If 43^x \times 47^y = (2021)^2, x \neq 0, y \neq 0, then what is the value of the following? \frac{4xy+x+y}{2xy-x-y}

Explanation

Notice that 2021 = 43 \times 47, so (2021)^2 = 43^2 \times 47^2. Equating the powers, x=2 and y=2. Substituting: \frac{4(2)(2)+2+2}{2(2)(2)-2-2} = \frac{16+4}{8-4} = \frac{20}{4} = 5.

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