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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.

Using the substitution method, what is the value of x? 10x - 2y = 40, 20x + 2y = 200

Learn how to solve the system of linear equations 10x - 2y = 40 and 20x + 2y = 200 using the substitution method. This detailed step-by-step guide helps 8th grade students understand algebraic substitution in linear equations.

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