1 / 28

Solving Simultaneous Equations - Methods and Examples

Learn different methods to solve pairs of simultaneous equations with step-by-step examples. Practice solving quadratic equations with linear equations.

tyokoyama
Download Presentation

Solving Simultaneous Equations - Methods and Examples

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1. Use a suitable method to solve the following pairs of simultaneous equations: x = y + 1 (i) x + y2 = 3 Substitute x = y + 1 into the quadratic equation: x + y2 = 3  (y + 1) + y2 = 3 y2 + y – 2 = 0 (y + 2)(y – 1) = 0 y + 2 = 0 and y – 1 = 0 y = – 2 and y = 1 x = y + 1 y = –2 x = –2 + 1 = –1  (– 1, – 2) y = 1 x = 1 + 1 = 2  (2, 1)

  2. 1. Use a suitable method to solve the following pairs of simultaneous equations: x = y (ii) x2 + 2xy = 3 Substitute x = y into the quadratic equation: x2 + 2xy = 3 x2 + 2x(x) = 3 x2 + 2x2 = 3 3x2 = 3 x2 = 1 x = – 1 and x = 1 x = y x = – 1 y = – 1  ( – 1, – 1) x = 1 y = 1  (1, 1)

  3. 1. Use a suitable method to solve the following pairs of simultaneous equations: x + y = 5 (iii) x2 + y2 = 13 Rearrange the linear equation: x + y = 5 y = 5 – x Substitute this value in for y, in the quadratic equation: x2 + y2 = 13 x2 + (5 – x)2 = 13

  4. 1. Use a suitable method to solve the following pairs of simultaneous equations: x + y = 5 (iii) x2 + y2 = 13 x2 + y2 = 13 x2 + (5 – x)2 = 13 x2 + (5 – x)(5 – x) = 13 x2 + [5(5 – x) – x(5 – x)] = 13 x2 + [25 – 5x – 5x + x2] = 13 2x2 – 10x + 25 = 13 2x2 – 10x + 12 = 0 x2 – 5x + 6 = 0 (x – 2)(x – 3) = 0

  5. 1. Use a suitable method to solve the following pairs of simultaneous equations: x + y = 5 (iii) x2 + y2 = 13 (x – 2)(x – 3) = 0 x – 2 = 0 and x – 3 = 0 x = 2 and x = 3 x + y = 5 y = 5 – x x = 2 y = 5 – 2 = 3  (2, 3) x = 3 y = 5 – 3 = 2  (3, 2)

  6. 2. Use a suitable method to solve the following pairs of simultaneous equations: a − b = −4 (i) a2 + 4b2 = 37 Rearrange the linear equation: a – b = – 4 a = b – 4 Substitute a = b – 4 into the quadratic equation: a2 + 4b2 = 37  (b – 4)2 + 4b2 = 37

  7. 2. Use a suitable method to solve the following pairs of simultaneous equations: a − b = −4 (i) a2 + 4 b2 = 37 a2 + 4b2 = 37  (b – 4)2 + 4b2 = 37 (b – 4)(b – 4) + 4b2 = 37 [b(b – 4) – 4(b – 4)] + 4b2 = 37 b2 – 4b – 4b + 16 + 4b2 = 37 5b2 – 8b + 16 = 37 5b2 – 8b – 21 = 0 (5b + 7)(b – 3) = 0

  8. 2. Use a suitable method to solve the following pairs of simultaneous equations: a − b = −4 (i) a2 + 4 b2 = 37 (5b + 7)(b – 3) = 0 5b + 7 = 0 and b – 3 = 0 5b = – 7 and b = 3

  9. 2. Use a suitable method to solve the following pairs of simultaneous equations: 2x − y = 5 (ii) 2 x2 − y2 = 7 Rearrange the linear equation: 2x – y = 5 y = 2x – 5 Substitute y = 2x – 5 into the quadratic equation: 2x2 – y2 = 7  2x2 – (2x – 5)2 = 7

