Two of the altitudes of an acute triangle divide the sides

Two of the altitudes of an acute triangle divide the sides into segments of lengths 5,3,2 and x units, as shown. What is the value of x?

x = 18

Call the apex angle  A

The  left base angle B

The right base angle C

Let the altitude meeting base (3 + 5)  =  8  be DC

Let the altitude meeting base 2 + x  be EB

Note  that right   triangles  AEB   and   ADC are similar

This implies that    AB/ AE  =  AC/ AD 

So

8 / 2 = (2 + x) /  5      

4  = (2 + x)  / 5

5*4  = 2 + x

x = 5*4  - 2

x   = 18    ( just as  jugoslav found  !!! )