#### Answer

$x=15$

#### Work Step by Step

$\dfrac{1}{x}+\dfrac{2}{5}=\dfrac{7}{x} \\
\dfrac{5+2x}{5x}=\dfrac{7}{x} \\
x(5+2x)=7(5x) \\
5x+2x^2=35x \\
2x^2+5x-35x=0 \\
2x^2 -30x=0 \\
x^2-15x=0 \\
x(x-15)=0 \\ $
But the solution is only $x=15$, because if $x=0$ then the original ecuation would be undefined