Let's solve this system of equations using the substitution method.

The given system is: $10x - 2y = 40 \quad \text{(Equation 1)}$ $20x + 2y = 200 \quad \text{(Equation 2)}$

Step 1: Solve one equation for one variable

We can take Equation 1 and solve for $y$: $10x - 2y = 40$ First, isolate $2y$: $-2y = 40 - 10x$ Now solve for $y$: $y = 5x - 20$

Step 2: Substitute the expression for $y$ into the other equation

Substitute $y = 5x - 20$ into Equation 2: $20x + 2(5x - 20) = 200$ Simplify: $20x + 10x - 40 = 200$ Combine like terms: $30x - 40 = 200$ Add 40 to both sides: $30x = 240$ Solve for $x$: $x = \frac{240}{30} = 8$

So, the value of $x$ is $8$.

Correct answer: Option B: $x = 8$.

Would you like a further breakdown or have any questions about this solution?

Here are 5 related questions for you:

1. How can you check if the solution $x = 8$ is correct for both equations?
2. What other methods can you use to solve a system of linear equations?
3. Why is substitution a valid method for solving this system of equations?
4. How do you solve a system of equations when both variables are present in both equations?
5. What happens if you substitute incorrectly? How can you avoid mistakes in substitution?

Tip: When using the substitution method, always simplify the equation as much as possible before substituting to avoid errors.