Elementary algebraic system with integer and positive rational unknowns The problem is to determine the solutions of the system $x+y+z=3$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=5$, given that $x$ is integer and $y,z$ are positive rationals.
From the hypotheses I have obtained that $y$ and $z$ are rational solutions of the quadratic equation
$$(5x-1)t^2-(5x-1)(3-x)t+(3-x)x=0$$
with integer coefficients, therefore its discriminant $\left( 5\,x-1 \right)  \left( x-3 \right)  \left( 5\,{x}^{2}-12\,x+3
 \right)$ has to be a perfect square.
But I cannot see a clear continuation from here.

(There was an old problem which looks similar.)
Solution:
First note that x<3, so if $x>0$ we can easily find $x=1$ is not good; and $x=2$ yields $(y,z)=(\frac 13, \frac 23)$ or $(\frac 23, \frac 13)$.
Now suppose $x<0, w=-x, w \in \mathbb N$. Your discriminant becomes $$(5w+1)(w+3)(5w^2+12w+3)=(5w^2+16w+3)(5w^2+12w+3)\\=(5w^2+14w+3)^2-(2w)^2$$
So we have a Pythagorean triple. There are only two possibilities:
Case 1: $5w^2+14w+3=k(m^2+n^2), 2w=2kmn,m>n, (m,n)=1, $ and not both odd.
But
$$k(m^2+n^2)>5w^2+14w=5k^2m^2n^2+14kmn>5k^2m^2+14kn^2>km^2+kn^2, \Rightarrow \Leftarrow.$$
Case 2: $5w^2+14w+3=k(m^2+n^2), 2w=k(m^2-n^2),m>n, (m,n)=1, $ and not both odd.
Again
$$k(m^2+n^2) > 5w^2 = 5 \frac{k^2}{4} (m+n)^2(m-n)^2 > k^2(m+n)^2 > k(m^2+n^2), \Rightarrow \Leftarrow.$$
Therefore the only solutions are $(2,\frac 13, \frac 23), (2, \frac 23, \frac 13).$
