We have

3(2x + y) = 7xy

And 3(x + 3y) = 11xy

Lets simplify these equations.

3(2x + y) = 7xy

Dividing throughout by xy, and assuming p = 1/x and q = 1/y,

⇒

⇒

6q + 3p = 7 …(i)

Also, 3(x + 3y) = 11xy

Dividing throughout by xy, and assuming p = 1/x and q = 1/y,

⇒

⇒

⇒ 3q + 9p = 11 …(ii)

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

Multiply equation (ii) by 2, so that the variable q in both the equations have same coefficient.

Recalling equations (i) and (ii),

6q + 3p = 7

3q + 9p = 11 [×2

⇒ - 15p = - 15

⇒ p = 1

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

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

6q + 3(1) = 7

⇒ 6q = 7 – 3

⇒ 6q = 4

⇒ q = 2/3

Thus, p = 1 and q = 2/3

As p = 1

⇒ 1 = 1/x

⇒ x = 1

And q = 1/y,

⇒

⇒ y = 3/2

Hence, we have x = 1 and y = 3/2