The system of equations presented is as follows:

$3x - 8y = 9 \quad \text{(Equation 1)}$ $-3x + 10y = -15 \quad \text{(Equation 2)}$

To solve this system, we can use either the substitution method or the elimination method. Let's proceed with the elimination method.

Step 1: Add both equations to eliminate $x$

By adding the two equations, the $x$-terms will cancel out:

$(3x - 8y) + (-3x + 10y) = 9 + (-15)$ $3x - 8y - 3x + 10y = -6$ $2y = -6$

Step 2: Solve for $y$

Now, divide both sides of the equation by 2:

$y = \frac{-6}{2} = -3$

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

Let's substitute $y = -3$ into the first equation $3x - 8y = 9$:

$3x - 8(-3) = 9$ $3x + 24 = 9$ $3x = 9 - 24$ $3x = -15$

Step 4: Solve for $x$

Now, divide both sides by 3:

$x = \frac{-15}{3} = -5$

Final Solution:

The solution to the system of equations is: $x = -5, \quad y = -3$

Would you like a more detailed explanation or further clarification on this process? Here are a few related questions you can explore:

1. What are the different methods for solving a system of linear equations?
2. How would the solution change if the system had no solution or infinitely many solutions?
3. Can the elimination method be used for non-linear equations?
4. How can you check if the solution is correct?
5. What would happen if the coefficients of $x$ and $y$ were both 0?

Tip: Always check your solution by substituting the values of $x$ and $y$ back into the original equations to verify the correctness.