1 / 4

Solving Systems Using Elimination

Solving Systems Using Elimination. ALGEBRA 1 LESSON 7-3. pages 356–359  Exercises 1. (1, 3) 2. (2, –2) 3. (5, –17) 4. (–3, 4) 5. (–9, ) 6. (– , 10) 7. a. x + y = 20, x – y = 4 b. 12 and 8 8. a. a + s = 456, 3.5 a + s = 1131 b. 270 adult, 186 student

bohanan
Download Presentation

Solving Systems Using Elimination

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. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 pages 356–359  Exercises 1. (1, 3) 2. (2, –2) 3. (5, –17) 4. (–3, 4) 5. (–9, ) 6. (– , 10) 7.a.x + y = 20, x – y = 4 b. 12 and 8 8.a.a + s = 456, 3.5a + s = 1131 b. 270 adult, 186 student 9. (–5, 1) 10. (11, –3) 11. (–2, – ) 12. (1, 14) 13. ( , 1) 14. (–2, 3) 15.a. 30w + = 17.65, 20w + 3 = 25.65 b. $.39 for a wallet size, $5.95 for an 8  10 16.a.x = burritos, y = tacos; 3x + 4y = 11.33, 9x + 5y = 23.56 b. $1.79 for a burrito, $1.49 for a taco 17. (–1, –3) 18. (2.5, 1) 19. (2, –2) 5 2 1 2 1 2 1 2 7-3

  2. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 20. (10, 8) 21. (–1, ) 22. (1, 5) 23–28.Methods may vary. Samples are given. 23. (–1, –2); substitution; both solved for y 24. (15, –10); elimination; equations not solved for y 25. (10, 2); substitution; one eq. solved for x 26. (–3, 11); elimination; eqs. not solved for a variable 27. (5, 1); substitution; one eq. solved for x 28. ( , 2 ); substitution; eqs. solved for y 29. one night: $81.25; one meal: $8.13 30.a. brass: $6; steel: $3 b. $99 31. She forgot to multiply –8 by 6. 32. Answers may vary. Sample: 2x – 3y = 6, x + 3y = 9; (5, ) 33. (10, –6) 34. (8, 12) 3 2 1 3 1 3 4 3 7-3

  3. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 35. (–15, –1) 36. (20, 12) 37. (–2, 16) 38. (33, –48) 39. 9 40. Answers may vary. Sample: You solve a system using the elimination method by adding or subtracting the eqs. to eliminate one of the variables. This sum or difference is one eq. with one variable that can be solved. Use addition: Use subtraction: Use multiplication: 3x + 2y = 6 5x + 3y = 15 4x + 5y = 20 –x – 2y = 4 5x – 2y = 10 2x – y = 10 41.B1 = 3 volts; B2 = 1.5 volts 42. (–3, 2) 43. ( , 0) (a = 0, b = 1) 44. (8, 13, 20) 45. CD: $3.40, cassette: $1.80 46. a. 2.8 g of gold b. about 2.7% 47. D 48. H c a / / 7-3

  4. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 51–53.Coordinates given in alphabetical order. 51. (6, 26) 52. (–1, 4) 53. (9, –5) 54. 55. 56. 57. 71 58. –18 59. –44 60. 22 61. 83 62. –6 49.[2]y – x = 13 7y + x = 11 8y = 24 y = 3  3 – x = 13 x = –10; (–10, 3) [1] no work shown 50.[4] A = h(b1 + b2) = (4)(4 + 2) = 2(6) = 12 The area is 12 square units. [3] graph and formula with one mathematical error [2] graph but error in formula [1] graph only 1 9 2 9 1 2 1 15 1 2 7-3

More Related