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Let's solve the system of equations using the echelon method.

The given system is:

\[
\begin{aligned}
1. \quad & 3x + 2y = -9 \\
2. \quad & 5x - 2y = -15
\end{aligned}
\]

### Step 1: Eliminate one variable
To eliminate one variable, we can add the two equations to cancel out the \( y \)-terms.

Add equation (1) and (2):

\[
(3x + 2y) + (5x - 2y) = -9 + (-15)
\]

This simplifies to:

\[
8x = -24
\]

Solving for \( x \):

\[
x = \frac{-24}{8} = -3
\]

### Step 2: Solve for \( y \)
Now substitute \( x = -3 \) into one of the original equations. Let's use equation (1):

\[
3(-3) + 2y = -9
\]

Simplifying:

\[
-9 + 2y = -9
\]

Add 9 to both sides:

\[
2y = 0
\]

Solve for \( y \):

\[
y = 0
\]

### Final Answer:
The solution is \( (x, y) = (-3, 0) \).

Therefore, the correct choice is:

- **A. The solution is (-3, 0)**.

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Would you like further details or have any questions?

Here are some related questions to consider:

1. How does the echelon method differ from substitution or elimination methods?
2. What other methods can be used to solve systems of equations?
3. How would the solution change if both equations were parallel?
4. How can systems of equations be applied in real-life scenarios?
5. What is the significance of having infinitely many solutions?

**Tip**: When solving a system of equations, always check your solution by substituting the values back into the original equations to ensure accuracy.

Use the echelon method to solve the following system of two equations in two unknowns. 3x + 2y = -9 and 5x - 2y = -15.

Learn how to solve a system of linear equations using the echelon method with the equations 3x + 2y = -9 and 5x - 2y = -15. This tutorial explains key algebraic concepts and provides a step-by-step solution suitable for students in Grades 9-12.

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