Question

The numerator of fraction is less than its denominator by '2'. When '1' is added to both the numerator and denominator we get another fraction. The sum of these two fractions is $\frac{19}{15}$. The fraction is $\frac{3}{q}$, then $q$ is

Answer 1

Let the numerator be $x$

Therefore, denominator is $x+2$

The fraction is $\frac{x}{x+2}$

Now, $\frac{x}{x+2}+\frac{x+1}{x+3}=\frac{19}{15}$

$=>\frac{x^2+3x+(x+1)(x+2)}{(x+2)(x+3)}=\frac{19}{15}$

$=>\frac{x^2+3x+x^2+3x+2}{(x+2)(x+3)}=\frac{19}{15}$

$=>15(x^2+3x+x^2+3x+2)=19(x+2)(x+3)$

$=>15(2x^2+6x+2)=19(x^2+5x+6)$

$=>30x^2+90x+30=19x^2+95x+114$

$=>11x^2-5x-84=0$

$=>11x^2-33x+28x-84=0$

$=>11x(x-3)+28(x-3)=0$

$=>(11x+28)(x-3)=0$

Therefore, $x=3$ or $x=-\frac{28}{11}$

For $x=-\frac{28}{11}$, denominator is :-

$-\frac{28}{11}+2=\frac{-28+22}{11}=-\frac{6}{11}$

So, the fraction is $\frac{-\frac{28}{11}}{-\frac{6}{11}}=\frac{28}{6}=\frac{14}{3}$

Now adding $1$ on both numerator and denominator we get $\frac{15}{4}$

But, $\frac{14}{3}+\frac{15}{4}\ne \frac{19}{15}$

$\therefore x=3$ and denominator is $x+2=3+2=5$

$\therefore$ the fraction is $\frac{3}{5}$