We have

and

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

⇒ 5z + 6y = 13 …(i)

⇒ 3z + 4y = 7 …(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 5, so that variable z in both the equations have same coefficient.

Recalling equations (i) & (ii),

5z + 6y = 13 [×3

3z + 4y = 7 [×5

⇒ - 2y = 4

⇒ y = - 2

Substitute y = - 2 in eq.(i)/eq.(ii), as per convenience of solving.

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

3z + 4( - 2) = 7

⇒ 3z – 8 = 7

⇒ 3z = 7 + 8

⇒ 3z = 15

⇒ z = 5

Thus, z = 5 and y = - 2

As z = 1/x,

⇒ 5 = 1/x

⇒ x = 1/5

Hence, we have x = 1/5 and y = - 2