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?

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 $k = 1; x = 3; y =8$.
