Solve the pair of equation for x and y respectively (2x^{-1} – 3y^{-1} = 4; 4x^{-1} + y^{-1} = 1)
-
A.
-1, 2 -
B.
1, 2 -
C.
2, 1 -
D.
2, -1
Correct Answer: Option D
Explanation
(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)