Three Diophantine equations in three variables, finding $x+y+z$

Question

We have 3 equations: $$9x+y-8z=0$$ $$4x-8y+7z=0$$ $$xy+yz+zx=47$$ $x, y$ and $z$ are positive integers; we have to find their sum.

I am not sure if the question is correct because I have not been able to solve it.

Answer 1

Add the first and second equations together: $$13x-7y-z=0$$ $$z=13x-7y$$ Now substitute back into the first equation to eliminate $z$: $$9x+y-8(13x-7y)=0$$ $$-95x+57y=0$$ $$3y=5x$$ Since all variables are positive integers and 3 is relatively prime to 5, $y$ must be a multiple of 5 and $x$ a multiple of 3. We can write $x=3k$ and $y=5k$, where $k$ is a positive integer. After substituting these into the last equation to eliminate $x$ and $y$ it becomes $$15k^2+8kz=47$$ $$k(15k+8z)=47$$ But note that 47 is a prime number, so one of $15k+8z$ and $k$ must be 1. It cannot be $15k+8z$ since the lowest value it can attain is 23, so $k=1$. The solution to the original system of equations then follows easily: $$x=3,y=5,z=4,x+y+z=12$$ You may verify that all three equations are satisfied by these values.

Answer 2

Hint: eliminating $y,z$ from the given system we obtain $$x^2=9$$