What is the value of x?

Let $\overline{AB}=a, \overline{AF}=b$

$\triangle ADB$ is similar to $\triangle AEF$ therefore $\dfrac{b}{x}=\dfrac{a}{60}$

$\triangle BCA$ is similar to $\triangle BEF$ therefore $\dfrac{a-b}{x}=\dfrac{a}{40}$

From the first equation $a=\dfrac {60b}{x}$

Substituting into second equation $\dfrac{1}{x}\left(\dfrac{60b}{x}-b\right)=\dfrac{60b}{40x}$

$\dfrac{60}{x}-1=\dfrac32, x=\boxed{24}$

We can use the relationship:

$\dfrac{1}{x}=\dfrac{1}{40}+\dfrac{1}{60}$

It follows that:

$\dfrac{1}{x}=\dfrac{100}{2400}$

Finally:

$\color{#3D99F6}\boxed{x=24}$