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Solving Systems Using Elimination

Solving Systems Using Elimination. ALGEBRA 1 LESSON 7-3. (For help, go to Lesson 7-2.). Solve each system using substitution. 1. y = 4 x – 3 2. y + 5 x = 4 3. y = –2 x + 2 y = 2 x + 13 y = 7 x – 20 3 x – 17 = 2 y. 7-3. Solving Systems Using Elimination. ALGEBRA 1 LESSON 7-3.

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Solving Systems Using Elimination

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  1. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 (For help, go to Lesson 7-2.) Solve each system using substitution. 1.y= 4x – 32.y+ 5x = 43. y= –2x + 2 y= 2x + 13 y= 7x – 20 3x – 17 = 2y 7-3

  2. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 Solutions 1.y= 4x – 3 y= 2x + 13 Substitute 4x – 3 for y in the second equation. y= 2x + 13 4x – 3 = 2x + 13 4x – 2x – 3= 2x – 2x + 13 2x – 3= 13 2x = 16 x = 8 y= 4x – 3 = 4(8) – 3 = 32 – 3 = 29 Since x = 8 and y = 29, the solution is (8, 29). 7-3

  3. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 Solutions(continued) 2.y+ 5x = 4 y= 7x – 20 Substitute 7x – 20 for y in the first equation. y+ 5x = 4 7x – 20 + 5x = 4 12x – 20 = 4 12x= 24 x = 2 y= 7x – 20 = 7(2) – 20 = 14 – 20 = –6 Since x = 2 and y = –6, the solution is (2, –6). 7-3

  4. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 Solutions(continued) 3.y= –2x + 2 3x– 17 = 2y Substitute –2x + 2 for y in the second equation. 3x– 17 = 2y 3x– 17 = 2(–2x + 2) 3x– 17 = –4x + 4 7x– 17 = 4 7x = 21 x = 3 y= –2x + 2 = –2(3) + 2 = –6 + 2  –4 Since x = 3 and y = –4, the solution is (3, –4). 7-3

  5. Benchmark Hints

  6. Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 pages 349–352  Exercises 1. D 2. C 3. B 4. A 5–16.Coordinates given in alphabetical order. 5. (9, 28) 6. (– , –4 ) 7. (6 , – ) 8. (2, 4 ) 9. (4, 20) 10. ( , 9 ) 11. (2, 0) 12. (7 , 11 ) 13. (6, –2) 14. (3, –2) 15. (8, –7) 16. (–3, 9.4) 17. 4 cm by 13 cm 18. 4 wk 19. (15, 15) 20. (9, 126) 21. (–4, 4) 3 4 3 8 22. 15 video rentals 23. 80 acres flax, 160 acres sunflowers 24. 9 yr 25. estimate: ( , 1);  ; ( , 1) 7 17 8 17 1 2 1 2 1 2 1 2 1 3 1 3 1 2 7-2

  7. Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 28. estimate: (–3.5, –3.5); (– , – ) 29. estimate: (– , 4 ); (– , ) 26. estimate: (–2, 3);  ; (–2, 3) 27. estimate: (–1, 1); (–1, 1) 10 3 11 3 3 4 3 4 2 3 14 3 7-2

  8. = / Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 38. (2, ) 39. (– , 0) 40. (4, –2) 41. inf. many solutions 42. no solution 43. 1 solution 44. a.g = b b. (b, g) = (572, 598) c. 26 45.a. (t, d) = (9, 79.2) b. yes 46. (r, s, t) = (7, 9, 4) 47. 29.8 48. 4803 49. 520 50.[2] 7(–7) – 4(–2) 29       –49 + 8 29            –41 29 No, (–2, –7) must satisfy both equations to be a solution of the system. [1] no explanation given 1 2 34.a. no solution b. c. Graphing shows 2 parallel lines. Substitution results in a false equation with no variables. 35. (2, 4) 36. (– , – ) 37. (2, –4) 1 2 23 22 1 2 1 2 7-2

  9. WB 7.2 Answers • Practice 7-2 First Column Only! • (1, 1) 4. (-3, 2) • 7. (5,-2) 10. no solution • 13. infinitely many solutions 16. (-2,-2) • 19. (4, 8) 22. (-3, 0) • 25. (3, -2/3) 28. infinitely many solutions • 31. (10.5, 8.2) 34. (28,-36) • 37. 88 cones 38. paint: $17/gal, brush: $5.50 OMIT

  10. WB 7.2 Answers • Practice 7-2 Multiples of 3 • 3. (5, 5) 6. (7, 4) • 9. (100, 50) 12. (-2,-3) • 15. (1,-2) 18. infinitely many solutions • 21. no solution 24. infinitely many solutions • 27. (6.7, 2.4) 30. no solution • 33. Infinitely many solutions 36. (-18,-30)

  11. Solving Systems Using Substitution ALGEBRA 1 LESSON 7-2 Solve each system using substitution. 1. 5x+ 4y = 52. 3x+ y = 43. 6m– 2n = 7 y = 5x 2x– y = 6 3m+ n = 4 (0.2, 1) (2, 2) (1.25, 0.25) 7-2

  12. 7.3 Solving Systems Using Elimination p. 353 12 Apr 13 • Obj: To solve systems by adding or subtracting. • Elimination Method: A method for solving a system of linear equations. You add or subtract the equations to eliminate a variable.

  13. 2x+ 3y = 11 –2x+ 9y =1 0+ 12y = 12 Addition Property of Equality y = 1 Solve for y. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 Solve by elimination. 2x+ 3y = 11 –2x+ 9y = 1 Step 1: Eliminate x because the sum of the coefficients is 0. Step 2: Solve for the eliminated variable x using either original equation. 2x+ 3y = 11 Choose the first equation. 2x+ 3(1) = 11 Substitute 1 for y. 2x+ 3 = 11 Solve for x. 2x= 8 x= 4 7-3

  14. Check: See if (4, 1) makes true the equation not used in Step 2. –2(4) + 9(1) 1 Substitute 4 for x and 1 for y into the second equation. –8 + 9 1 1 = 1 Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 (continued) Since x= 4 and y = 1, the solution is (4, 1). 7-3

  15. Define: Let a= number of adults Let s = number of students Relate: total number at the game total amount collected Write: a + s = 1139 5 a + s = 3067 Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 On a special day, tickets for a minor league baseball game were $5 for adults and $1 for students. The attendance that day was 1139, and $3067 was collected. Write and solve a system of equations to find the number of adults and the number of students that attended the game. Solve by elimination. 7-3

  16. Solving Systems Using Elimination ALGEBRA 1 LESSON 7-3 (continued) Step 1: Eliminate one variable. a+ s = 1139 5a+ s = 3067 –4a+ 0 = –1928 Subtraction Property of Equality a = 482 Solve for a. Step 2: Solve for the eliminated variable using either of the original equations. a+ s = 1139 Choose the first equation. 482+ s = 1139 Substitute 482 for a. s = 657 Solve for s. There were 482 adults and 657 students at the game. Check: Is the solution reasonable? The total number at the game was 482 + 657, or 1139. The money collected was $5(482), or $2410, plus $1(657), or $657, which is $3067. The solution is correct. 7-3

  17. ASSIGNMENTS: • CW: p. 356 # 1-8 all • HW: WB 7.3 #1,2,3,7,8,9,11, & 12

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