Question 1110118
<pre>
Use the first equation to substitute 3a+b for 2c
in the second equation:

3b = 2c+a 
3b = 3a+b+a
3b = 4a+b
2b = 4a
 b = 2a
Divide both sides by 2b, to get a/b on the right

{{{(b)/(2b) = (2a)/(2b)}}}
{{{1/2=a/b}}}
1:2 = a:b
a:b = 1:2

Substitute 2a for b in

3a+b  = 2c
3a+2a = 2c
   5a = 2c

Divide both sides by 5c to get a/c on the left:

{{{(5a)/(5c) = (2c)/(5c)}}}

{{{a/c=2/5}}}

a:c = 2:5

So now we have:

a:b = 1:2
a:c = 2:5

Since 'a' corresponds to 1 in the first equation and
to 2 in the second equation, get them so that a corresponds
to the same number in both by multiplying both parts of the 
first ratio 1:2 by 2, getting 2:4

a:b = 2:4
a:c = 2:5

Now that a corresponds to the same number, 2, in both,
we can conclude:

a:b:c = 2:4:5

Edwin</pre>