Question 1178678: Let a, b, and c be integers that satisfy 2a + 3b = 52, 3b + c = 41, and bc = 60. Find a+b+ c.
Found 3 solutions by MathLover1, greenestamps, josgarithmetic:
Answer by MathLover1(20849)   (Show Source):
You can put this solution on YOUR website!
 ...........eq.1
 ...........eq.2
 ...........eq.3
-----------------------------
Find 
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...........eq.3, solve for
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solutions:
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other solution is:  ,  ,
then 
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
(1) The fastest way to the answer: trial and error with simple mental arithmetic.
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.
Then use 2a+3b=52 with b=12 to find a=8.
And last a+b+c = 8+12+5 = 25.
ANSWER: 25
(2) Using formal algebra....
3b+c=41
c = 41-3b
bc = b(41-3b) = 60
41b-3b^2 = 60
3b^2-41b+60 = 0
Factor into the form
(3b-m)(b-n) = 0
3b^2-(m+3n)+mn=0
We need mn=60 and m+3n=41....
Note that is exactly what we needed in solving the problem using trial and error -- so using the formal algebra didn't make solving the problem easier.
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.
Continuing then...
(3b-5)(b-12)=0
b = 5/3 (not an integer) or b=12; so b=12
Then, as before, c=5 and a=8.
Answer by josgarithmetic(39617)  (Show Source):
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