Question

One positive integer is five more than the other. When the reciprocal of the larger is subtracted

from the reciprocal of the smaller the result is 5/14. Find the two integers

If you could, please show me the steps so that I will know how to solve a problem similar to this one in the future. Thanks!!!

Answer 1

Let x be a first positive integer number, the second number is 5 more then x and you can write that the second number is x+5.
The recipracals of numbers x and x+5 are
`(1)/(x)` and
`(1)/(x+5)`, respectively. Since x<x+5, then
`(1)/(x) \ \textgreater \ (1)/(x+5)` and

`(1)/(x)- (1)/(x+5)=(5)/(14)`.
To solve this equation you have to find common denominator:

`((x+5)-x)/(x(x+5))= (5)/(14)`

`(5)/(x(x+5))= (5)/(14)`

`(1)/(x(x+5))= (1)/(14)`, then

`x(x+5)=14` and

`x^(2) +5x-14=0`

`D=5^2-4\cdot(-14)=25+56=81,\ √(D)=9` and solutions are

`x_1= (-5+9)/(2) =2` and
`x_2= (-5-9)/(2) =-7`.
Since x is positive integer, the right solution is x=2 and incorrect x=-7.
Answer: x=2, x+5=7