Answer:
[tex]x+y+z \Rightarrow 2+2+6 = 10[/tex]
Step-by-step explanation:
First we have to factor the equation:
[tex]6xyz+30xy+21xz+2yz+105x+10y+7z=812[/tex]
The commom factor of first three terms and [tex]105x[/tex] is [tex]3x[/tex]. Note that [tex]105x=3x(35)[/tex]
[tex]3x(2yz+10y+7z+35)+2yz+10y+7z=812[/tex]
[tex]3x(2yz+10y+7z+35)+(2yz+10y+7z)=812[/tex]
In order to have two equal factors add 35 both sides:
[tex]3x(2yz+10y+7z+35)+(2yz+10y+7z+35)=847[/tex]
Let's factor [tex](2yz+10y+7z+35)[/tex]
[tex](2yz+10y+7z+35) \Rightarrow 2y\left(z+5\right)+7\left(z+5\right) \Rightarrow \left(z+5\right)\left(2y+7\right)[/tex]
Now we have:
[tex]3x(z+5)\left(2y+7\right)+\left(z+5\right)\left(2y+7\right)=847[/tex]
[tex](3x + 1) (2y + 7) (z + 5) = 847[/tex]
Now it is the interesting part: You have to figure out that 3 numbers multiplied by each other will result in 847.
We have [tex]847 = 7 \cdot 11 \cdot 11[/tex]
Let's try
[tex]2y+7=7\\2y=0[/tex]
I will stop right here. This is not true.
Let's try then
[tex]2y+7=11\\2y=4\\y=2[/tex]
Now let's try
[tex]z+5 = 11\\z=6[/tex]
Now let's try
[tex]3x+1=7\\3x=6\\x=2[/tex]
Note that in this case I didn't considered [tex]3x+1=11[/tex] because it would not be an integer.
Therefore,
[tex]x+y+z \Rightarrow 2+2+6 = 10[/tex]
