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We are asked to solve the system of equations given by:

\[
\begin{cases}
-3x + 14y = -28 \\
-2(x - 2y) + 3y = -14
\end{cases}
\]

### Step 1: Simplify the second equation
We start by simplifying the second equation.

\[
-2(x - 2y) + 3y = -14
\]

Distribute the \(-2\):

\[
-2x + 4y + 3y = -14
\]

Combine like terms:

\[
-2x + 7y = -14
\]

### Step 2: Solve using the substitution or elimination method
Now, we have the following system:

\[
\begin{cases}
-3x + 14y = -28 \\
-2x + 7y = -14
\end{cases}
\]

We can use the elimination method. Let's multiply the second equation by 2 to align the coefficients of \(y\):

\[
2(-2x + 7y = -14) \Rightarrow -4x + 14y = -28
\]

Now, subtract the first equation from this result:

\[
(-4x + 14y) - (-3x + 14y) = (-28) - (-28)
\]

Simplify:

\[
-4x + 3x = 0 \quad \text{and} \quad -28 + 28 = 0
\]

This gives:

\[
-x = 0 \quad \Rightarrow \quad x = 0
\]

### Step 3: Solve for \(y\)
Substitute \(x = 0\) into one of the original equations, for example, the second one:

\[
-2(0) + 7y = -14 \quad \Rightarrow \quad 7y = -14 \quad \Rightarrow \quad y = -2
\]

### Solution:
The solution to the system is \(x = 0\) and \(y = -2\), or \((0, -2)\).

Thus, the correct answer is:
\[
\boxed{A) \ (0, -2)}
\]

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Would you like any further details or explanations on this?

Here are 5 related questions for further practice:
1. How can we solve a system of equations using substitution instead of elimination?
2. What are the conditions for a system of equations to have no solution or infinite solutions?
3. How can we check the solution of a system of equations graphically?
4. Can you explain how to use matrices to solve systems of equations?
5. How do linear transformations relate to systems of linear equations?

**Tip:** When solving systems of equations, always check if simplifying one equation makes the system easier to solve.

Solve the system of equations: -3x + 14y = -28 and -2(x - 2y) + 3y = -14

Learn how to solve the system of linear equations -3x + 14y = -28 and -2(x - 2y) + 3y = -14 using the elimination method. Suitable for students in Grades 8-10. Detailed step-by-step solution provided.

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