Question 1186592
let  the number of coins in Box X be {{{x}}} and  the number of coins in Box Y be {{{y}}}

total number {{{x+y}}}

if there were 218 more coins Box X than in Box Y, we have

{{{x=y+218}}}.....eq.1



After {{{76}}} coins were removed from Box Y, we have {{{y-76}}} 

and {{{34}}} coins were removed from Box X,  we have {{{x-34}}}


the number of coins in Box Y was 1/6 of the total number of coins in the two boxes.

{{{y-76=(1/6)((x-34)+(y-76))}}} ..........substitute {{{x}}} from eq.1

{{{y-76=(1/6)(y+218-34+y-76)}}}

{{{y-76=(1/6)(2y+108)}}}.........both sides multiply by {{{6}}}

{{{6y-456=2y+108}}}
{{{6y-2y=456+108}}}
{{{4y=564}}}

{{{y=141}}}



answer: {{{highlight(141)}}}  coins were there in Box Y at first 

check:

{{{x=141+218=359}}} coins were there in Box X

total: {{{359+141=500}}}

After {{{76}}} coins were removed from Box Y, we have {{{141-76=65}}} 

and {{{34}}} coins were removed from Box X,  we have {{{359-34=325}}}
total: {{{325+65=390}}}

now in  Box Y is {{{65}}} and it should be {{{1/6}}} of {{{390}}}: {{{390/6=65}}} ->true