To solve the system of equations:

\[
\begin{align*}
3x + 2y & = 6 \quad \text{(Equation 1)} \\
2x - 2y & = -4 \quad \text{(Equation 2)}
\end{align*}
\]

### Step 1: Add the two equations to eliminate \(y\).

Add Equation 1 and Equation 2:

\[
(3x + 2y) + (2x - 2y) = 6 + (-4)
\]

Simplifying this:

\[
3x + 2x + 2y - 2y = 2
\]

This reduces to:

\[
5x = 2
\]

### Step 2: Solve for \(x\).

\[
x = \frac{2}{5}
\]

### Step 3: Substitute \(x\) back into one of the original equations to solve for \(y\).

Using Equation 1:

\[
3\left(\frac{2}{5}\right) + 2y = 6
\]

Simplify:

\[
\frac{6}{5} + 2y = 6
\]

Subtract \(\frac{6}{5}\) from both sides:

\[
2y = 6 - \frac{6}{5}
\]

Convert 6 to a fraction with a denominator of 5:

\[
2y = \frac{30}{5} - \frac{6}{5} = \frac{24}{5}
\]

Solve for \(y\):

\[
y = \frac{24}{5 \times 2} = \frac{24}{10} = \frac{12}{5}
\]

### Solution:
\[
x = \frac{2}{5}, \quad y = \frac{12}{5}
\]

Would you like more details or have any questions?

Here are some related questions:
1. How can substitution be used to solve systems of equations?
2. What is the graphical interpretation of the solution to this system?
3. How can the elimination method be applied to different types of systems?
4. What happens if a system of equations has no solution?
5. How can you verify the solution of a system of equations?

**Tip:** Always check your solution by substituting the values back into the original equations.

Learn how to solve the system of equations 3x + 2y = 6 and 2x - 2y = -4 using the substitution and elimination methods. This problem covers systems of linear equations and involves step-by-step solutions for finding the values of x and y. Suitable for students in Grades 9-12.

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