We have the system of equations:

\[
1) \ 4x + 5y = -45
\]
\[
2) \ 2x - 5y = 15
\]

### Step 1: Add the equations

We can eliminate \( y \) by adding the two equations, since the \( 5y \) terms will cancel out.

\[
(4x + 5y) + (2x - 5y) = -45 + 15
\]
\[
6x = -30
\]

### Step 2: Solve for \( x \)

\[
x = \frac{-30}{6} = -5
\]

### Step 3: Substitute \( x = -5 \) into one of the original equations

Let's substitute it into the second equation:

\[
2x - 5y = 15
\]
\[
2(-5) - 5y = 15
\]
\[
-10 - 5y = 15
\]
\[
-5y = 25
\]
\[
y = \frac{25}{-5} = -5
\]

### Final Solution:

\[
x = -5, \quad y = -5
\]

Would you like further details or clarifications?

Here are 5 related questions to explore further:
1. How would the solution change if we altered one of the coefficients in the system?
2. Can this system be solved using substitution instead of elimination?
3. How can we represent this system graphically?
4. What is the determinant of the coefficient matrix in this system?
5. How do we solve this system using matrix methods?

**Tip**: When solving systems of equations, elimination is often useful if the coefficients of one variable are already opposites.

Discover how to solve the system of equations 4x + 5y = -45 and 2x - 5y = 15 using the elimination method. Learn step-by-step instructions to find the values of x and y. Ideal for students in Grades 8-10.

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