Question

Sum of two natural numbers is 8 and the sum of their reciprocal is 8/15 . Find the numbers.​

Answer 1

Answer:

$x + y = 8$

$\frac{1}{x} + \frac{1}{y} = \frac{8}{15}$

$\implies \frac{x + y}{xy} = \frac{8}{15}$

$\implies xy = 15$

so

$x + \frac{15}{x} = 8$

${x}^{2} - 8x + 15 = 0$

or x=3

so numbers are 3 and 5

Answer 2

Since the sum of two numbers is 8.

So, Let the two natural numbers be x and 8-x.

Now,Since the sum of their reciprocals=8/15.

$= > \frac{1}{x} + \frac{1}{8 - x} = \frac{8}{15} \\ = > \frac{8 - x + x}{x(8 - x)} = \frac{8}{15} \\ = > \frac{8}{8x - {x}^{2} } = \frac{8}{15} \\ dividing \: both \:the \: sides \: by \: 8. \\ = > \frac{1}{8x - {x}^{2} } = \frac{1}{15} \\ = > 8x - {x}^{2} = 15 \\ = > {x}^{2} - 8x + 15 = 0 \\ = > {x}^{2} - 3x - 5x + 15 = 0 \\ = > x(x - 3) - 5(x - 3) = 0 \\ = > (x - 3)(x - 5) = 0 \\ = > x = 3 \: and \: x = 5$

=>The two numbers are 3 and 5 or 5 and 3.

Hope it is helpful.