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Answer is D.

$x\times 7 + 3 = 5 \times y + 4 \implies 7x = 5y + 1$.

Only option satisfying this is D.

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  1. Convert both numbers to base 10: \[73_x = 7x^1 + 3x^0 = 7x + 3\] \[54_y = 5y^1 + 4y^0 = 5y + 4\]

  2. Equate the base 10 values: \[7x + 3 = 5y + 4\]

  3. Solve for \(x\) in terms of \(y\): \[7x = 5y + 1\] \[x = \frac{5y + 1}{7}\]

  4. Check the answer choices:

    1. Option (A): \(x = 8, y = 16 \) \(\rightarrow x = \frac{5 \times 16 + 1}{7} = \frac{81}{7}\) (not an integer)

    2. Option (B): \(x = 10, y = 12 \) \(\rightarrow x = \frac{5 \times 12 + 1}{7} = \frac{61}{7}\) (not an integer)

    3. Option (C): \(x = 9, y = 13 \) \(\rightarrow x = \frac{5 \times 13 + 1}{7} = \frac{66}{7}\) (not an integer)

    4. Option (D): \(x = 8, y = 11 \) \(\rightarrow x = \frac{5 \times 11 + 1}{7} = \frac{56}{7} = 8\) (integer)

  5. Therefore, the only possible values of \(x\) and \(y\) are 8 and 11, respectively.

Answer:

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