To solve the system of equations
\[ \begin{cases} 7x - y = 7 \quad (1) \ x + 2y = 6 \quad (2) \end{cases} \]
we can use substitution or elimination. Here, we'll use substitution.
From equation (1), we can express \(y\) in terms of \(x\):
\[ y = 7x - 7 \quad (3) \]
Now, substitute equation (3) into equation (2):
\[ x + 2(7x - 7) = 6 \]
This simplifies to:
\[ x + 14x - 14 = 6 \]
Combining the terms gives:
\[ 15x - 14 = 6 \]
Now, add 14 to both sides:
\[ 15x = 20 \]
Divide both sides by 15:
\[ x = \frac{20}{15} = \frac{4}{3} = 1 \frac{1}{3} \]
Now substitute \(x = 1 \frac{1}{3}\) back into equation (3) to find \(y\):
\[ y = 7(1 \frac{1}{3}) - 7 \]
Convert \(1 \frac{1}{3}\) to an improper fraction:
\[ 1 \frac{1}{3} = \frac{4}{3} \]
So \(y\) becomes:
\[ y = 7 \cdot \frac{4}{3} - 7 = \frac{28}{3} - 7 = \frac{28}{3} - \frac{21}{3} = \frac{7}{3} = 2 \frac{1}{3} \]
Thus, the solution to the system is:
\[ x = 1 \frac{1}{3}, \quad y = 2 \frac{1}{3} \]
Looking at the choices provided, the correct answer is:
Choice A: \( x = 1 \frac{1}{3}, y = 2 \frac{1}{3} \)
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