Sure! Let's solve the system of equations step-by-step.
Given:
[tex]\[
\begin{cases}
x = 3y + 1 \quad \text{(Equation 1)}\\
2x + 4y = 12 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
First, from Equation 1, we have an expression for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = 3y + 1 \][/tex]
Next, we'll substitute this expression for [tex]\(x\)[/tex] into Equation 2. So wherever we see [tex]\(x\)[/tex] in Equation 2, we replace it with [tex]\(3y + 1\)[/tex]:
[tex]\[ 2(3y + 1) + 4y = 12 \][/tex]
Now, let's simplify and solve this equation for [tex]\(y\)[/tex]:
[tex]\[ 2(3y) + 2(1) + 4y = 12 \][/tex]
[tex]\[ 6y + 2 + 4y = 12 \][/tex]
[tex]\[ 10y + 2 = 12 \][/tex]
Subtract 2 from both sides to isolate terms involving [tex]\(y\)[/tex]:
[tex]\[ 10y = 10 \][/tex]
Divide both sides by 10 to solve for [tex]\(y\)[/tex]:
[tex]\[ y = 1 \][/tex]
Now that we have [tex]\(y = 1\)[/tex], we can substitute this value back into Equation 1 to find the corresponding [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].
Therefore, the correct answer is:
a. [tex]\((4, 1)\)[/tex]
