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Let's solve this system of equations step-by-step. The system of equations provided is:

\[
8x = 5y + 24
\]
\[
-9y = 40 - 8x
\]

### Step 1: Rearrange the Equations
Let's rearrange each equation to express \( y \) in terms of \( x \), or vice versa.

#### Rearranging the first equation:
\[
8x = 5y + 24
\]
Subtract 5y from both sides:
\[
8x - 5y = 24
\]

#### Rearranging the second equation:
\[
-9y = 40 - 8x
\]
Add \( 8x \) to both sides:
\[
8x - 9y = 40
\]

Now we have:
1. \( 8x - 5y = 24 \)
2. \( 8x - 9y = 40 \)

### Step 2: Eliminate \( x \) by Subtraction
To eliminate \( x \), subtract the second equation from the first:

\[
(8x - 5y) - (8x - 9y) = 24 - 40
\]

This simplifies to:
\[
4y = -16
\]

### Step 3: Solve for \( y \)
Divide both sides by 4:
\[
y = -4
\]

### Step 4: Substitute \( y = -4 \) into One of the Original Equations
Let's substitute \( y = -4 \) into the first equation:
\[
8x = 5(-4) + 24
\]
\[
8x = -20 + 24
\]
\[
8x = 4
\]
Now, divide by 8:
\[
x = \frac{1}{2}
\]

### Solution
The solution to the system is:
\[
x = \frac{1}{2}, \quad y = -4
\]

Would you like more details or have any questions?

### Additional Questions to Explore
1. How can we verify the solution by substituting values back into the original equations?
2. What method would you use to solve this system if you chose substitution instead of elimination?
3. How does changing a constant (e.g., 24 to 25) in one equation affect the solution?
4. Can this system be solved graphically? What would the intersection point represent?
5. What would happen if the coefficients of \( x \) in both equations were not equal?

### Tip
When using the elimination method, always align terms carefully to avoid calculation errors, especially when subtracting equations.

Solve the system of equations: 8x = 5y + 24 and -9y = 40 - 8x

Learn how to solve a system of equations with 8x = 5y + 24 and -9y = 40 - 8x using the elimination method. This solution includes step-by-step instructions for eliminating variables and solving for x and y, covering key algebra concepts. Suitable for Grades 9-11.

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