Question

the sum of two number is 8 and the sum of their reciprocal is 8/15 . find the number

Answer 1

Let numbers are x and ( 8 - x ),

Given, Sum of their reciprocal = 8 / 15

ATQ,

1 / x + 1 / ( 8 - x ) = 8 / 15

$\frac{8 - x + x }{x( 8 - x)} = \frac{8}{15}\\ \\ \rightarrow \frac{8}{8x - x^2} = \frac{8}{15} \\ \\ \rightarrow \frac{1}{8x - x^2} = \frac{1}{15} \\ \\ \rightarrow 15 = 8x - x^2 \\ \\ \rightarrow x^2 - 8x + 15 = 0 \\ \\ \rightarrow x^2 - ( 5 + 3 ) x + 15 = 0 \\ \\ \rightarrow x^2 - 5x - 3x + 15 = 0 \\ \\ \rightarrow x( x - 5 ) - 3( x - 5 ) = 0 \\ \\ \rightarrow ( x - 5 ) ( x - 3 ) = 0$

Hence, x - 5 = 0 Or x - 3 = 0

x = 5 Or x = 3

Answer 2

Let the one number be x
and other number be y
x+y =8 (given)
x = 8-y ....(1)
$\frac{1}{x} + \frac{1}{y} = \frac{8}{15} \\ \frac{y + x}{xy} = \frac{8}{15} \\ 15y + 15x = 8xy \\ 15x + 15y = 8xy \\ 15(x + y) = 8xy \\ 15(8) = 8(8 - y)y \: \: \: \: \:[from \: equaton \: (1)] \\ 120 = 64y - 8y ^{2} \\ 8 {y }^{2} - 64y + 120 = 0 \\ {y}^{2} - 8y + 15 = 0 \\ {y}^{2} - 5y - 3y + 15 = 0 \\ y(y - 5) - 3(y - 5) = 0 \\ (y - 3)(y - 5) = 0 \\ y - 3 = 0 \\ y = 3 \\ and \\ y - 5 = 0 \\ y = 5 \\we \: know \: that \: = > x + y = 8 \\ when \: \: y = 3, \: x = 5 \\ when \: y = 5 \:, \: x = 3 \\ thus , the \: two \: numbers \: are \: 3 \: and \: 5$