Solve the pair of equation for x and y respectively \(2x^{-1} – 3y^{-1} = 4; 4x^{-1} +…

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

\(y^{-1} = -1, y = -1\)