Solution:

From the first condition, we have:

 \(\displaystyle \frac {x \times 3}{y - 3} = \frac {18}{11}\)

∴ \(\displaystyle \frac {3x}{y - 3} = \frac {18}{11}\)

By cross multiplication,
 3x × 11 = 18(y − 3)
∴ 33x = 18y − 54
∴ 33x − 18y = − 54 ... (i)
∴ 33x − 18y + 54 = 0
Dividing both sides by 3, we get,
 11x − 6y + 18 = 0 ...(Equation I)

From the second condition, we have:

 \(\displaystyle \frac {x + 8}{y \times 2} = \frac {1}{2}\)

∴ \(\displaystyle \frac {x + 8}{2y} = \frac {1}{2}\)

By cross multiplication,
 2(x + 8) = 2y
∴ 2x + 16 = 2y
∴ 2x − 2y = − 16 ... (ii)
∴ 2x − 2y + 16 = 0
Dividing both sides by 2, we get,
 x − y + 8 = 0 ...(Equation II)

Multiplying equation (ii) by 9, we get,
 18x − 18y = − 144 ... (iii)

Subtracting equation (iii) from equation (i), we get,

		33x	−	18y	=	−	54	... (i)
−	⊕	18x	⊖	18y	=	⊖	144	... (iii)
	−		+			+
		15x			=		90

∴ \(\displaystyle x = \frac {90}{15} = 6\)

∴ x = 6 ... (iv)

Substituting the value of x from equation (iv) into equation (ii), we get,
  2(6) − 2y = − 16
∴ 12 − 2y = − 16
∴ − 2y = − 28
∴ \(\displaystyle y = \frac {28}{2} = 14\)
∴ y = 14 ... (v)

∴ Given fraction = \(\displaystyle \frac {6}{14}\).

Verification:

Substituting the values of x and y from equations (iv) and (v) into equation (i), we get,
  33x − 18y = − 54 ... (i)
∴ 33(6) − 18(14) = − 54
∴ 198 − 252 = − 54
∴ − 54 = − 54

Substituting the values of x and y from equations (iv) and (v) into equation (ii), we get,
 2x − 2y = − 16 ... (ii)
∴ 2(6) − 2(14) = − 16
∴ 12 − 28 = − 16
∴ − 16 = − 16