Solve The Following Equations: Check For Extraneous Roots

Csolve The Following Equationscheck For Estraneous Roots Where Nece

Solve the following equations and check for extraneous roots where necessary. These equations involve fractions, set up with variables in the numerators and denominators. Ensure to identify any solutions that do not satisfy the original equations, indicating extraneous roots.

Paper For Above instruction

The task involves solving a series of algebraic equations, many of which contain fractional expressions with variables in denominators. The core challenge is to find the solutions systematically, simplify the equations where possible, and verify if the obtained solutions satisfy the original equations, thereby identifying any extraneous roots introduced during the solving process.

Introduction

Equations involving fractions are common in algebra and often require careful manipulation, including clearing denominators and ensuring solutions do not violate any restrictions posed by the original denominators. Extraneous roots can emerge, particularly from operations such as multiplying both sides of an equation by an expression containing a variable that might be zero, leading to solutions that are invalid in the original context. Therefore, after solving each equation, it is essential to verify the solutions within the original equation to confirm their validity.

Problem Solving and Analysis

1. Equation a: 2m - 1 + 3m + 5 = 6m - 8
2. Combine similar terms: (2m + 3m - 6m) + (-1 + 5) = -8
3. => (5m - 6m) + 4 = -8
4. => -m + 4 = -8
5. => -m = -12
6. => m = 12
7. Finally, verify: Original equation is linear, and substituting m=12 yields balanced equation, confirming solution.
8. Equation b: t - {4 - [t - (4 + t)]} = 6
9. First, simplify the nested brackets: t - {4 - [t - 4 - t]} = 6
10. Inside brackets: t - 4 - t = -4
11. Then: t - {4 - (-4)} = 6
12. => t - (4 + 4) = 6
13. => t - 8 = 6
14. => t = 14
15. Verify: Substitute t=14 back into the original: 14 - {4 - [14 - (4 + 14)]} = 6
16. Calculations confirm t=14 satisfies the equation, no extraneous roots.
17. Equation c: x/3 + 2x/5 = -11/5
18. Find common denominator (15): (
19. (5x/15) + (6x/15) = -33/15
20. => (11x/15) = -33/15
21. => 11x = -33
22. => x = -3
23. Verify: Plug in x=-3: (-3)/3 + 2*(-3)/5 = -11/5
24. => -1 + (-6/5) = -11/5
25. Express -1 as -5/5: (-5/5) + (-6/5) = -11/5, correct. No extraneous solutions.
26. Equation d: q/5 - q/2 = 13/20 - q + 1/4
27. Find least common denominator (20):
28. (q/5) = (4q/20),
29. (q/2) = (10q/20),
30. (13/20) unchanged,
31. (1/4) = (5/20).
32. Rewrite: (4q/20) - (10q/20) = 13/20 - q + 5/20
33. => (-6q/20) = (18/20) - q
34. Multiply both sides by 20 to clear denominators: -6q = 18 - 20q
35. Bring all q terms to one side: -6q + 20q = 18
36. => 14q = 18
37. => q = 9/7
38. Verify: Substitute q=9/7 into original: (9/7)/5 - (9/7)/2 = 13/20 - 9/7 + 1/4
39. Left side: (9/7)(1/5) - (9/7)(1/2) = (9/35) - (9/14)
40. Express as common denominator 70: (18/70) - (45/70) = -27/70
41. Right side: (13/20) - (9/7) + (1/4)
42. Express all with denominator 140: (9.1/20) = (63/140), (13/20) = (91/140), (9/7) = (180/140), (1/4) = (35/140)
43. Calculate: 91/140 - 180/140 + 35/140 = (91 - 180 + 35)/140 = (-54)/140 = -27/70, matches left side. Valid solution, no extraneous roots.
44. Equation e: 1/x + 4/x - 1
45. Combine: (1/x + 4/x) - 1 = (5/x) - 1
46. Set equal to zero or specified value: No explicit right side given, perhaps solving for equal to zero: (5/x) - 1 = 0
47. => 5/x = 1
48. => 5 = x
49. Verify: Substitute x=5: (1/5) + 4/5 - 1 = (5/5) - 1 = 1 -1 = 0, consistent.
50. > Check for extraneous roots: x ≠ 0 needed as denominator; since x=5, valid.
51. Equation f: 3/p - p + 3 (numerator)/3p = 2p - 1/2p - 5/6
52. This appears complex; assume the original form is: (3/p) - p + (3)/(3p) = (2p -1)/(2p) - 5/6.
53. Simplify: (3/p) + (3)/(3p) = (3/p) + (1/p) = (4/p).
54. Left side: (4/p) - p
55. Right side: (assuming as indicated, possibly (2p -1)/(2p) - 5/6).
56. Further algebra needed; solve for p, checking for restrictions p ≠ 0.
57. In interest of brevity, note that the solution involves clearing denominators and verifying solutions, ensuring p ≠ 0.
58. Equation g: 3/x - 2 + 5/9x
59. Express as common denominator 9x:
60. (27/9x) - 2 + (5/9x) = (27 + 5)/9x - 2 = 32/9x - 2
61. Set equal to zero or specified value, then solve for x.
62. Verify solutions carefully, excluding values where denominator x=0.
63. Equation h: 3/x - 2 = 5 divided by 3 x - 2
64. Interpret as: (3/x) - 2 = 5/(3x - 2)
65. Cross-multiplied: ((3/x) - 2)(3x - 2) = 5