SOLUTION: Let a, b, and c be integers that satisfy 2a + 3b = 52, 3b + c = 41, and bc = 60. Find a+b+ c.

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
...........eq.2, solve for
...........eq1)
...........eq.3, solve for
................2)

from 1) and 2) we have

........solve for

....factor

solutions:

or

go to
...........eq1), substitute

or

so we have two pairs of solutions for and
,
and
,
go to
...........eq.1, substitute

one solution is: , ,

then

...........eq.1, substitute

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): You can put this solution on YOUR website!

Taking just E2 and E3,

.
.

.
.

(Method may not have been the intention of the exercise)
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