Answer :

To solve the system of equations:

[tex]\[ \begin{array}{l} x = 3y + 1 \quad \text{(Equation 1)} \\ 2x + 4y = 12 \quad \text{(Equation 2)} \end{array} \][/tex]

we follow these steps:

1. Express [tex]\( x \)[/tex] from Equation 1:

[tex]\[ x = 3y + 1 \][/tex]

2. Substitute [tex]\( x \)[/tex] into Equation 2:

Substituting the expression [tex]\( x = 3y + 1 \)[/tex] in Equation 2, we get:

[tex]\[ 2(3y + 1) + 4y = 12 \][/tex]

3. Simplify and solve for [tex]\( y \)[/tex]:

Distribute and combine like terms:

[tex]\[ 6y + 2 + 4y = 12 \][/tex]

Combine the [tex]\( y \)[/tex]-terms:

[tex]\[ 10y + 2 = 12 \][/tex]

Subtract 2 from both sides:

[tex]\[ 10y = 10 \][/tex]

Divide by 10:

[tex]\[ y = 1 \][/tex]

4. Solve for [tex]\( x \)[/tex] using the value of [tex]\( y \)[/tex]:

Substitute [tex]\( y = 1 \)[/tex] back into Equation 1:

[tex]\[ x = 3(1) + 1 = 3 + 1 = 4 \][/tex]

So, the solution to the system of equations is [tex]\( x = 4 \)[/tex] and [tex]\( y = 1 \)[/tex]. Therefore, the point [tex]\( (x, y) \)[/tex] is [tex]\( (4, 1) \)[/tex].

Thus, the correct answer is:

a. [tex]\( (4, 1) \)[/tex]