Question

Numerator of fraction is 3 less than its denominator. If 1 is added to the denominator, fraction is decreased by $\dfrac{1}{{15}}$. Find the fraction.

Answer 1

Hint— Since we have to find the fraction then let the value of numerator or denominator and accordingly find the other value. So, here we will let the denominator of the fraction be $x$ and we are given that the numerator of fraction is 3 less than its denominator. So, we get the numerator as $x - 3$. Hence, our fraction becomes $\dfrac{{x - 3}}{x}$. Simplify the fraction and find the value of $x$ and thus, find the fraction.

Complete step-by-step solution
First let the denominator of the required fraction be $x$ and we know from the question that the numerator is 3 less than its denominator so, we get the numerator as $x - 3$.
Thus, the expression of the fraction becomes $\dfrac{{x - 3}}{x}$. ---(1)
Now, we are given the question that if 1 is added to the denominator, the required fraction will get decreased by $\dfrac{1}{{15}}$.
So, according to this, we get,
$\dfrac{{x - 3}}{{x + 1}} = \left( {\dfrac{{x - 3}}{x}} \right) - \dfrac{1}{{15}}$
Simplify the right-hand side of the equation by taking the L.C.M.,
Thus, we get,

$$\Rightarrow \dfrac{{x - 3}}{{x + 1}} = \dfrac{{15\left( {x - 3} \right) - 1\left( x \right)}}{{15x}} \\\ \Rightarrow \dfrac{{x - 3}}{{x + 1}} = \dfrac{{15x - 45 - x}}{{15x}} \\\ \Rightarrow \dfrac{{x - 3}}{{x + 1}} = \dfrac{{14x - 45}}{{15x}} \\\$$

Now, we will cross multiply the expression and simplify further,

$$\Rightarrow 15x\left( {x - 3} \right) = \left( {14x - 45} \right)\left( {x + 1} \right) \\\ \Rightarrow 15{x^2} - 45x = 14{x^2} + 14x - 45x - 45 \\\ \Rightarrow 15{x^2} - 14{x^2} = 14x + 45x - 45x - 45 \\\ \Rightarrow {x^2} - 14x + 45 = 0 \\\$$

Next, we will find the roots of the equation using middle term splitting method,
Thus, we get,

$$\Rightarrow {x^2} - 9x - 5x + 45 = 0 \\\ \Rightarrow x\left( {x - 9} \right) - 5\left( {x - 9} \right) = 0 \\\ \Rightarrow \left( {x - 9} \right)\left( {x - 5} \right) = 0 \\\$$

Further, we will apply the zero-factor property,
$\Rightarrow \left( {x - 9} \right) = 0$ or $\left( {x - 5} \right) = 0$
$\Rightarrow x = 9$ or $x = 5$
So, here, we get the two values for $x$ which shows that there are two possibilities of the fraction.
So, consider the first value, $x = 9$ and we will substitute the value in the expression $\dfrac{{x - 3}}{x}$.
Thus, the required fraction is $\dfrac{{9 - 3}}{9} = \dfrac{6}{9}$
Next, consider the second value, $x = 5$ and we will substitute the value in the expression $\dfrac{{x - 3}}{x}$.
Thus, the required fraction is $\dfrac{{5 - 3}}{5} = \dfrac{2}{5}$
Thus, we get the two fractions after solving,
Hence, the required fractions are $\dfrac{6}{9}$ or $\dfrac{2}{5}$.

Note: When we are having the quadratic equation then it's necessary to have 2 roots of the equation. After finding the roots of the above question, substitute it in the original fraction in terms of $x$ and not the equation we have found. We can find the roots of the equation by applying the quadratic formula also to find the roots of the equation.