Answered By Sujalakshmy V

a) X represents a positive integer

b) X+1 represents the consecutive positive integer to X

c) 1/X represents the reciprocal of the first integer, X

d) 1/X+1 represents the reciprocal of the second integer, X+1

e) 1/X +1/X+1 represents the sum of the reciprocals of the two positive consecutive integers

f)   $\frac{1}{X}+\frac{1}{X+1}=\frac{31}{240}$1X +1X+1 =31240   

$\frac{X+1}{X\left(X+1\right)}+\frac{X}{X\left(X+1\right)}=\frac{31}{240}$X+1X(X+1) +XX(X+1) =31240  

 $\frac{X+1+X}{X\left(X+1\right)}=\frac{31}{240}$X+1+XX(X+1) =31240  

  $\frac{2X+1}{X\left(X+1\right)}=\frac{31}{240}$2X+1X(X+1) =31240  

  $2X+1=\frac{31}{240}X\left(X+1\right)$‍2X+1=31240 X(X+1) 

Multiply both sides by 240,

  $240\left(2X+1\right)=31X\left(X+1\right)$240(2X+1)=31X(X+1)

  $480X+240=31X^2+31$480X+240=31X²+31  

Rearranging  

31x² - 480x - 209 =0

on solving, using quadriatic formula,

x =   $\frac{480+\sqrt{\left(480^2\right)-\left(4\times31\times\left(-209\right)\right)}}{2\times31}$480+√(480²)−(4×31×(−209))2×31   and   $\frac{480-\sqrt{\left(480^2\right)-\left(4\times31\times\left(-209\right)\right)}}{_{2\times31}}$480−√(480²)−(4×31×(−209))_2×31  

x = 16 and -0.4

Since the question specifies X is a positive integer, the value of X = 16.

The answer to the question:

The two positive consecutive integers are 16 and 17

Answered By Leonardo F

We can solve this question by understanding the meaning of reciprocal. If we consider x to be an integer, then x+1 is its positive conscutive integer. Hence:

a) x is a positive integer

b) x+1 is its positive conscutive integer

c) 1/x is the reciprocal of the original positive integer

d) 1/(x+1) is the reciprocal of the consecutive integer

e) 1/x + 1/(x+1) is the sum of the reciprocals of the two positive consecutive integers

To solve the problem, we need to equal the expression on letter e to 31/240:

 $\frac{1}{x}+\frac{1}{x+1}=\frac{31}{240}$1x +1x+1 =31240  

Multiplying both sides of the equation by a common factor of  $x\left(x+1\right)$x(x+1) to get rid of the denominators, we will have:

 $\left(x+1\right)+x=\frac{31}{240}\left(x+1\right)x$(x+1)+x=31240 (x+1)x 

Rearranging the equation: 

 $2x+1=\frac{31}{240}\left(x^2+x\right)$2x+1=31240 (x²+x) 

 $2x+1=\frac{31}{240}x^2+\frac{31}{240}x$2x+1=31240 x²+31240 x 

 $\frac{31}{240}x^2+\frac{31}{240}x-2x-1=0$31240 x²+31240 x−2x−1=0 

 $\frac{31}{240}x^2+\left(\frac{31}{240}-2\right)x-1=0$31240 x²+(31240 −2)x−1=0 

Applying the quadratic formula to solve this equation:

 $a=\frac{31}{240}$a=31240  

 $b=\frac{31}{240}-2$b=31240 −2 

 $c=-1$c=−1 

 $x1=\frac{-\left(\frac{31}{240}-2\right)+\sqrt{\left(\frac{31}{240}-2\right)^2-4\left(\frac{31}{240}\right)\left(-1\right)}}{2\left(\frac{31}{240}\right)}$x1=−(31240 −2)+√(3124‍0 −2)²−4(31240 )(−1)2(31240 )  

 $x1=15$x1=15 

 $x2=\frac{-\left(\frac{31}{240}-2\right)-\sqrt{\left(\frac{31}{240}-2\right)^2-4\left(\frac{31}{240}\right)\left(-1\right)}}{2\left(\frac{31}{240}\right)}$x2=−(31240 −2)−√(31240 −2)²−4(31240 )(−1)2(31240 )  

 $x2\approx-0.516$x2≈−0.516

The only possible positive integer is 15. That means that its immediate successor is 16. Hence, the two numbers are 15 and 16 such that:

 $\frac{1}{15}+\frac{1}{16}=\frac{31}{240}$115 +116 =31240  

Answered By Leonardo F

The answer to this problem is 15 and 16, if we calculate the sum of the reciprocals using 16 and 17 it's not going to work, because: $\frac{1}{16}+\frac{1}{17}=\frac{33}{272}\ne\frac{31}{240}$116 +117 =33272 ≠31240