  10. 2. Use a suitable method to solve the following pairs of simultaneous equations: 2x − y = 5 (ii) 2 x2 − y2 = 7 2x2 – y2 = 7  2x2 – (2x – 5)2 = 7 2x2 – [(2x – 5)(2x – 5)] = 7 2x2 – [2x(2x – 5) – 5(2x – 5)] = 7 2x2 – [4x2 – 10x – 10x + 25] = 7 2x2 – 4x2 + 10x + 10x – 25 = 7 – 2x2 + 20x – 25 = 7 – 2x2 + 20x – 32 = 0 – x2 + 10x – 16 = 0

  11. 2. Use a suitable method to solve the following pairs of simultaneous equations: 2x − y = 5 (ii) 2 x2 − y2 = 7 x2 – 10x + 16 = 0 (x – 2)(x – 8) = 0 x – 2 = 0 and x – 8 = 0 x = 2 and x = 8 2x – y = 5 y = 2x – 5 x = 2 y = 2(2) – 5 = 4 – 5 = – 1  (2, – 1) x = 8 y = 2(8) – 5 = 16 – 5 = 11  (8, 11)

  12. 2. Use a suitable method to solve the following pairs of simultaneous equations: p − 2q = −4 (iii) p2 + q2 = 16 Rearrange the linear equation: p – 2q = – 4 p = 2q – 4 Substitute p = 2q – 4 into the quadratic equation: p2 + q2 = 16  (2q – 4)2 + q2 = 16

  13. 2. Use a suitable method to solve the following pairs of simultaneous equations: p − 2q = −4 (iii) p2 + q2 = 16 p2 + q2 = 16  (2q – 4)2 + q2 = 16 (2q – 4)(2q – 4) + q2 = 16 2q(2q – 4) – 4(2q – 4) + q2 = 16 4q2 – 8q – 8q + 16 + q2 = 16 5q2 – 16q + 16 = 16 5q2 – 16q = 0 q(5q – 16) = 0

  14. 2. Use a suitable method to solve the following pairs of simultaneous equations: p − 2q = −4 (iii) p2 + q2 = 16 q(5q – 16) = 0 q = 0 and 5q – 16 = 0 5q = 16

  15. 3. Use a suitable method to solve the following pairs of simultaneous equations: x − y = 6 (i) x2 − 2x − y = 10 Rearrange the linear equation: x – y = 6 y = x – 6 Substitute y = x – 6 into the quadratic equation: x2 – 2x – y = 10 x2 – 2x – (x – 6) = 10 x2 – 2x – x + 6 = 10 x2 – 3x – 4 = 0 (x + 1)(x – 4) = 0 x + 1 = 0 and x – 4 = 0 x = – 1 and x = 4

  16. 3. Use a suitable method to solve the following pairs of simultaneous equations: x − y = 6 (i) x2 − 2x − y = 10 x – y = 6 y = x – 6 x = – 1 y = – 1 – 6 = –7  ( – 1, – 7) x = 4 y = 4 – 6 = –2  (4, – 2)

  17. 3. Use a suitable method to solve the following pairs of simultaneous equations: a − 2b = − 5 (ii) a2 + ab− b2 = 5 Rearrange the linear equation: a – 2b = – 5 a = 2b – 5 Substitute a = 2b – 5 into the quadratic equation: a2 + ab – b2 = 5  (2b – 5)2 + (2b – 5)b – b2 = 5

  18. 3. Use a suitable method to solve the following pairs of simultaneous equations: a − 2b = − 5 (ii) a2 + ab− b2 = 5 a2 + ab – b2 = 5  (2b – 5)2 + (2b – 5)b – b2 = 5 (2b – 5)(2b – 5) + 2b2 – 5b – b2 = 5 2b(2b – 5) – 5(2b – 5) + 2b2 – 5b – b2 = 5 4b2 – 10b – 10b + 25 + 2b2 – 5b – b2 = 5 5b2 – 25b + 25 = 5 5b2 – 25b + 20 = 0 b2 – 5b + 4 = 0 (b – 4)(b – 1) = 0

