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The denominator of a fraction is $2$ more than the numerator. If $2$ is added to both its numerator and denominator, then the sum of the new fraction and the original fraction is $\frac{46}{35}$. Find the original fraction.
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Let the fraction be $\frac{x}{x+2}$
Therefore, $\frac{x}{x+2} + \frac{x+2}{x+4} = \frac{46}{35}$
$\Rightarrow 24x^2 + 4x - 228 = 0$ or $6x^2 + x - 57 = 0$
$\Rightarrow (6x + 19)(x - 3) = 0$
$x \neq -\frac{19}{6}$
$\therefore x = 3$
So, the required fraction is $\frac{3}{5}$
Therefore, $\frac{x}{x+2} + \frac{x+2}{x+4} = \frac{46}{35}$
$\Rightarrow 24x^2 + 4x - 228 = 0$ or $6x^2 + x - 57 = 0$
$\Rightarrow (6x + 19)(x - 3) = 0$
$x \neq -\frac{19}{6}$
$\therefore x = 3$
So, the required fraction is $\frac{3}{5}$