To solve the given system of equations using the elimination method, we start by writing the system:

$4a + 0b = 28 \quad \text{(or just } 4a = 28\text{)}$ $0a + 4b = -8 \quad \text{(or just } 4b = -8\text{)}$

Step 1: Solve for $a$ and $b$ individually

From the first equation $4a = 28$, divide both sides by 4 to solve for $a$:

$a = \frac{28}{4} = 7$

From the second equation $4b = -8$, divide both sides by 4 to solve for $b$:

$b = \frac{-8}{4} = -2$

Step 2: Write the solution as an ordered pair

The solution to the system is:

$(a, b) = (7, -2)$

Conclusion:

The correct choice is:

A. The solution is $(7, -2)$.

Would you like further details or have any questions?

Related Questions:

1. How does the elimination method work when both variables are present in both equations?
2. Can we solve this system by substitution instead? How would it compare?
3. What happens if the system had coefficients that resulted in a contradiction (e.g., no solution)?
4. How would the system look if it had infinitely many solutions?
5. How can we check if the solution $(7, -2)$ is correct?

Tip: Always check your solution by substituting the values back into the original equations!