The denominator of a fraction exceeds the numerator by 3 The sum of the fraction and its multiplicative inverse is $2 \frac{9}{28}$ . If this fraction is denoted by $\frac{p}{q}$ , then find the value of $p + q$

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Solution

Let the fraction be $\frac{x}{x + 3}$
Its multiplicative inverse $= \frac{x + 3}{x}$
Given, $= \frac{x}{x + 3} + \frac{x + 3}{x} = 2 \frac{9}{28}$
$\Rightarrow x^{2} + {(x + 3)}^{2} = \frac{65}{28} x (x + 3) \Rightarrow 56 x^{2} + 168 x$
$+ 252 = 65 x^{2} + 195 x \Rightarrow 9 x^{2} + 27 x - 252 = 0$
$\Rightarrow x^{2} + 3 x - 28 = 0 \Rightarrow (x + 7) (x - 4) = 0$
$\Rightarrow x = 4$ or $- 7$
$∴$ The fraction is 4/7

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The denominator of a fraction exceeds the numerator by 3 The sum of the fraction and its multiplicative inverse is 2928. If this fraction is denoted by pq, then find the value of p+q

Let the fraction be xx+3Its multiplicative inverse =x+3xGiven, =xx+3+x+3x=2928⇒x2+(x+3)2=6528x(x+3)⇒56x2+168x+252=65x2+195x⇒9x2+27x−252=0⇒x2+3x−28=0⇒(x+7)(x−4)=0⇒x=4 or −7∴ The fraction is 4/7

The denominator of a fraction exceeds the numerator by 3 The sum of the fraction and its multiplicative inverse is $2 \frac{9}{28}$ . If this fraction is denoted by $\frac{p}{q}$ , then find the value of $p + q$