Answer:
[tex]\huge\boxed{14\ \text{and}\ 7}[/tex]
Step-by-step explanation:
[tex]n,\ m-\text{positive integer}\\\\n=2m-\text{a positive integer is twice another}\\\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}-\text{the sum of the reciprocal of the two positive integer is }\ \dfrac{3}{14}\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}n=2m&(1)\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}&(2)\end{array}\right[/tex]
[tex]\text{Substitute (1) to (2):}\\\\\dfrac{1}{2m}+\dfrac{1}{m}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{1\cdot2}{m\cdot2}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{2}{2m}=\dfrac{3}{14}\\\\\dfrac{1+2}{2m}=\dfrac{3}{14}\\\\\dfrac{3}{2m}=\dfrac{3}{14}\Rightarrow2m=14\qquad\text{divide both sides by 2}\\\\\dfrac{2m}{2}=\dfrac{14}{2}\\\\\boxed{m=7}[/tex]
[tex]\text{Substitute it to (1):}\\\\n=2\cdot7\\\\\boxed{n=14}[/tex]
