-1, 2
1, 2
2, 1
2, -1
Correct answer is D
\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\)
Let \(x^{-1}\) = a and \(y^{-1}\)= b
2a - 3b = 4 .......(i)
4a + b = 1 .........(ii)
(i) x 3 = 12a + 3b = 3........(iii)
2a - 3b = 4 ...........(i)
(i) + (iii) = 14a = 7
∴ a = \(\frac{7}{14}\) = \(\frac{1}{2}\)
From (i), 3b = 2a - 4
3b = 1 - 4
3b = -3
∴ b = -1
From substituting, \(2^{-1} = x^{-1}\)
∴ x = 2
\(y^{-1} = -1, y = -1\)
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