Find $x+y$, if integers $x$ and $y$ satisfy the equation $y+1/x=25/3$ Find $x+y$, if $x$ and $y$ are integers and satisfy the equation $y+1/x=25/3$
so I got to the answer by placing $3$ in the $x$ cause it looked like a fraction and $8$ was left for $y$, 
My question is: Is there any other/better way to solve this algebraically?
 A: Hint:
Then $$y = {25x-3\over 3x}\implies 3x\mid 25x-3\implies x\mid 25x-3$$ $$\implies x\mid -3\implies x\in\{\pm 1,\pm3\}$$
Since $3\mid 25x-3 \implies 3\mid 25x\implies 3\mid x$, so $x=\pm 3$.
A: If 
$$
\frac{x y +1}{x} = \frac{25}{3} \Rightarrow \left\{\begin{array}{rcl}x y +1 & = & 25 k\\
x & = & 3 k\end{array}\right.
$$
hence
$25k-3ky = 1\Rightarrow k(25-3y)=1\Rightarrow y = 8$ etc.
A: We have: $y = 8+\dfrac{x-3}{3x}\implies \dfrac{x-3}{3x}\in \mathbb{Z}\implies x-3=3nx\implies x-3nx=3\implies(3n-1)x=-3\implies x\mid 3 \implies x = \pm 1, \pm 3$. Since $y$ is an integer, $x = 3$. Thus $y = 8$, and $x+y = 3+8 = 11$.
A: We can solve it algebraicly by considering divisiblity by 3 (see below).
But we can also justify why your intuition yields the only possible answer.
No $\frac {25}3 = 8\frac 13$ and $8< 8\frac 13 < 8+1$.
If $x > 1$ then $y < y+ \frac 1x< y + 1$.  
So $y = 8; \frac 1x = \frac 13; x=3$.
If $x < -1$ then $y-1 < (y-1) + (1-\frac 1{|x|}) < y$.
So $y-1 = 8$ and $1 -\frac 1{|x|} =\frac 13$ so $\frac 1x = -\frac 23; x = -\frac 32$ which is not an integer.
If $x =\pm 1$ then $y+\frac 1x= y\pm 1$ is an integer and can not equal $8\frac 13$.
And obviously $x \ne 0$.
==== number theory divisibility by $3$ argument below =====
$y+1/x=25/3$
$xy +1 = \frac {25x}3 $
$3$ is prime.  $3\not \mid 25$ so $3|x$.
Let $x = 3k$
$3ky + 1 = 25k$.
RHS is a multiple of $K$.  LHS is one more than a multiple of $k$.  That's only possible if $k =\pm 1$.
$\pm 3y +1 = \pm 25$ or
$3y \pm 1 = 25$
$3y = 24, 26$.  Only $24$ is divisible by $3$.
So $k = 1; x = 3; y =8$.
A: Correct me if wrong .
Pedestrian approach.
$y+1/x=25/3= 8+1/3$
Note: $y, x \not = 0$
1) Let $x,y$ be positive integers.
$x>0$, since 
$y+1/x =  (8+1/3$) $\not \in \mathbb{Z^+}$,
hence $1/x <1$;  $y \in \mathbb{Z^+}$ , 
$\rightarrow$ $y=8$, and $x=3$.
2) $x,y <0$, ruled out .
3) $y <0, x >0$ , 
ruled out , recall $1/x <1$.
4) Remains: $y>0$ , $x <0$, i, e.
$y-1/|x| = 8 +1/3.$
Recall: $1/|x| <1$.
Check:
$y= 9$, ruled out, $y >9$, ruled out.
Remains option 1).
A: Given: $y+1/x=25/3$, multiply both sides by $3$:
$$3y+\frac3x=25 \Rightarrow x=\{\pm1;\pm3\}.$$
Hence:
$$y=8+\frac13-\frac1x \Rightarrow (x,y)=(3,8).$$
