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To solve the system of equations:

\[
\begin{aligned}
    3a - 2b &= 2 \quad \text{(Equation 1)} \\
    3a + 2b &= -2 \quad \text{(Equation 2)}
\end{aligned}
\]

We can add the two equations together to eliminate \(b\). 

### Step 1: Add the two equations
\[
(3a - 2b) + (3a + 2b) = 2 + (-2)
\]
Simplifying this:
\[
6a = 0
\]
\[
a = 0
\]

### Step 2: Substitute \(a = 0\) into one of the original equations (let's use Equation 1):
\[
3(0) - 2b = 2
\]
\[
-2b = 2
\]
\[
b = -1
\]

### Final Solution:
\[
a = 0, \quad b = -1
\]

Would you like more details, or do you have any questions about the solution?

### Related Questions:
1. How can you solve this system using substitution instead of elimination?
2. What happens if the system of equations had no solution or infinitely many solutions?
3. How would the solution change if the coefficients of \(a\) and \(b\) were different?
4. Can you graph these two equations and show their point of intersection?
5. How do we solve systems of equations with three variables?

### Tip:
Always check your solution by substituting the values of \(a\) and \(b\) back into both original equations to verify they satisfy both.

Solve the system of equations: 3a - 2b = 2 and 3a + 2b = -2

Learn how to solve the system of linear equations 3a - 2b = 2 and 3a + 2b = -2 using the elimination method. Follow a step-by-step solution for eliminating variables and finding values for both 'a' and 'b'. Ideal for students in Grades 8-10.

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