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Let's go through the problem step by step.

### Problem 9: Express the equation 
\[
\frac{(x-6)^2}{16} + \frac{y^2}{12} = 1
\]
to its general form.

This equation is the standard form of an ellipse. To convert it to the general quadratic form \( Ax^2 + By^2 + Cx + Dy + E = 0 \), we'll follow these steps:

1. **Clear the denominators** by multiplying through by the least common denominator (LCD). The denominators are 16 and 12, so the LCD is 48.
   \[
   48 \times \left( \frac{(x-6)^2}{16} + \frac{y^2}{12} = 1 \right)
   \]
   This simplifies to:
   \[
   3(x-6)^2 + 4y^2 = 48
   \]

2. **Expand** the squares:
   \[
   3(x^2 - 12x + 36) + 4y^2 = 48
   \]
   \[
   3x^2 - 36x + 108 + 4y^2 = 48
   \]

3. **Move all terms to one side** of the equation:
   \[
   3x^2 + 4y^2 - 36x + 108 - 48 = 0
   \]
   Simplifying the constants:
   \[
   3x^2 + 4y^2 - 36x + 60 = 0
   \]

This matches option (a):
\[
\boxed{3x^2 + 4y^2 - 36x + 60 = 0}
\]

### Problem 10: Transform the equation 
\[
3x^2 + 2y^2 - 6x - 12y + 15 = 0
\]
to its standard form.

This is the equation of an ellipse, and we need to complete the square for both \(x\) and \(y\).

1. **Group the \(x\)-terms and the \(y\)-terms**:
   \[
   3(x^2 - 2x) + 2(y^2 - 6y) = -15
   \]

2. **Complete the square**:
   For the \(x\)-terms: \(x^2 - 2x\). Take half of \(-2\), which is \(-1\), and square it to get 1. Add and subtract inside the parentheses:
   \[
   3((x - 1)^2 - 1)
   \]

   For the \(y\)-terms: \(y^2 - 6y\). Take half of \(-6\), which is \(-3\), and square it to get 9. Add and subtract inside the parentheses:
   \[
   2((y - 3)^2 - 9)
   \]

3. Substitute back into the equation:
   \[
   3((x - 1)^2 - 1) + 2((y - 3)^2 - 9) = -15
   \]
   Simplify:
   \[
   3(x - 1)^2 - 3 + 2(y - 3)^2 - 18 = -15
   \]
   Combine constants:
   \[
   3(x - 1)^2 + 2(y - 3)^2 - 21 = -15
   \]
   \[
   3(x - 1)^2 + 2(y - 3)^2 = 6
   \]

4. **Divide by 6** to get the standard form:
   \[
   \frac{(x - 1)^2}{2} + \frac{(y - 3)^2}{3} = 1
   \]

This matches option (a):
\[
\boxed{\frac{(x - 1)^2}{2} + \frac{(y - 3)^2}{3} = 1}
\]

Would you like further details or clarifications? Here are five related questions you might consider:

1. What is the general form of a conic section?
2. How do you identify whether a conic section represents an ellipse, hyperbola, or circle?
3. Can you explain the difference between completing the square and expanding binomials?
4. How do you find the center, vertices, and foci of an ellipse from its standard form?
5. How would the process change if the equation represented a hyperbola?

**Tip**: When completing the square, always double-check the constants added to ensure correct balancing on both sides of the equation.

Express the equation (x-6)^2/16 + y^2/12 = 1 to its general form and transform the equation 3x^2 + 2y^2 - 6x - 12y + 15 = 0 to standard form.

Learn how to express the equation (x-6)^2/16 + y^2/12 = 1 to its general form and transform 3x^2 + 2y^2 - 6x - 12y + 15 = 0 to standard form. This problem involves algebraic techniques such as completing the square and working with conic sections, particularly ellipses. Suitable for students in Grades 10-12.

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