To solve the system of equations using elimination, we have the following system:

$2x + 3y = 13 \quad \text{(1)}$ $-2x + y = -1 \quad \text{(2)}$

Step 1: Add the equations

To eliminate $x$, we add both equations (1) and (2):

$(2x + 3y) + (-2x + y) = 13 + (-1)$ $0x + (3y + y) = 12$ $4y = 12$

Step 2: Solve for $y$

$y = \frac{12}{4} = 3$

Step 3: Substitute $y = 3$ into one of the original equations

Let's substitute $y = 3$ into equation (2):

$-2x + 3 = -1$

Step 4: Solve for $x$

$-2x = -1 - 3$ $-2x = -4$ $x = \frac{-4}{-2} = 2$

Thus, the value of $x$ is 2.

Would you like further details or explanations on any steps?

Here are five related questions to expand your understanding:

1. What happens if we use substitution instead of elimination?
2. Can you verify the solution by plugging $x = 2$ and $y = 3$ back into both equations?
3. How does elimination differ from substitution in terms of strategy?
4. What would happen if the system had no solution or infinitely many solutions?
5. How can you graphically interpret the solution of this system?

Tip: When using elimination, always check if you can cancel out one variable by adding or subtracting the equations!