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Solving Simultaneous Equations

Solving Simultaneous Equations. Algebraically. What is to be learned. How to solve simultaneous equations by using algebra . 2x + 2 = 8 x = 3 2x + y = 8 x = 3 y = 2 or x = 4, x = 2, etc Need More Information. y = 0. y = 4. 2x + y = 8

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Solving Simultaneous Equations

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  1. Solving Simultaneous Equations Algebraically

  2. What is to be learned • How to solve simultaneous equations by using algebra

  3. 2x + 2 = 8 x = 3 2x + y = 8 x = 3 y = 2 or x = 4, x = 2, etc Need More Information y = 0 y = 4

  4. 2x + y = 8 and x + y = 5 x = 3, y = 2 Only solution that fits both equations

  5. Apples and Bananas 4 apples and 3 bananas cost 81p 2 apples and 2 bananas cost 46p Cost of 6 apples and 5 bananas? 127p (81 + 46) Cost of 2 apples and 1 banana 35p (81 – 46)

  6. As Equations 4a + 3b = 81 2a + 2b = 46 Adding 6a + 5b = 127 4a + 3b = 81 2a + 2b = 46 Subtracting 2a + b = 35 You can add or subtract equations

  7. Some Revision! 7 + (-5) = 7 – 5 6b + (-4b) = 6b – 4b = 2b 5r + (-5r) = 5r – 5r = 0r ↑ Zero ↑

  8. Solving Simultaneous Equations The Elimination Method 3a + 2b = 19 a – 2b = 1 Need to eliminate a or b To do so Add or Subtract the equations

  9. Adding a 3 +2b = 19 a – 2b = 1 Add 4a b has been eliminated! → a = 5 Need to find value of b 15 + 2b = 19 2b = 4 0b = 20 X5 b = 2

  10. 5t + 2r = 22 2t – 2r = 6 Add 7t = 28 t = 4 5t + 2r = 22 → 5X4 +2r = 22 → 20 + 2r = 22 → 2r = 2 r = 1 t=4, r=1

  11. Solving Simultaneous EquationsAlgebraically The Elimination Method Will have two different letters Must eliminate one of them

  12. Elimination by Adding * Ex 2x + 3y = 19 3x – 3y = 6 Add 5x = 25 x = 5 Sub x = 5 into equation*  2X5 + 3y = 19 10 + 3y = 19 3y = 9 y = 3 x = 5, y = 3 “y”s eliminated

  13. You Try 1. x + y = 8 x – y = 2 2. 2x + 3y = 13 5x – 3y = 1 3. 4x + 2y = 16 3x – 2y = 5 4. 5x + 3y = 2 2x – 3y = 5 x = 5, y =3 x = 2 , y = 3 x = 3, y =2 x = 1 , y = -1

  14. Subtracting 4y + 3d = 17 2y + 3d = 13 Subtract 2y = 4 y = 2 → 4X2 + 3d = 17 → 8 + 3d = 17 3d = 9 d = 3 Solution y = 2, d = 3

  15. You Try 1. 3x + y = 8 x + y = 4 2. 5x + 3y = 17 2x + 3y = 14 3. 4x + 2y = 18 3x + 2y = 14 4. 5x + 3y = 9 3x + 3y = 3 x = 2, y =2 x = 1 , y = 4 x = 4, y =1 x = 3 , y = -2

  16. Adding or Subtracting? 2m + 3t = 18 4f + 3p = 10 8m – 3t = 22f + 3p = 8 Add 10m = 20 Same Sign → Subtract Subtract 2f = 2

  17. Solving Simultaneous EquationsAlgebraically (continued) Can add or subtract the equations Same Sign  2m – 3t = 7 4f – 3p = 10 8m – 3t = 2 2f + 3p = 8 subtract Subtract Add

  18. A New Problem! 5t + 4r = 22 3t + 2r = 12  Same Sign → Subtract BUT 2t + 2r = 10  Can only eliminate if the values are the same (same number in front of letter)

  19. More Apples and Bananas 2 apples and 3 bananas cost 57p Cost of 4 apples and 6 bananas? 114 p (2 X 57) We are allowed to multiply!

  20. Multiplying to Eliminate X2 → 6t + 4r = 24 5t + 4r = 22 3t + 2r = 12 5t + 4r = 22 Must make the r values equal subtract t = 2 then continue as before

  21. And Finally Usually you need to multiply both equations 3s + 2r = 17 2s - 5r = 5

  22. And Finally Usually you need to multiply both equations 3s + 2r = 17 2s - 5r = 5 X5→ 15s + 10r = 85 X2→ 4s - 10r = 10 Add 19s = 95

  23. Solving Simultaneous EquationsAlgebraically (last bit) To eliminate a letter the amounts must be equal in both equations Multiply one (or both) of equations before eliminating

  24. Ex 3s - 3r = 9 7s - 4r = 18 X4→ 12s - 12r = 36 X3→ 21s - 12r = 54 Subtract 9s = 18 (bottom – top) s = 2 etc

  25. Constructing Simultaneous Equations length 92cm 4b + 3s = 92 length 1.4m 6b + 5s = 1.4 140 b = 20, s = 4 68cm length?

  26. Group of adults and children go to theatre 20 in group. Adults cost £5 and children £3 Total cost is £76 How many adults in group? a adults c children

  27. Group of adults and children go to theatre 20 in group. Adults cost £5 and children £3 Total cost is £76 How many adults in group? a + c = 20 a adults c children

  28. Group of adults and children go to theatre 20 in group. Adults cost £5 and children £3 Total cost is £76 How many adults in group? a + c = 20 a adults c children 10 adults  10X5 10 children  10X3 a adults  aX5 c children  cX3 children cost 3c adults cost 5a

  29. Group of adults and children go to theatre 20 in group. Adults cost £5 and children £3 Total cost is £76 How many adults in group? a + c = 20 5a + 3c = 76 a adults c children children cost 3c adults cost 5a 8 adults 12 children

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