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We are given the system of equations:

\[
6x + 2y = 4 \quad \text{(1)}
\]
\[
2x + 3y = 5 \quad \text{(2)}
\]

We need to use the substitution method to solve for \(x\). Let's proceed step by step:

### Step 1: Solve one equation for one variable
From equation (2), we can solve for \(x\) in terms of \(y\):
\[
2x = 5 - 3y
\]
\[
x = \frac{5 - 3y}{2} \quad \text{(3)}
\]

### Step 2: Substitute into the other equation
Now, substitute equation (3) into equation (1) to solve for \(y\):
\[
6\left(\frac{5 - 3y}{2}\right) + 2y = 4
\]
Simplify:
\[
3(5 - 3y) + 2y = 4
\]
\[
15 - 9y + 2y = 4
\]
\[
15 - 7y = 4
\]
\[
-7y = 4 - 15
\]
\[
-7y = -11
\]
\[
y = \frac{11}{7}
\]

### Step 3: Substitute back to find \(x\)
Now substitute \(y = \frac{11}{7}\) back into equation (3) to find \(x\):
\[
x = \frac{5 - 3\left(\frac{11}{7}\right)}{2}
\]
\[
x = \frac{5 - \frac{33}{7}}{2}
\]
\[
x = \frac{\frac{35}{7} - \frac{33}{7}}{2}
\]
\[
x = \frac{\frac{2}{7}}{2}
\]
\[
x = \frac{1}{7}
\]

Thus, the value of \(x\) is \( \frac{1}{7} \), which corresponds to option **A**.

Would you like more details or have any questions?

### Related Questions:
1. How do you apply the substitution method to more complex systems of equations?
2. What is the advantage of using substitution over elimination in this case?
3. How would the process change if the system involved three variables instead of two?
4. Can you graphically interpret the solution of this system?
5. How does the method of solving linear equations change when dealing with inequalities?

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

Using the substitution method, what is the value of x? 6x + 2y = 4, 2x + 3y = 5.

Learn how to solve a system of linear equations using the substitution method. This problem involves the equations 6x + 2y = 4 and 2x + 3y = 5. Step-by-step instructions help students in Grades 9-12 master the concept of substitution in algebra.

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