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The sum of a fraction and its reciprocal is $\dfrac{13}{6}$. Find the fraction if its numerator is 1 less than the denominator.

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Last updated date: 20th Jun 2024
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Answer
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Hint: We are given that the numerator is one less than the denominator, thus we shall the numerator and denominator as two separate variables. We will then express the numerator variable in terms of the denominator variable and solve the quadratic equation thus formed by the factoring method.

Complete step by step solution:
Let the numerator of the fraction be $x$ and let the denominator of the fraction be $y$.
Given that sum of fraction and its reciprocal is $\dfrac{13}{6}$.
$\Rightarrow \dfrac{x}{y}+\dfrac{y}{x}=\dfrac{13}{6}$ ………….. (1)
We have also been given the relation that the numerator is 1 less than the denominator.
$\Rightarrow x=y+1$
Thus, the fraction formed is $\dfrac{x}{y}=\dfrac{y+1}{y}$. ………….. (2)
We shall make changes accordingly to equation (1) by substituting $x=y+1$.
$\Rightarrow \dfrac{y+1}{y}+\dfrac{y}{y+1}=\dfrac{13}{6}$
Taking LCM on the left-hand side of the equation, we get
$\begin{align}
  & \Rightarrow \dfrac{y+1\left( y+1 \right)+y\left( y \right)}{y\left( y+1 \right)}=\dfrac{13}{6} \\
 & \Rightarrow \dfrac{{{\left( y+1 \right)}^{2}}+{{y}^{2}}}{y\left( y+1 \right)}=\dfrac{13}{6} \\
\end{align}$
Using the algebraic property ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we shall expand ${{\left( y+1 \right)}^{2}}$.
$\begin{align}
  & \Rightarrow \dfrac{{{y}^{2}}+1+2y+{{y}^{2}}}{y\left( y+1 \right)}=\dfrac{13}{6} \\
 & \Rightarrow \dfrac{2{{y}^{2}}+2y+1}{{{y}^{2}}+y}=\dfrac{13}{6} \\
\end{align}$
Cross-multiplying both sides with each other we get,
$\begin{align}
  & \Rightarrow \left( 2{{y}^{2}}+2y+1 \right)6=13\left( {{y}^{2}}+1 \right) \\
 & \Rightarrow 12{{y}^{2}}+12y+6=13{{y}^{2}}+13y \\
 & \Rightarrow {{y}^{2}}+y-6=0 \\
\end{align}$
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, ${{y}^{2}}+y-6=0$, $a=1,$ $b=1$ and $c=-6$.
We will find numbers by hit and trial whose product is equal to $-6\times 1=-6$ and whose sum is equal to 1
Such two numbers are 3 and $-2$ as $3-2=1$ and $3\times -2=-6$.
Now, factoring the equation:
$\Rightarrow {{y}^{2}}+3y-2y-6=0$
Taking common, we get:
$\begin{align}
  & \Rightarrow y\left( y+3 \right)-2\left( y+3 \right)=0 \\
 & \Rightarrow \left( y-2 \right)\left( y+3 \right)=0 \\
\end{align}$
Hence, $y-2=0$ or $y+3=0$
On transposing the constant terms to the other side, we get
$\Rightarrow y=-3$ or $y=2$
Therefore, the roots of the quadratic equation are $y=-3,2$.
 Substituting $y=2$ in equation (2), we get
$\dfrac{x}{y}=\dfrac{2+1}{2}$
$\Rightarrow \dfrac{x}{y}=\dfrac{3}{2}$

Therefore, the fraction is $\dfrac{3}{2}$.

Note: Another method of solving the quadratic equation formed was by first finding the discriminant of the equation and then calculating the roots by the predefined method as per given in algebraic mathematics.