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# A fraction becomes $\dfrac{1}{3}$, if 1 is subtracted from both its numerator and denominator. If 3 is added to both the numerator and the denominator, it becomes $\dfrac{1}{2}$. Find the fraction.  Verified
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Hint- Here, we will be assuming the original fraction which will consist of two unknowns i.e., numerator and denominator. According to the problem, we will obtain two equations in two unknowns and then will use elimination method.

Let us suppose that the original fraction is $\dfrac{x}{y}$ where x is the numerator of the original fraction and y is the denominator of the original fraction.
Now, when 1 is subtracted from both its numerator and denominator of the original fraction, the fraction reduces to $\dfrac{{x - 1}}{{y - 1}}$.
According to problem statement, $\dfrac{{x - 1}}{{y - 1}} = \dfrac{1}{3}$
By applying cross multiplication in the above equation, we get
$\Rightarrow 3\left( {x - 1} \right) = 1\left( {y - 1} \right) \Rightarrow 3x - 3 = y - 1 \\ \Rightarrow 3x - y - 2 = 0{\text{ }} \to {\text{(1)}} \\$
Now, when 3 is added to both the numerator and the denominator of the original fraction, the fraction reduces to $\dfrac{{x + 3}}{{y + 3}}$.
According to problem statement, $\dfrac{{x + 3}}{{y + 3}} = \dfrac{1}{2}$
By applying cross multiplication in the above equation, we get
$\Rightarrow 2\left( {x + 3} \right) = 1\left( {y + 3} \right) \Rightarrow 2x + 6 = y + 3 \\ \Rightarrow 2x - y + 3 = 0{\text{ }} \to {\text{(2)}} \\$
Following the approach of the Elimination method.
By subtracting equation (2) from equation (1), we get
$\Rightarrow 3x - y - 2 - \left( {2x - y + 3} \right) = 0 - 0 \Rightarrow 3x - y - 2 - 2x + y - 3 = 0 \Rightarrow x - 5 = 0 \Rightarrow x = 5$
By putting $x = 5$ in equation (1), we get
$\Rightarrow \left( {3 \times 5} \right) - y - 2 = 0 \Rightarrow 15 - 2 = y \Rightarrow y = 13$
Therefore, the numerator of the original fraction (x) is 5 and the denominator of the original fraction (y) is 13.
Hence, the original fraction is $\dfrac{5}{{13}}$.

Note- In this particular problem, we can also use substitution method instead of elimination method in order to obtain the values of two variables assumed (x and y). Here, we can also verify that the obtained original fraction is correct or not by simply subtracting 1 from both numerator and denominator which gives $\dfrac{4}{{12}} = \dfrac{1}{3}$ and by adding 3 to both numerator and denominator which gives $\dfrac{8}{{16}} = \dfrac{1}{2}$.