Question

The denominator of a fraction is 2 more than its numerator if the sum of the fraction and its reciprocal is 34 by 15 find the fraction

Answer 1

Answer:

$\frac{3}{5}$ ,  $\frac{5}{3}$

Step-by-step explanation:

Let x be the numerator .

Since we are given that denominator is 2 more than the numerator .

So, denominator = x+2

So Fraction =  $\frac{x}{x+2}$

Reciprocal of fraction =  $\frac{x+2}{x}$

Now we are given that the sum of fraction and its reciprocal is 34/15.

⇒$\frac{x}{x+2}+\frac{x+2}{x} =\frac{34}{15}$

⇒$\frac{(x*x)+[(x+2)*(x+2)]}{(x+2)(x)}=\frac{34}{15}$

⇒$\frac{2x^{2}+4x+4}{x^{2}+2x}=\frac{34}{15}$

⇒$15*(2x^{2}+4x+4)=34*(x^{2}+2x)$

⇒$30x^{2}+60x+60)=34x^{2}+68x$

⇒$4x^{2}+8x-60=0$

⇒$x^{2}+2x-15=0$

⇒$x^{2}+5x-3x-15=0$

⇒$x(x+5)-3(x+5)=0$

⇒$(x-3)(x+5)=0$

⇒$x-3=0, x+5=0$

⇒ x = 3 , x= -5

So Fraction when x = 3

$\frac{x}{x+2} =\frac{3}{3+2}=\frac{3}{5}$

Fraction when x = -5

$\frac{x}{x+2} =\frac{-5}{-5+2}=\frac{5}{3}$

Answer 2

Answer: $\bf\huge\frac{3}{5}$

Step-by-step explanation:

Let numerator be p.

Denominator is p + 2.

⇒ Fraction = $\bf\huge\frac{p}{p + 2}$

$\bf\huge\frac{p}{p + 2} + \frac{p + 2}{p} = \frac{34}{15}$

⇒ 15(p^2+ p^2 + 4p + 4) = 34(p^2 + 2p)

⇒ 30p^2 + 60p + 60 = 34p^2 + 68p  

⇒ 4p^2 + 8p - 60 = 0

⇒ p^2 + 2p - 15 = 0

⇒ p^2 + 5p - 3p - 15 = 0

⇒ p(p + 5) - 3(p + 5) = 0

⇒ (p + 5)(p - 3) = 0

⇒ p = 3

Therefore

Fraction = $\bf\huge\frac{3}{5}$