Question 1098822
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The condition says that after all exchanges there were an equal number of sweets in all 3 boxes.

{{{900/3}}} = 300, which means that finally there were 300 sweets in each box.


Further, the condition says


a - 18 + {{{(1/3)*(c+5)}}} = 300,    (1)

b + 18 - 5 = 300,           (2)

(c+5) - {{{(1/3)*(c+5)}}} = 300.     (3)


From (3),  {{{(2/3)*(c+5)}}} = 300  ====>  c+5 = {{{(300*3)/2}}} = {{{900/2}}} = 450  ====>  c = 450-5 = 445.

From (2),  b = 300 + 5 - 18 = 287.

From (1), a = 300 - {{{1/3)*(445+5)}}} + 18 = 300 - 150 + 18 = 168.



<U>Answer</U>.  Originally, there were 168 sweets in A, 287 sweets in B  and 445 sweets in C.
</pre>

Solved.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;There is NO NEED to solve system in 3 equations in 3 unknowns.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This problem is for young students (5 - 6 grades) who have no any notion on systems of equations.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution and the approach by @josgarithmetic go entirely and totally out of the target.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;For your safety, simply ignore his writing.