The given equation is:

$2 \begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$

We can break this into steps.

Step 1: Expand the scalar multiplication

Multiply the first matrix by 2:

$2 \begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} = \begin{bmatrix} 2x & 10 \\ 14 & 2(y-3) \end{bmatrix}$ Which simplifies to: $\begin{bmatrix} 2x & 10 \\ 14 & 2y - 6 \end{bmatrix}$

Step 2: Add the second matrix

Now add the matrices element by element:

$\begin{bmatrix} 2x & 10 \\ 14 & 2y - 6 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2x + 3 & 10 - 4 \\ 14 + 1 & (2y - 6) + 2 \end{bmatrix}$

This results in: $\begin{bmatrix} 2x + 3 & 6 \\ 15 & 2y - 4 \end{bmatrix}$

Step 3: Set equal to the right-hand side matrix

Now equate this matrix to the given right-hand side matrix:

$\begin{bmatrix} 2x + 3 & 6 \\ 15 & 2y - 4 \end{bmatrix} = \begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}$

Step 4: Solve for $x$ and $y$

From the first element of the first row: $2x + 3 = 7 \quad \Rightarrow \quad 2x = 4 \quad \Rightarrow \quad x = 2$

From the second element of the second row: $2y - 4 = 14 \quad \Rightarrow \quad 2y = 18 \quad \Rightarrow \quad y = 9$

Thus, the values of $x$ and $y$ are: $x = 2, \quad y = 9$

Let me know if you'd like further details on any step or have any questions!

Here are 5 related questions to expand your knowledge:

1. What would the solution look like if the scalar was a fraction instead of 2?
2. How do matrix addition and scalar multiplication behave with larger matrices?
3. Can this method of solving matrices be applied to 3x3 matrices?
4. What are the real-world applications of solving matrix equations like this?
5. How does the determinant of a matrix influence the solution?

Tip: Always check if the matrices you're adding or multiplying are of compatible dimensions to avoid mistakes!