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The given equation is \( 2x + 3y = 12 \), and you are asked to find \( \frac{dx}{dt} \) given that \( \frac{dy}{dt} = -2 \).

To solve this, we will implicitly differentiate both sides of the equation with respect to \( t \).

### Step 1: Differentiate \( 2x + 3y = 12 \)
- The derivative of \( 2x \) with respect to \( t \) is \( 2 \frac{dx}{dt} \) (using the chain rule).
- The derivative of \( 3y \) with respect to \( t \) is \( 3 \frac{dy}{dt} \) (also using the chain rule).
- The derivative of the constant \( 12 \) is 0.

So, differentiating both sides of the equation:

\[
\frac{d}{dt}(2x + 3y) = \frac{d}{dt}(12)
\]

\[
2 \frac{dx}{dt} + 3 \frac{dy}{dt} = 0
\]

### Step 2: Substitute the given value \( \frac{dy}{dt} = -2 \)

\[
2 \frac{dx}{dt} + 3(-2) = 0
\]

Simplifying:

\[
2 \frac{dx}{dt} - 6 = 0
\]

### Step 3: Solve for \( \frac{dx}{dt} \)

\[
2 \frac{dx}{dt} = 6
\]

\[
\frac{dx}{dt} = \frac{6}{2} = 3
\]

### Final Answer:

\[
\frac{dx}{dt} = 3
\]

Would you like further details on this? Here are a few related questions to expand on this concept:

1. How do you apply implicit differentiation to functions of multiple variables?
2. What is the chain rule, and how is it used in this context?
3. Can you explain how to differentiate implicit equations with respect to time in general?
4. What does the result \( \frac{dx}{dt} = 3 \) mean in practical terms?
5. How would the solution change if \( \frac{dy}{dt} \) were positive instead of negative?

**Tip**: Always remember to apply the chain rule when differentiating variables that depend on another variable like time \( t \).

Assume that 2x + 3y = 12 and dy/dt = -2. Find dx/dt.

Learn how to find dx/dt using implicit differentiation, given 2x + 3y = 12 and dy/dt = -2. This problem demonstrates how to apply the chain rule in calculus, suitable for high school students studying derivatives.

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