Let's solve the system of equations step by step:

1. **First equation:**
   \[
   -2 = 3y + 6
   \]
   Subtract 6 from both sides:
   \[
   -2 - 6 = 3y
   \]
   Simplify:
   \[
   -8 = 3y
   \]
   Now, divide both sides by 3:
   \[
   y = \frac{-8}{3}
   \]

2. **Second equation:**
   \[
   3y = -3x
   \]
   Divide both sides by 3:
   \[
   y = -x
   \]

Now we have two expressions for \(y\):
\[
y = \frac{-8}{3} \quad \text{and} \quad y = -x
\]

Equating the two:
\[
\frac{-8}{3} = -x
\]

Multiply both sides by -1:
\[
\frac{8}{3} = x
\]

Thus, the solution is:
\[
x = \frac{8}{3}, \quad y = \frac{-8}{3}
\]

Would you like further clarification or details on the solution?

Here are 5 related questions to consider:
1. How do you solve a system of linear equations with substitution?
2. What is the significance of the slope in the equation \( y = -x \)?
3. Can this system of equations be solved using elimination?
4. What is the graphical representation of the system?
5. How do fractions in the solution affect interpretation in real-life contexts?

**Tip:** When solving equations with fractions, it's often easier to clear the fractions early by multiplying both sides by the denominator.

Learn how to solve the system of linear equations -2 = 3y + 6 and 3y = -3x using the substitution method. This problem covers essential algebraic concepts and is ideal for students in Grades 8-10.

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