Question

Please help

Answer 1

Answer:

$x+y+z \Rightarrow 2+2+6 = 10$

Step-by-step explanation:

First we have to factor the equation:

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

The commom factor of first three terms and $105x$ is $3x$. Note that $105x=3x(35)$

$3x(2yz+10y+7z+35)+2yz+10y+7z=812$

$3x(2yz+10y+7z+35)+(2yz+10y+7z)=812$

In order to have two equal factors add 35 both sides:

$3x(2yz+10y+7z+35)+(2yz+10y+7z+35)=847$

Let's factor $(2yz+10y+7z+35)$

$(2yz+10y+7z+35) \Rightarrow 2y\left(z+5\right)+7\left(z+5\right) \Rightarrow \left(z+5\right)\left(2y+7\right)$

Now we have:

$3x(z+5)\left(2y+7\right)+\left(z+5\right)\left(2y+7\right)=847$

$(3x + 1) (2y + 7) (z + 5) = 847$

Now it is the interesting part: You have to figure out that 3 numbers multiplied by each other will result in 847.

We have $847 = 7 \cdot 11 \cdot 11$

Let's try

$2y+7=7\\2y=0$

I will stop right here. This is not true.

Let's try then

$2y+7=11\\2y=4\\y=2$

Now let's try

$z+5 = 11\\z=6$

Now let's try

$3x+1=7\\3x=6\\x=2$

Note that in this case I didn't considered $3x+1=11$ because it would not be an integer.

Therefore,

$x+y+z \Rightarrow 2+2+6 = 10$