituting them into the second equation, a b = 22: = 22 This is true and so our solution is correct.

52 52 of 43 Copy into your notes 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.

53 53 of 43 Copy into your notes Solving problems example 2 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, 4a + 3c = Dividing equation 2 by 2 gives, 2a + 6c = 44 2 a + 3c = 22 3 We can now subtract equation 3 from equation 1 to eliminate the terms containing c.

54 Copy into your notes 54 of , divide both sides by 3: 4a + 3c = a + 3c = a = a = 8.50 Substitute a = 8.50 in 3 : subtract 8.50 from both sides: divide both sides by 3: c = 22 3c = c = 4.50 The cost of an adult s ticket is $8.50 and the cost of a child s ticket is $4.50.

55 55 of 43 Questions to do from the books Word problems Gamma Achieve Merit Excellence P86 Ex6.08 P86 Ex6.09 CAT 1.2 P57 Q