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Let's re-examine the problem with the correct setup and conventions to ensure clarity and consistency.

Given:

1. Initial depth: 15 meters below sea level.
2. Change in depth during the second day is one less unit than three times the change in depth during the first day.
3. If you quadruple the change in depth during the first day and add the change in depth during the second day, you get a result of going down 8 meters.

Goal:

Determine the current depth of the submarine.

Approach:

1. Define variables:

   • Let $x$ be the change in depth during the first day (positive for going down, negative for going up).
   • Let $y$ be the change in depth during the second day (positive for going down, negative for going up).

2. Formulate the equations based on the given conditions:

   • From the second condition: $y = 3x - 1$
   • From the third condition: $4x + y = 8$ (since "going down" means a positive change in depth)

3. Solve the system of equations to find $x$ and $y$.

Solution:

Let's solve the system of equations:

1. Substitute $y = 3x - 1$ into $4x + y = 8$: $4x + (3x - 1) = 8$

2. Simplify and solve for $x$: $4x + 3x - 1 = 8$ $7x - 1 = 8$ $7x = 9$ $x = \frac{9}{7}$

3. Substitute $x = \frac{9}{7}$ back into $y = 3x - 1$: $y = 3\left(\frac{9}{7}\right) - 1$ $y = \frac{27}{7} - 1$ $y = \frac{27}{7} - \frac{7}{7}$ $y = \frac{20}{7}$

Final Depth Calculation:

• Initial depth: 15 meters below sea level.
• Change in depth during the first day: $x = \frac{9}{7}$ meters (going down).
• Change in depth during the second day: $y = \frac{20}{7}$ meters (going down).

Thus, the final depth is: $15 + \frac{9}{7} + \frac{20}{7} = 15 + \frac{29}{7} = \frac{105}{7} + \frac{29}{7} = \frac{134}{7} \approx 19.14 \text{ meters below sea level}$

Conclusion:

The correct depth of the submarine is approximately 19.14 meters below sea level, or exactly $\frac{134}{7}$ meters.

Summary:

• The correct interpretation of the conditions leads to the equations $y = 3x - 1$ and $4x + y = 8$.
• Solving these equations gives $x = \frac{9}{7}$ and $y = \frac{20}{7}$.
• The final depth of the submarine is $\frac{134}{7}$ meters below sea level, which is approximately 19.14 meters.

Your colleagues' interpretation of the sign convention leads them to a different setup, but as long as the conventions are consistent, both methods are valid. However, using the depth-positive-down convention, the final depth is indeed approximately 19.14 meters.