The given system of equations is:

$-3x + 3y = 15$ $-4x - 5y = -16$

We will solve this system using the elimination method.

Step 1: Align the equations

Write the equations in the given form:

1. $-3x + 3y = 15$
2. $-4x - 5y = -16$

Step 2: Eliminate one variable

To eliminate one variable, we aim to make the coefficients of either $x$ or $y$ equal in both equations. Here, we will eliminate $y$.

Make the $y$-coefficients equal:

• Multiply the first equation by $5$ and the second equation by $3$: $5(-3x + 3y) = 5(15) \implies -15x + 15y = 75$ $3(-4x - 5y) = 3(-16) \implies -12x - 15y = -48$

Now we have:

1. $-15x + 15y = 75$
2. $-12x - 15y = -48$

Step 3: Add the equations

Add the two equations to eliminate $y$: $(-15x + 15y) + (-12x - 15y) = 75 + (-48)$ $-15x - 12x = 27$ $-27x = 27$ Solve for $x$: $x = -1$

Step 4: Solve for $y$

Substitute $x = -1$ into one of the original equations. Using the first equation: $-3x + 3y = 15$ Substitute $x = -1$: $-3(-1) + 3y = 15$ $3 + 3y = 15$ $3y = 12$ $y = 4$

Final Solution:

The solution to the system of equations is: $x = -1, \, y = 4$

Would you like a detailed explanation of a specific step?
Here are some related questions:

1. How can we eliminate $x$ instead of $y$?
2. How does substitution compare to elimination in solving this system?
3. Can this method solve non-linear systems of equations?
4. What are other ways to check the solution's correctness?
5. How does elimination work with three equations and three variables?

Tip: Always double-check your solution by substituting the values back into the original equations.