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The system of equations provided is:

\[
3c + 4m = 12.50
\]
\[
3c + 2m = 8.50
\]

To solve for \( c \) and \( m \), we can use elimination or substitution. I'll use the elimination method here:

### Step 1: Eliminate \( c \) by subtracting the second equation from the first.
\[
(3c + 4m) - (3c + 2m) = 12.50 - 8.50
\]
This simplifies to:
\[
2m = 4
\]
Dividing both sides by 2:
\[
m = 2
\]

### Step 2: Substitute \( m = 2 \) back into one of the original equations to solve for \( c \).
Using the second equation:
\[
3c + 2(2) = 8.50
\]
\[
3c + 4 = 8.50
\]
Subtracting 4 from both sides:
\[
3c = 4.50
\]
Dividing by 3:
\[
c = 1.50
\]

### Solution
The values of \( c \) and \( m \) are:
\[
c = 1.50 \quad \text{and} \quad m = 2
\]

Would you like further details or have any questions?

Here are 5 related questions:
1. How can we verify the solution by plugging \( c \) and \( m \) back into both equations?
2. What other methods can solve this system of linear equations?
3. How does the elimination method compare to the substitution method?
4. How would the solution change if the constants in the equations were different?
5. What happens if the two equations represent parallel lines?

**Tip:** Always check solutions by substituting back into the original equations to ensure accuracy.

Solve the system of equations: 3c + 4m = 12.50 and 3c + 2m = 8.50.

Learn how to solve the system of linear equations 3c + 4m = 12.50 and 3c + 2m = 8.50 using the elimination method. This step-by-step solution covers key algebraic concepts and is suitable for students in Grades 8-10.

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