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]
