SOLUTION: Sewer #1 contains 9 less rats, than sewer #2. Sewer #2 contains 5 fewer rats the sewer #3. If there are 56 gigantic rats in total, how many rats are in sewer #1?

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Question 279281: Sewer #1 contains 9 less rats, than sewer #2. Sewer #2 contains 5 fewer rats the sewer #3. If there are 56 gigantic rats in total, how many rats are in sewer #1?
Found 2 solutions by JBarnum, Stitch:
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
Sewer #1=a
Sewer #2=b
sewer #3=c
a=b-2
b=c-5
a%2Bb%2Bc=56
use substitution and salve for c
a%2Bb%2Bc=56
%28b-2%29%2B%28c-5%29%2Bc=56
%28%28c-5%29-2%29%2B%28c-5%29%2Bc=56
3c-12=56
3c=68
c=68%2F3=22.66
b=%2868%2F3%29-5
b=%2868%2F3%29-15%2F3
b=53%2F3=17.66
a=b-2
a=%2853%2F3%29-2
a=%2853%2F3%29-6%2F3
a=47%2F3=15.66 id say 15 rats as there arent partial rats in a sewer

Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
I think that his first equation was wrong
it should have been: A+=+B+-+9
So then you would have:
A+=+B+-+9
B+=+C+-+5
A+%2B+B+%2B+C+=+56
Use substitution and solve for C
A+%2B+B+%2B+C+=+56
%28B+-+9%29+%2B+%28C+-+5%29%2B+C+=+56
%28%28C+-+5%29-+9%29+%2B+%28C+-+5%29%2B+C+=+56
3C+-+19+=+56
3c+=+75
highlight%28C+=+25%29
B+=+C+-+5
B+=+25+-+5
highlight%28B+=+20%29
A+=+B+-+9
A+=+20+-+9
highlight%2811+=+A%29
There are 11 rats in sewer #1