Let the numerator be \(x\)
If the numerator is 5 less then denominator, then we write
\(\frac{x}{x+5}\) or \(\frac{x-5}{x}\)dding 6 to the numerator and 4 to the denominator, we write
\(\frac{x+6}{x+5+4}=\frac{x+6}{x+9}\)
\begin{array}{l}
\Rightarrow \frac{x+6}{x+9}=2\left(\frac{x}{x+5}\right) \\
\frac{x+6}{x+9}=\frac{2 x}{x+5} \\
\text { We cross - multiply } \\
(x+6)(x+5)=2 x(x+9) \\
x^2+11 x+30=2 x^2+18 x \\
2 x^2-x^2+18 x-11 x-30=0 \\
\Rightarrow x^2+7 x-30=0
\end{array}
by the formula method,
\begin{array}{l}
a=1, b=7, c=-30 \\
x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
x=\frac{-7 \pm \sqrt{7^2-4(1)(-30)}}{2(1)}
\end{array}
\begin{array}{l}
\Longrightarrow x=\frac{-7 \pm \sqrt{49+120}}{2} \\
x=\frac{-7 \pm \sqrt{169}}{2} \\
x=\frac{-7 \pm 13}{2} \\
x=\frac{-7-13}{2} \text { and } \frac{-7+13}{2} \\
x=\frac{-20}{2} \text { and } \frac{6}{2} \\
x=-10 \text { and } 3 \\
\text { therefore, the fraction is } \\
\frac{-10}{-10+5}=\frac{-10}{-5}=2 \\
\text { or with } x=3 \\
\frac{3}{3+5}=\frac{3}{8}
\end{array}
Note: \(x=-10\) will not satisfy the given condition that the numerator is 5 less than the denominator, hence
\(x=3\) with the numerator \(=3 / 8\)