SOLUTION: Divide 46 into two parts such that the sum of the quotients obtained by dividing one part by 7 and the other part by 3may be equal to 10.

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Question 938084: Divide 46 into two parts such that the sum of the quotients obtained by dividing one part by 7 and the other part by 3may be equal to 10.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let x and y be the two parts of 46 .
x%2By=46
The quotients (if there is no remainder) would be
x%2F7 and y%2F3 such that
x%2F7%2By%2F3=10 ---> 3x%2B7y=210
So, we have the system of linear equations
system%28x%2By=46%2C3x%2B7y=210%29-->system%28x=46-y%2C3x%2B7y=210%29-->system%28x=46-y%2C3%2846-y%29%2B7y=210%29-->system%28x=46-y%2C138-3y%2B7y=210%29-->system%28x=46-y%2C138%2B4y=210%29-->system%28x=46-y%2C4y=210-138%29-->system%28x=46-y%2C4y=72%29-->system%28x=46-y%2Cy=18%29-->system%28x=46-18%2Cy=18%29-->highlight%28system%28x=28%2Cy=18%29%29

NOTE: If the divisions have a quotient and a remainder,
the problem gets complicated.
We would have positive integer quotients a and b
and positive integer remainders R and r .
The two parts of 46 would be
x=7a%2BR with 1%3C=R%3C=6, so R= 1, or 2, or 3, or 4, or 5, or 6, and
y=3b%2Br with 1%3C=r%3C=2, so r= 1, or 2.
Then a%2Bb=10 and %287a%2BR%29%2B%283b%2Br%29=46--->7a%2B3b=46-%28R%2Br%29 .
We would solve system%28a%2Bb=10%2C7a%2B3b=46-%28R%2Br%29%29 to get
system%28a=4-%28R%2Br%29%2F4%2Cb=10-a%29 .
For {{a}} and b to be integers, with 2%3C=R%2Br%3C%2B6%2B2=8 ,
R%2Br must be a multiple of 4
If R%2Br=8 it must be
system%28R=6%2Cr=2%2Ca=4-2=2%2Cb=10-2=8%29 ---> system%28x=7%2A2%2B6%2Cy=3%2A8%2B2%29 ---> system%28x=20%2Cy=26%29
If R%2Br=4 it could be
system%28R=2%2Cr=2%2Ca=4-1=3%2Cb=10-3=7%29 ---> system%28x=7%2A3%2B2%2Cy=3%2A7%2B2%29 ---> system%28x=23%2Cy=23%29 , or
system%28R=3%2Cr=1%2Ca=4-1=3%2Cb=10-3=7%29 ---> system%28x=7%2A3%2B3%2Cy=3%2A7%2B1%29 ---> system%28x=24%2Cy=22%29