To solve the system of equations by elimination, we proceed as follows:

Given system:

1. $2x = 8 - 4y$
2. $5x + 5y = 12$

Step 1: Rewrite equations in standard form

1. From $2x = 8 - 4y$, rearrange to:
   $2x + 4y = 8$
2. The second equation is already in standard form:
   $5x + 5y = 12$

So, the system becomes: 2x + 4y = 8 \tag{1} 5x + 5y = 12 \tag{2}

Step 2: Eliminate one variable

To eliminate $y$, we make the coefficients of $y$ the same by multiplying equations appropriately:

• Multiply equation (1) by $5$: 10x + 20y = 40 \tag{3}
• Multiply equation (2) by $4$: 20x + 20y = 48 \tag{4}

Now subtract equation (3) from equation (4): $(20x + 20y) - (10x + 20y) = 48 - 40$ $10x = 8$ $x = \frac{8}{10} = \frac{4}{5}$

Step 3: Substitute $x = \frac{4}{5}$ into one of the original equations

Substitute into $2x + 4y = 8$: $2\left(\frac{4}{5}\right) + 4y = 8$ $\frac{8}{5} + 4y = 8$ $4y = 8 - \frac{8}{5}$ $4y = \frac{40}{5} - \frac{8}{5}$ $4y = \frac{32}{5}$ $y = \frac{32}{5} \div 4 = \frac{32}{5} \times \frac{1}{4}$ $y = \frac{8}{5}$

Final Answer:

$x = \frac{4}{5}, \; y = \frac{8}{5}$

The correct choice is: A. The solution to the system is $x = \frac{4}{5}, y = \frac{8}{5}$.

Let me know if you'd like a detailed explanation or have any questions!

Questions for deeper understanding:

1. How does elimination compare to substitution for solving systems of equations?
2. What happens if the system has no solution or infinitely many solutions?
3. Can this problem be solved using matrices or graphing? How?
4. How can you check the solution of a system of equations?
5. What are common pitfalls when solving systems by elimination?

Tip:

Always simplify the coefficients before elimination to make calculations easier!