Question

x,y and z are non-negative integers where
\(\large xy+3x+2y=2\)
and
\(\large yz+4y+3z=52\)
Find the value of \(\large x+y+z+xy+yz+xz+xyz\)

Answer 1

Any idea?

Answer 2

I would start out substituting

Answer 3

yeah, I think too

Answer 4

So what does xy equal?

Answer 5

\(\large xy=2-3x-2y\)

Answer 6

Exactly same for yz then plug into equation I am working on it but am stuck at xyz

Answer 7

Unless you convert and multiply. Say x = (-xy+2y-2)/3 then do the same for y and z.

Answer 8

xy+3x+2y=2
x(y+3)=2(1-y)
x=(1-y)/(y+3)
Since x,y,z are non negative integers
only solution of x and y is when y=1 and x=0

Answer 9

and when y=1,
1*z+4*1+3z=52
4z=48
z=12

Answer 10

so, x+y+z+xy+yz+xz+xyz=25

Answer 11

Thanks @sauravshakya . Your solution is concise and nice

And Here is my solution

We have \(\large \begin{cases}\large xy+3x+2y=2\ \ \ (1)\\ \large yz+4y+3z=52\ \ \ (2) \end{cases} \Rightarrow \begin{cases} y=\frac{2-3x}{x+2}\\ y=\frac{52-3z}{z+4} \end{cases}\)
\(\large \Rightarrow \frac{2-3x}{x+2}=\frac{52-3z}{z+4} \Leftrightarrow z-8x-12=0 \Rightarrow z=8x+12\ \ \ (3)\)

Now, substituting \(\large (3)\) into \(\large(2)\) give
\(\large y(8x+12)+4y+3(8x+12)=52\\\large \Leftrightarrow 2xy+3x+2y=2\ \ \ (4)\)

Then, subtract \(\large (1)\) from \(\large (4)\) got:\(\large xy=0\)
\(\large \Rightarrow x=0\ Or\ y=0\)

\(\Large \color{red}{*x=0} \Rightarrow y=1\ \ and \ z=12\) (satisfy x,y and z are non-negative integers)
\(\Large \color{red}{*y=0} \Rightarrow x= \frac{2}{3}\) (not satify )


Thus,
\(\large x+y+z+xy+yz+xz+xyz = (x+1)(y+1)(z+1)-1=25\)