Find equivalent resistance ?

Explanation

Let potential at point B is x and E is Y
`R_(eq)=(V)/(i)`
Applying KCl at point B
`(x-v)/(10)+(x-y)/(2)+(x-0)/(5)=0`
`8x-5y=v` .... (1)
Applying KCL at point E
`(y-v)/(5)+(y-x)/(2)+(y-0)/(10)=0`
`rArr8y-5x=2v` .... (2)
solving & y `x=(6v)/(13),y=(7v)/(13)`
current from branches BC & EF adds up to total current (i) flowing in the circuit.
`i=i_(3)+i_(4)=(x-0)/(5)+(y-0)/(10)=(19v)/(130)`
`becausei=(V)/(R_(eq)) " " thereforeR_(eq)=(130)/(19)`