We are given the following system of linear equations:

$3x + 4y = -10 \quad \text{(Equation 1)}$ $2x - 4y = 0 \quad \text{(Equation 2)}$

Step 1: Solve the system by substitution or elimination.

First, let's solve Equation 2 for one variable. We can solve for $x$:

$2x - 4y = 0$ $2x = 4y$ $x = 2y$

Step 2: Substitute into Equation 1.

Now, substitute $x = 2y$ into the first equation:

$3(2y) + 4y = -10$ $6y + 4y = -10$ $10y = -10$ $y = -1$

Step 3: Solve for $x$.

Using $y = -1$, substitute it back into $x = 2y$:

$x = 2(-1) = -2$

Step 4: Verify the solution.

We have $x = -2$ and $y = -1$. Let's check if this solution satisfies both equations:

• Equation 1: $3(-2) + 4(-1) = -6 - 4 = -10$ ✔️
• Equation 2: $2(-2) - 4(-1) = -4 + 4 = 0$ ✔️

The solution is correct. Therefore, the ordered pair is $(-2, -1)$, which corresponds to option C.

Would you like any more details on the method? Here are some related questions:

1. What is the elimination method for solving systems of equations?
2. Can systems of equations have no solutions or infinitely many solutions?
3. How can you solve a system using matrix methods?
4. What is substitution, and how does it differ from elimination?
5. How would this system be solved graphically?

Tip: Always check your solution by substituting it back into both equations.