To solve the given system of equations, let's go through the step-by-step process:
Given the system of equations:
[tex]\[
\begin{aligned}
1) \quad & x = 3y + 1 \\
2) \quad & 2x + 4y = 12
\end{aligned}
\][/tex]
### Step 1: Substitute [tex]\(x\)[/tex] from Equation 1 into Equation 2
From Equation 1, we know:
[tex]\[ x = 3y + 1 \][/tex]
We'll substitute this expression for [tex]\(x\)[/tex] in Equation 2:
[tex]\[ 2(3y + 1) + 4y = 12 \][/tex]
### Step 2: Simplify the Equation
Distribute and combine like terms:
[tex]\[ 6y + 2 + 4y = 12 \][/tex]
[tex]\[ 10y + 2 = 12 \][/tex]
### Step 3: Solve for [tex]\(y\)[/tex]
Subtract 2 from both sides of the equation:
[tex]\[ 10y = 10 \][/tex]
Divide both sides by 10:
[tex]\[ y = 1 \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Now that we have [tex]\(y = 1\)[/tex], we can substitute it back into Equation 1 to find [tex]\(x\)[/tex]:
[tex]\[ x = 3(1) + 1 \][/tex]
[tex]\[ x = 3 + 1 \][/tex]
[tex]\[ x = 4 \][/tex]
So, the solution to the system of equations is [tex]\((x, y) = (4, 1)\)[/tex].
### Conclusion
The correct pair [tex]\((x, y)\)[/tex] is:
[tex]\[
(a) \quad (4, 1)
\][/tex]
Thus, the answer is:
[tex]\[
\boxed{(4, 1)}
\][/tex]
