If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+21xz+2yz+105x+10y+7z=812$, find $x+y+z$.
1 answer:
Answer:
10
Step-by-step explanation:
When we simplify we get
Then we continue to factor to get:
We then see that we can factor into
we then do the prime factorization of 847, which i think is, . we have to find the numbers that multiply to 847 and then plug them into z+5, 3x=1,2y+7.
It has to be a positive, non-negative integer, right?
We also see that 3x+1=11 so we see that x=10/3 (which wont work).
So 3x+1=7, so x=2.
So 11 has to be in another term. It has to be in 2y+7=11 so y=2
for the last term we get z+5=11 so z=6
2+2+6=10
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