To solve the given system of equations by substitution, let's follow these steps carefully:
The system of equations is:
[tex]\[
\begin{array}{l}
1. \quad x - 3y = 4 \\
2. \quad 2x - 6y = 8
\end{array}
\][/tex]
Step 1: Solve one of the equations for one of the variables. Let's solve the first equation for [tex]\( x \)[/tex].
[tex]\[
x = 3y + 4
\][/tex]
Step 2: Substitute this expression for [tex]\( x \)[/tex] into the second equation.
[tex]\[
2(3y + 4) - 6y = 8
\][/tex]
Step 3: Distribute and combine like terms.
[tex]\[
6y + 8 - 6y = 8
\][/tex]
This simplifies to:
[tex]\[
8 = 8
\][/tex]
Notice that [tex]\( 8 = 8 \)[/tex] is always true. This indicates that the second equation does not provide new information beyond what is already available from the first equation. This implies that the two equations are essentially the same (one is a multiple of the other).
Step 4: Determine the implications. Since the second equation is not providing new information, it means that there are infinitely many solutions (each corresponding to a point on the line defined by [tex]\( x = 3y + 4 \)[/tex]).
However, let's verify if there is one specific solution that matches within the provided choices.
Recall the substitution and simplification:
[tex]\[
x = 3y + 4
\][/tex]
Substitute [tex]\( y = 0 \)[/tex] (because there's no restriction yet):
[tex]\[
x = 3(0) + 4 = 4
\][/tex]
So, one specific solution point is [tex]\( x = 4 \)[/tex] and [tex]\( y = 0 \)[/tex], which satisfies both equations.
This specific solution corresponds to:
[tex]\[
x = 4, y = 0
\][/tex]
Therefore, the answer is:
[tex]\[
B. \quad x = 4, y = 0
\][/tex]
