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 [tex] \frac{1}{x} [/tex] and [tex] \frac{1}{x+5} [/tex], respectively. Since x<x+5, then [tex] \frac{1}{x} \ \textgreater \ \frac{1}{x+5} [/tex] and
[tex] \frac{1}{x}- \frac{1}{x+5}=\frac{5}{14} [/tex].
To solve this equation you have to find common denominator:
[tex] \frac{(x+5)-x}{x(x+5)}= \frac{5}{14} [/tex]
[tex] \frac{5}{x(x+5)}= \frac{5}{14} [/tex]
[tex] \frac{1}{x(x+5)}= \frac{1}{14} [/tex], then
[tex]x(x+5)=14[/tex] and
[tex] x^{2} +5x-14=0[/tex]
[tex]D=5^2-4\cdot(-14)=25+56=81,\ \sqrt{D}=9 [/tex] and solutions are
[tex]x_1= \frac{-5+9}{2} =2[/tex] and [tex]x_2= \frac{-5-9}{2} =-7[/tex].
Since x is positive integer, the right solution is x=2 and incorrect x=-7.
Answer: x=2, x+5=7
