Question

If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+21xz+2yz+105x+10y+7z=812$, find $x+y+z$.

Answer 1

Answer:

10

Step-by-step explanation:

When we simplify we get $z(6xy+21x+2y+7)+30xy+105x+10y=812.$

Then we continue to factor to get: $(z+5)(6xy+21x+2y+7)&=847.$

We then see that we can factor $(6xy+21x+2y+7)$ into $(z+5)(3x+1)(2y+7)&=847.$

we then do the prime factorization of 847, which i think is,  $7*11*11$. 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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Answer 2

The value of x+y+z is 8.83.

Given equation,

$6xyz+30xy+21xz+2yz+105x+10y+7z=812$.

We have to calculate the value of $x+y+z$.

Here,

$6xyz+30xy+21xz+2yz+105x+10y+7z=812$

Adding 35 both sides, we get

$6xyz+21xz+2yz+7z+30xy+105x+10y+35=812+35$

Taking z common we get,

$z(6xy+21x+2y+7)+30xy+105x+10y+35=847$

On further simplifying, we get

$(z+5)(6xy+21x+2y+7+6xy+21x+2y+7)=847$

$(z+5)(12xy+42x+4y+14)=847$

$(z+5)[6x(2y+7)+2(2y+7)]=847$

$(z+5)(2y+7)(6x+2)=847$

Now the prime factorization of 847 is $11\times11\times7.$

So,

$(z+5)(2y+7)(6x+2)=11\times11\times7$

Here if $z+5=11$

$z=6$

And if $2y+7=11$

$2y=4\\y=2$

Again,

$6x+2=7\\x=\frac{5}{6}$

Now,

$x+y+z=\frac{5}{6} +2+6$

$x+y+z=0.83+8\\x+y+z=8.83$

Hence the value of x+y+z is 8.83.

For more details follow the link:

https://brainly.com/question/11897796