To solve the system using the substitution method, we start with the two equations provided:
- \( x = y - 4 \)
- \( x + 8y = 2 \)
We can substitute the expression for \( x \) from the first equation into the second equation.
Substituting \( x = y - 4 \) into the second equation:
\[ (y - 4) + 8y = 2 \]
Now, combine like terms:
\[ y - 4 + 8y = 2 \] \[ 9y - 4 = 2 \]
Next, we solve for \( y \):
Add 4 to both sides:
\[ 9y = 6 \]
Now divide by 9:
\[ y = \frac{6}{9} = \frac{2}{3} \]
Now that we have \( y \), we can substitute it back into the first equation to find \( x \):
\[ x = y - 4 \] \[ x = \frac{2}{3} - 4 \] \[ x = \frac{2}{3} - \frac{12}{3} \] \[ x = \frac{2 - 12}{3} = \frac{-10}{3} \]
So we have:
\[ (x, y) = \left( \frac{-10}{3}, \frac{2}{3} \right) \]
Now, let's find out what these values correspond to in the answer options given:
- A: \( \left( \frac{2}{3}, -3 \frac{1}{3} \right) \) which is \( \left( \frac{2}{3}, -\frac{10}{3} \right) \) → Not a match.
- B: \( \left( -3 \frac{1}{3}, \frac{2}{3} \right) \) is \( \left( -\frac{10}{3}, \frac{2}{3} \right) \) → This is a match!
- C: Infinite solutions
- D: No solutions
Therefore, the correct answer is:
B: \( \left( -3 \frac{1}{3}, \frac{2}{3} \right) \).