Question 1043772
Let {{{ x }}} = ml of water in x bottle initially
Let {{{ y }}} = ml of water in y bottle initially
-----------------------------------
(1) 
After 1st step:
Bottle x has: {{{ x - 2y }}} ml
Bottle y has: {{{ y + 2y }}} ml
-------------------------
(2)
After 2nd step:
Bottle x has: {{{ ( x - 2y ) + ( x - 2y ) }}}
Bottle y has: {{{ ( y + 2y ) - ( x - 2y ) }}}
-----------------------------------
I am given that 
{{{ ( x - 2y ) + ( x - 2y ) = 200 }}}
{{{ ( y + 2y ) - ( x - 2y ) = 200 }}}
----------------------------
(1) {{{ 2x - 4y = 200 }}}
(2) {{{ -x + 5y = 200 }}}
-----------------------
Multiply both sides of (2) by {{{ 2 }}}
(1) {{{ 2x - 4y = 200 }}}
(2) {{{ -2x + 10y = 400 }}}
----------------------
{{{ 6y = 600 }}}
{{{ y = 100 }}}
---------------
If both bottle end up with 200 ml, there has to 
be 400 ml total, so
{{{ 400 - y = x }}}
{{{ 400 - 100 = x }}}
{{{ x = 300 }}}
--------------------
Initially, x has 300 ml
y has 100 ml
------------------------
check answer:
(1) 
After 1st step:
Bottle x has: {{{ x - 2y = 300 - 2*100 }}} 
{{{ 300 - 200 = 100 }}} ml
Bottle y has: {{{ y + 2y = 100 + 2*100 }}} 
{{{ 100 + 200 = 300 }}} ml
--------------------------------
(2)
After 2nd step:
Bottle x has: {{{ ( x - 2y ) + ( x - 2y ) = ( 325 - 2*75 ) + ( 325 - 2*75) }}}
{{{ 300 - 200 + 300 - 200 = 600 - 400 }}}
{{{ 600 - 400 = 200 }}} ml
Bottle y has: {{{ ( y + 2y ) - ( x - 2y ) = 3*100 - 300 + 2*100 }}}
{{{ 300 - 300 + 200 = 200 }}} ml
----------
Both bottles end up with 200 ml
OK