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We are given the equation \( y^2 + xy - 3x = 17 \), along with information that \( \frac{dy}{dt} = 3 \) when \( x = 1 \) and \( y = -5 \). The task is to find \( \frac{dx}{dt} \) under these same conditions.

### Step-by-Step Process:

1. **Differentiate both sides of the equation with respect to \( t \):**
   \[
   \frac{d}{dt}(y^2 + xy - 3x) = \frac{d}{dt}(17)
   \]
   Using the chain rule and product rule, we get:
   \[
   2y \frac{dy}{dt} + \left( x \frac{dy}{dt} + y \frac{dx}{dt} \right) - 3 \frac{dx}{dt} = 0
   \]
   Simplifying, this becomes:
   \[
   2y \frac{dy}{dt} + x \frac{dy}{dt} + y \frac{dx}{dt} - 3 \frac{dx}{dt} = 0
   \]

2. **Substitute the given values:**
   We are given \( x = 1 \), \( y = -5 \), and \( \frac{dy}{dt} = 3 \). Substituting these into the differentiated equation:
   \[
   2(-5)(3) + (1)(3) + (-5) \frac{dx}{dt} - 3 \frac{dx}{dt} = 0
   \]
   Simplifying the constants:
   \[
   -30 + 3 - 5 \frac{dx}{dt} - 3 \frac{dx}{dt} = 0
   \]
   This simplifies to:
   \[
   -27 - 8 \frac{dx}{dt} = 0
   \]

3. **Solve for \( \frac{dx}{dt} \):**
   \[
   -8 \frac{dx}{dt} = 27
   \]
   \[
   \frac{dx}{dt} = -\frac{27}{8}
   \]

Thus, the value of \( \frac{dx}{dt} \) is \( -\frac{27}{8} \).

Would you like more details on any of the steps or have any further questions?

### Related Questions:
1. How does the product rule apply when differentiating terms like \( xy \)?
2. What does it mean physically if \( \frac{dx}{dt} \) is negative?
3. How can you verify if the differentiation was done correctly in this problem?
4. How would this problem change if \( \frac{dy}{dt} \) were a different value?
5. Could you solve this problem using implicit differentiation?

### Tip:
When solving related rates problems, always ensure that every variable dependent on time is differentiated with respect to time, using the chain rule!

If y^2 + xy - 3x = 17, and dy/dt = 3 when x = 1 and y = -5, what is dx/dt when x = 1 and y = -5?

This related rates problem involves differentiating the equation y^2 + xy - 3x = 17 using implicit differentiation. Given that dy/dt = 3 when x = 1 and y = -5, the task is to find dx/dt. The problem applies the chain rule and product rule, providing a detailed solution for students at a Grade 12 or early college calculus level.

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