  19. 3. Use a suitable method to solve the following pairs of simultaneous equations: a − 2b = − 5 (ii) a2 + ab− b2 = 5 (b – 4)(b – 1) = 0 b – 4 = 0 and b – 1 = 0 b = 4 and b = 1 a – 2b = – 5 a = 2b – 5 b = 4 a = 2(4) – 5 = 8 – 5 = 3  (3, 4) b = 1 a = 2(1) – 5 = 2 – 5 = – 3  (– 3, 1)

  20. 3. Use a suitable method to solve the following pairs of simultaneous equations: x + y = −2 (iii) x2 + 2xy + 2y2 = 8 Rearrange the linear equation: x + y = – 2 x = – 2 – y Substitute x = – 2 – y into the quadratic equation: x2 + 2xy = 2y2 = 8  (– 2 – y)2 + 2(– 2 – y)y + 2y2 = 8

  21. 3. Use a suitable method to solve the following pairs of simultaneous equations: x + y = −2 (iii) x2 + 2xy + 2y2 = 8 x2 + 2xy = 2y2 = 8  (– 2 – y)2 + 2(– 2 – y)y + 2y2 = 8 (– 2 – y)(– 2 – y) + (– 4 – 2y)y + 2y2 = 8 – 2(– 2 – y) – y(– 2 – y) – 4y – 2y2 + 2y2 = 8 4 + 2y + 2y + y2 – 4y = 8 y2 + 4 = 8 y2 – 4 = 0 (y + 2)(y – 2) = 0 y = – 2 and y = 2

  22. 3. Use a suitable method to solve the following pairs of simultaneous equations: x + y = −2 (iii) x2 + 2xy + 2y2 = 8 x + y = – 2 x = – 2 – y y = – 2 x = – 2 – (–2) = – 2 + 2 = 0  (0, – 2) y = 2 x = – 2 – (2) = –2 – 2 = – 4  (– 4, 2).

  23. 4. (i) Solve the following simultaneous equations: 3x − y = −2 and y = x2 + 3x + 2 Rearrange the linear equation: 3x – y = – 2 y = 3x + 2 Substitute y = 3x + 2 into the quadratic equation: y = x2 + 3x + 2  3x + 2 = x2 + 3x + 2 0 = x2 0 = x 3x – y = – 2 y = 3x + 2 x = 0 y = 3(0) + 2 = 0 + 2 = 2  (0, 2)

  24. 4. (ii) Which of the sketches below represents the graphs of the equations in part (i)? Give a reason for your answer. (a) Graph(C) (b) There is only one solution to the simultaneous equations, which means the line must be a tangent to the curve (they only have one point of intersection). Also, the point of intersection, (0, 2) is on the y-axis. (c)

  25. 5. (i) Solve the following simultaneous equations: x − 3y = − 5 and x2 + y2 = 25 Rearrange the linear equation: x – 3y = – 5 x = 3y – 5 Substitute x = 3y – 5 into the quadratic equation: x2 + y2 = 25  (3y – 5)2 + y2 = 25 (3y – 5)(3y – 5) + y2 = 25 3y(3y – 5) – 5(3y – 5) + y2 = 25 9y2 – 15y – 15y + 25 + y2 = 25 10y2 – 30y + 25 = 25 10y2 – 30y = 0 y2 – 3y = 0 y(y – 3) = 0

  26. 5. (i) Solve the following simultaneous equations: x − 3y = − 5 and x2 + y2 = 25 y(y – 3) = 0 y = 0 and y – 3 = 0 y = 3 x – 3y = – 5 x = 3y – 5 y = 0 x = 3(0) – 5 = 0 – 5 = –5  (– 5, 0) y = 3 x = 3(3) – 5 = 9 – 5 = 4  (4, 3)

  27. 5. (ii) Which of the sketches below represents the graphs of the equations in part (i)? Give a reason for your answer. (a) Graph (b) The two solutions from the simultaneous equations are the points where the line and the circle intersect. (b) Only one of the diagrams appears to have a (– 5, 0) and (4, 3) as the points of intersection of the line and circle. (c)

More Related