Question: If the numerator of a fraction is multiples by 4 and the denominator is reduced by 2, the result is 2. If the numerator of the fraction is increased by 15 and 2 is subtracted from double of the denominator, the result is 9/7. Find the fraction.

Solution: 
Given,

Let the numerator of the required fraction be 'x' and the denominator be'y'. So, the required fraction would be $\frac{x}{y}$

According to the question,

Condition I,
If the numerator of a fraction is multiples by 4 and the denominator is reduced by 2, the result is 2
$or, \dfrac{4x}{y-2} = 2$

$or, 4x = 2(y-2)$

$or, x = \dfrac{2(y-2)}{4}$

$or, x = \dfrac{y-2}{2}$ - (i)

Condition II,
If the numerator of the fraction is increased by 15 and 2 is subtracted from double of the denominator, the result is 9/7
$or, \dfrac{x +15}{2y-2} = \dfrac{9}{7}$

$or, 7(x +15) = 9(2y -2)$

$or, 7x + 105 = 18y -18$ 

$or, 7x + 105 +18 = 18y$ 

$or, 7x + 123 = 18y$

$or, 7x = 18y -123$- (ii)

Put value of x from equation (i) in equation (ii), we get,

$or, 7 \left ( \dfrac{y-2}{2} \right )  = 18y -123$

$or, \dfrac{7(y-2) }{2} = 18y -123$

$or, 7y - 14 = 2(18y -123)$

$or, 7y -14  = 36y -246$

$or, 246 -14 = 36y -7y$

$or, 232 = 29y$

$or, y = \dfrac{232}{29}$

$\therefore y = 8$

Put value of y in equation (i), we get,

$or, x = \dfrac{8-2}{2}$

$or, x = \dfrac{6}{2}$

$\therefore x = 3$

So, (x,y) = (3,8)

Therefore, the required fraction is $\frac{3}{8}$.