The denominator of a fraction is 3 more than its numerator. The fraction is
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Answer:
Hope the attachment helps you...
Let the numerator of the fraction be x.
Given that the denominator of a fraction is 3 more than its numerator = > x + 3.
The original fraction = x/x + 3 ----- (1)
Given that the sum of the fraction and its reciprocal is 29/10.
= \ \textgreater \ \frac{x}{x + 3} + \frac{x + 3}{x} = \frac{29}{10}= \textgreater
x+3
x
+
x
x+3
=
10
29
= \ \textgreater \ \frac{x}{x + 3} 10(x + 3) + \frac{x+ 3}{x} * 10x(x + 3) = \frac{29}{10}* 10x(x + 3)= \textgreater
x+3
x
10(x+3)+
x
x+3
∗10x(x+3)=
10
29
∗10x(x+3)
= > 10x^2 + 10(x + 3)^2 = 29x(x + 3)
= > 10x^2 + 10(x^2 + 9 + 6x) = 29x^2 + 87x
= > 10x^2 + 10x^2 + 90 + 60x = 29x^2 + 87x
= > 20x^2 + 90 + 60x = 29x^2 + 87x
= > -9x^2 - 27x + 90 = 0
= > -9(x^2 + 3x - 10) = 0
= > x^2 + 3x - 10 = 0
= > x^2 - 2x + 5x - 10 = 0
= > x(x - 2) + 5(x - 2) = 0
= > x = -5, x = 2.
Since x cannot be -ve, so x = 2.
Substitute in (1), we get
The original fraction = 2/5.
Hope this helps!