\(2x^{-1} - 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1\) 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 \(y^{-1} = -1, y = -1\)