We have

and

where y≠0

Lets simplify these equations. Assuming 1/y = z, we can rewrite them,

⇒ 2x – 3z = 9 …(i)

⇒ 3x + 7z = 2 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.(i) by 3 and eq.(ii) by 2, so that variable x in both the equations have same coefficient.

Recalling equations (i) & (ii),

2x – 3z = 9 [×3

3x + 7z = 2 [×2

⇒ - 23z = 23

⇒ z = - 1

Substitute z = - 1 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

2x – 3( - 1) = 9

⇒ 2x + 3 = 9

⇒ 2x = 6

⇒ x = 3

Thus, z = - 1 and x = 3

As z = 1/y,

⇒ - 1 = 1/y

⇒ y = - 1

Hence, we have x = 3 and y = - 1