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The numerator of a fraction is $3$ less than its denominator. If $2$ is added to both numerator and denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$. Find the original fraction.
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Let denominator be $x$
$\therefore$ Numerator $= (x - 3)$
Therefore, fraction $= \frac{x-3}{x}$ ($1$)
ATQ
$\frac{x-3}{x} + \frac{x-3+2}{x+2} = \frac{29}{20}$ ($1$)
$\Rightarrow 11x^2 - 98x - 120 = 0$ ($1$)
$\Rightarrow (x-10)(11x + 12) = 0$ ($1/2$)
So, $x = 10$ ($1/2$)
$\therefore$ Fraction $= \frac{7}{10}$ ($1$)
$\therefore$ Numerator $= (x - 3)$
Therefore, fraction $= \frac{x-3}{x}$ ($1$)
ATQ
$\frac{x-3}{x} + \frac{x-3+2}{x+2} = \frac{29}{20}$ ($1$)
$\Rightarrow 11x^2 - 98x - 120 = 0$ ($1$)
$\Rightarrow (x-10)(11x + 12) = 0$ ($1/2$)
So, $x = 10$ ($1/2$)
$\therefore$ Fraction $= \frac{7}{10}$ ($1$)