Question

The denominator of an irreducible fraction is greater than the numerator by $2$. If we reduce the numerator of the reciprocal fraction by $3$ and subtract the given fraction from the resulting one, we get $\frac{1}{15}$. Find the given fraction.

Answer 1

Let $x=$ numerator of given fraction.

Then $x+2=$ denominator.

Given fraction$:\frac{x}{x+2}$

Reciprocal fraction$=\frac{x+2}{x}$

New Fraction:

$\frac{(x+2)-3}{x}-\frac{x}{x+2}=\frac{1}{15}$

$\Rightarrow \frac{x-1}{x}-\frac{x}{x+2}=\frac{1}{15}$

$\Rightarrow 15(x+2)(x-1)-15x\left(x\right)=x(x+2)$

$\Rightarrow 15x^2+15x-30-15x^2=x^2+2x$

$\Rightarrow x^2-13+30=0$

$\Rightarrow (x-10)(x-3)=0$

$\Rightarrow x=10$ or $x=3$

If $x=10$, the fraction can be reduced.

If $x=3$, the given fraction is $\frac{3}{5}$ and this works

$\frac{2}{3}-\frac{3}{5}=\frac{1}{15}$