Question 1105696
paul put some beads in jars A B C.
<pre>
Suppose in the beginning,
 
jar A contains "a" beads,
jar B contains "b" beads
jar C contains "c" beads
</pre>
the ratio of the number of beads in A to B 2 to 3. 
<pre>
{{{a/b=2/3}}}
{{{3a=2b}}}
</pre>
The ratio of beads in B to C is 2 to 1. 
<pre>
{{{b/c=2/1}}}
{{{b=2c}}}
</pre>
If paul transfers an equal number of beads from B to A 
and C he will have an equal number of beads in A and B 
<pre>
Let's do this in steps so that we don't get mixed up:

1. He takes n beads out of B, so B now contains only b-n beads.
2. He then puts those n beads in A, so A now contains a+n beads.
3. He then takes n more beads out of B, so B now contains only 
   b-2n beads. (He took another n beads out of B).
4. He puts those n beads in C, so C now contains c+n beads.

Now A and B have the same number of beads, so we set them
equal:

{{{a+n=b-2n}}}
{{{a+3n=b}}}
</pre>
and the total of beads in C will increase to 297.
<pre>
{{{c+n=297}}}

So we have 4 equations in 4 unknown

{{{system(3a=2b,b=2c,a+3n=b,c+n=297)}}}
</pre>
find the number of beads in all 3 jars.
<pre>
b appears in 3 of the equations, so we substitute
2c for b in the 1st, 3rd and 4th:

{{{system(3a=2(2c),a+3n=2c,c+n=297)}}}

{{{system(3a=4c,a+3n=2c,c+n=297)}}}

c occurs in all three equations, so we solve the second 
for c = 297-n and substitute 297-n for c in the first two
equations:

{{{system(3a=4(297-n),a+3n=2(297-n))}}}

{{{system(3a=1188-4n,a+3n=594-2n))}}}

{{{system(3a=1188-4n,a+5n=594))}}}

Solve the second equation for a = 594-5n and
substitute 594-5n for a in the first equation

{{{3(594-5n)=1188-4n)}}}

{{{1782-15n=1188-4n)}}}

{{{-11n=-594)}}}

{{{n=54}}}

Substitute 54 for n in

{{{a+5n=594))}}}
{{{a+5(54)=594))}}}
{{{a+270=594}}}
{{{a=324}}}

Also substitute 54 for n in 

{{{c+n=297}}}
{{{c+54=297}}}
{{{c=243}}}

Substitute 243 for c in

{{{b=2c}}}

{{{b=2(243)}}}
{{{b=486}}}

So A contained a=324 beads\
B contained b=486 beads
C contained c=243 beads

Checking:
 
1. When he took n=54 beads from B, B then contained only 
   486-54=432 beads.
2. When he put those 54 beads in A, A then contained 
   324+54=378 beads. 378.
3. When he took n=54 more beads out of B, B then contained 
   only 432-54=378 beads.

Aha!  That checks for A and B then both contained the same
number of beads,  

To finish,

4. When he put those n=54 beads in C, C then contained 
   243+54=297 beads.

That checks.  So we're right.  

Edwin</pre>