We have the system of equations:

$-3x + 8y = 15$

$2x - 3y = -10$

Step 1: Solve for one variable using elimination or substitution.

We will use the elimination method. Our goal is to eliminate one variable by making the coefficients of either $x$ or $y$ equal.

Multiply the first equation by 2 and the second equation by 3 to align the coefficients of $x$:

$(2)(-3x + 8y) = (2)(15) \quad \Rightarrow \quad -6x + 16y = 30$

$(3)(2x - 3y) = (3)(-10) \quad \Rightarrow \quad 6x - 9y = -30$

Step 2: Add the two equations.

$(-6x + 16y) + (6x - 9y) = 30 + (-30)$

$-6x + 6x + 16y - 9y = 0$

$7y = 0$

$y = 0$

Step 3: Solve for $x$.

Substituting $y = 0$ into the first equation:

$-3x + 8(0) = 15$

$-3x = 15$

$x = -5$

Final Answer:

$\boxed{x = -5, \quad y = 0}$

Would you like a more detailed explanation or have any questions?

Related Questions:

1. How does the elimination method work in solving systems of equations?
2. Can we solve this system using the substitution method?
3. What are the applications of solving linear systems in real life?
4. How do we determine if a system of equations has no solution or infinite solutions?
5. Can we graph these two equations to verify the solution?

Tip: Always check your solution by substituting the values of $x$ and $y$ back into the original equations!