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The problem presented involves determining which of the provided solutions is equivalent to the given parametric solution for a three-variable system. The solution is expressed in terms of a parameter \( t \) as follows:

\[
(x, y, z) = (t+1, 3t-10, 2t)
\]

### Problem Statement:
Select all that apply. The solution to a three-variable system is the following and it is defined as the parameter for \( t \):

\[
(x, y, z) = (t+1, 3t-10, 2t)
\]

Which of the following solutions is equivalent to the solution above?

(A) \((x, y, z) = (s-1, 3s + 3, 2s-20)\)

(B) \((x, y, z) = (r + 3, 3r - 7, 2r + 2)\)

(C) \((x, y, z) = (k+3, 3k - 7, 2k)\)

(D) \((x, y, z) = (m + 3, 3m + 16, 2m - 20)\)

(E) None of the above

### Solution:

We need to check which of these forms can be equivalent to the given parametric solution by substituting an appropriate parameter transformation.

**Option (A):**  
Given \((x, y, z) = (s-1, 3s + 3, 2s-20)\), we compare it to \((t+1, 3t-10, 2t)\).

- Comparing \( x \): \( t + 1 = s - 1 \) → \( t = s - 2 \).
- Substituting \( t = s - 2 \) in the expression for \( y \):
  \[
  y = 3t - 10 = 3(s - 2) - 10 = 3s - 6 - 10 = 3s - 16
  \]
  But the expression in the problem is \( 3s + 3 \), so Option (A) is **not equivalent**.

**Option (B):**  
Given \((x, y, z) = (r + 3, 3r - 7, 2r + 2)\):

- Comparing \( x \): \( t + 1 = r + 3 \) → \( t = r + 2 \).
- Substituting \( t = r + 2 \) in the expression for \( y \):
  \[
  y = 3t - 10 = 3(r + 2) - 10 = 3r + 6 - 10 = 3r - 4
  \]
  But the expression in the problem is \( 3r - 7 \), so Option (B) is **not equivalent**.

**Option (C):**  
Given \((x, y, z) = (k+3, 3k - 7, 2k)\):

- Comparing \( x \): \( t + 1 = k + 3 \) → \( t = k + 2 \).
- Substituting \( t = k + 2 \) in the expression for \( y \):
  \[
  y = 3t - 10 = 3(k + 2) - 10 = 3k + 6 - 10 = 3k - 4
  \]
  But the expression in the problem is \( 3k - 7 \), so Option (C) is **not equivalent**.

**Option (D):**  
Given \((x, y, z) = (m + 3, 3m + 16, 2m - 20)\):

- Comparing \( x \): \( t + 1 = m + 3 \) → \( t = m + 2 \).
- Substituting \( t = m + 2 \) in the expression for \( y \):
  \[
  y = 3t - 10 = 3(m + 2) - 10 = 3m + 6 - 10 = 3m - 4
  \]
  But the expression in the problem is \( 3m + 16 \), so Option (D) is **not equivalent**.

Since none of the options are equivalent to the given parametric solution, the correct answer is:

**(E) None of the above.**

The mark on the exam indicates the student correctly selected option (E), but there seems to have been confusion with the other choices. 

Would you like more details on this solution, or do you have any questions? Here are some related questions you might find interesting:

1. How do you determine if two parametric solutions are equivalent?
2. What are common mistakes students make when comparing parametric equations?
3. How can one simplify the process of comparing different forms of parametric solutions?
4. What role do parameter transformations play in equivalence of solutions?
5. Why is it important to thoroughly test each option in multiple-choice problems?

**Tip:** When comparing parametric forms, always consider a potential shift in the parameter that can align two equivalent expressions.

Which of the following solutions is equivalent to the given parametric solution for a three-variable system expressed as (x, y, z) = (t+1, 3t-10, 2t)?

This problem involves determining the equivalence of a parametric solution for a three-variable system expressed as (x, y, z) = (t+1, 3t-10, 2t). The solution uses substitution methods and explores the concepts of parametric equations and equivalence.

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