 ## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 26 of Exercise 10E

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26
##### The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is . Find the fraction.

Let the numerator be x

Denominator = x + 3

Original number =  On taking the LCM   { using (a + b)2 = a2 + 2ab + b2} 29x2 + 87x = 20x2 + 60x + 90

9x2 + 27x – 90 = 0

9(x2 + 3x – 10) = 0

x2 + 3x – 10 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 3 c = – 10

= 1. – 10 = – 10

And either of their sum or difference = b

= 3

Thus the two terms are 5 and – 2

Difference = 5 – 2 = 3

Product = 5. – 2 = – 10

x2 + 5x – 2x – 10 = 0

x(x + 5) – 2(x + 5) = 0

(x + 5)(x – 2) = 0

(x + 5) = 0 or (x – 2) = 0

x = 2 or x = – 5

x = 2 (rejecting the negative value)

So numerator is 2

Denominator = x + 3 = 2 + 3 = 5

So required fraction is 2/5

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