<PRE><B>Question 1178678</B>
&lt;br&gt;
(1) The fastest way to the answer: trial and error with simple mental arithmetic.&lt;br&gt;
Find pairs of integers b and c for which the product bc is 60, then find the pair for which 3b+c=41.  Answer b=12, c=5.&lt;br&gt;
Then use 2a+3b=52 with b=12 to find a=8.&lt;br&gt;
And last a+b+c = 8+12+5 = 25.&lt;br&gt;
ANSWER: 25&lt;br&gt;
(2) Using formal algebra....&lt;br&gt;
3b+c=41
c = 41-3b
bc = b(41-3b) = 60
41b-3b^2 = 60
3b^2-41b+60 = 0&lt;br&gt;
Factor into the form&lt;br&gt;
(3b-m)(b-n) = 0&lt;br&gt;
3b^2-(m+3n)+mn=0&lt;br&gt;
We need mn=60 and m+3n=41....&lt;br&gt;
Note that is exactly what we needed in solving the problem using trial and error -- so using the formal algebra didn&#39;t make solving the problem easier.&lt;br&gt;
However, assuming this question comes from a formal math course, you should know how to set up and solve the problem using the formal algebra.&lt;br&gt;
Continuing then...&lt;br&gt;
(3b-5)(b-12)=0
b = 5/3 (not an integer) or b=12; so b=12&lt;br&gt;
Then, as before, c=5 and a=8.&lt;br&gt;
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