Question

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

Answer 1

Let the fraction be $\frac{a}{b}$

$b-a=2⟹b=a+2$

$\frac{(b-3)}{\left(a\right)}-\frac{\left(a\right)}{\left(b\right)}=\frac{1}{15}$

$\frac{(a-1)}{\left(a\right)}-\frac{\left(a\right)}{(a+2)}=\frac{1}{15}$

$15(a^2+a-2-a^2)=a^2+2a$

$⟹a^2-13a+30=0$

$⟹a=10,3$

$⟹b=12,6$

$\frac{a}{b}=\frac{10}{12}=\frac{5}{6}$