When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes $\dfrac{1}{4}$. And when 6 is added to the numerator and the denominator is multiplied by 3, it becomes $\dfrac{2}{3}$. Find the fraction.

Hint- Here we will proceed by assuming the numerator and denominator be x and y respectively. Then we will use given conditions to form linear equations in 2 variables using a substitution method so that we will get the required numerator and denominator.

Complete step-by-step solution -
Let the numerator be $x$
And let the denominator be $y$
Now applying first condition i.e. When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes $\dfrac{1}{4}$,
We get-
$\dfrac{{x - 2}}{{y + 3}} = \dfrac{1}{4}$
$\Rightarrow 4\left( {x - 2} \right) = \left( {y + 3} \right)$
$\Rightarrow 4x – 8 = y + 3$
$\Rightarrow 4x – 11 = y $………… (1)
And applying second condition i.e. when 6 is added to numerator and the denominator is multiplied by 3, it becomes $\dfrac{2}{3}$ ,
We get-
$\dfrac{{x + 6}}{{3y}} = \dfrac{2}{3}$
Solving equation 1 and equation 2,
$\Rightarrow \dfrac{{3\left( {x + 6} \right)}}{3} = 2y $
$\Rightarrow x + 6 = 2y $…………… (3)
Now put value of equation 1 in equation 3,
We get-
$ x + 6 = 2(4x – 11) $
$\Rightarrow x + 6 = 8x – 22 $
$\Rightarrow 6 + 22 = 8x – 22 $
$\Rightarrow 6 + 22 = 8x – x $
$\Rightarrow 28 = 7x $
$\Rightarrow x = 4 $
Now substituting the value of x in equation 1,
We get-
$\Rightarrow y = 4 (4) – 11 $
$\Rightarrow y = 5 $
This implies-
Required fraction $ = \dfrac{x}{y} = \dfrac{4}{5}$

Note- While solving this question, we can assume any variables instead of x and y. As here we used a substitution method to solve these linear equations in 2 variables, we can also solve these linear equations in 2 variables using elimination method.