Question

If 2 is added to the numerator of a fraction, it reduces to $\dfrac{1}{2}$ and if 1 is subtracted from the denominator, it reduces to $\dfrac{1}{3}$. Find the fraction.

Answer 1

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 $x$.
Let the denominator be $y$.
According to first condition i.e. If 2 is added to the numerator of a fraction, it reduces to $\dfrac{1}{2}$-
$\dfrac{{x + 2}}{y} = \dfrac{1}{2}$…………. (1)
And according to second condition i.e. if 1 is subtracted from denominator, it reduces to $\dfrac{1}{3}$-
We have-
$\dfrac{x}{{y - 1}} = \dfrac{1}{3}$…………… (2)
Now simplifying first equation,
We get-
$\dfrac{{x + 2}}{y} = \dfrac{1}{2}$
$\Rightarrow 2(x + 2) = y$
 $\Rightarrow y = 2x + 4$ …………… (3)
And simplifying second equation,
We get-
$\dfrac{x}{{y - 1}} = \dfrac{1}{2}$
$\Rightarrow 3x = y – 1$ ………… (4)
We will solve these linear equations (3 and 4) in 2 variables by using substitution method-
Substituting equation 3 in equation 4,
We get-
$\Rightarrow 3x = 2x + 4 – 1$
$\Rightarrow x = 3$
Substituting the value of x in equation 3,
We get-
$\Rightarrow y = 2x + 4$
$\Rightarrow y = 2(3) + 4$
$\Rightarrow y = 6 + 4 $
$\Rightarrow y = 10 $
Hence the numerator required fraction be 3 and denominator be 10.
Thus, the required fraction be $\dfrac{x}{y} = \dfrac{3}{{10}}$.

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.