Question

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

Answer 1

Suppose that the dfraction $\frac{x}{y}$

According to the given question,

$\frac{x}{y}+\frac{y}{x}=\frac{13}{6}$

$\Rightarrow x^2+y^2=\frac{13}{6}xy$

$\Rightarrow 6x^2+6y^2=13xy$

$\Rightarrow 6x^2+6y^2-13xy=0.......\left(1\right)$

Now, According to the given question,

$x=y-1........\left(2\right)$

Put the value of y by equation (2) in (1) and we get,

$6{(y-1)}^2+6y^2-13(y-1)y=0$

$\Rightarrow 6(y^2+1^2-2y)+6y^2-13y^2+13y=0$

$\Rightarrow 6y^2+6-12y+6y^2-13y^2+13y=0$

$\Rightarrow -y^2+y+6=0$

$\Rightarrow y^2-y-6=0$

$\Rightarrow y^2-(3-2)y-6=0$

$\Rightarrow y^2-3y+2y-6=0$

$\Rightarrow y(y-3)+2(y-3)=0$

$\Rightarrow (y-3)(y+2)=0$

$\Rightarrow y-3=0y+2=0$

$\Rightarrow y=3y=-2$

Put the value of y in equation (2) and we get,

$x=y-1........\left(2\right)$

$x=3-1=2$

Then, the fraction is $=\frac{x}{y}=\frac{2}{3}$

Hence, this is the answer.