Find integer solution in equation $3xy + 10x - 13y - 35=0$
Solve in integers. 
to be honest I don’t know what to do because it’s the only way to factor like 

$x (3 y + 10) = 13 y + 35$ 
   or    $(3 x - 13) y = 35 - 10 x$

 A: Hint: Simon's favorite factoring trick, especially that 2nd example which explains how to deal with cases where the coefficient of the cross term is not 1 (like in this scenario).  

 $0 = 9xy + 30x - 39y - 105 = (3x-13)(3y+10) +25 $.
$(3x-13)(3y+10) = - 25 $.
 So we just check the 6 factors of -25 to see if they are of that form.
 We get $(3x-13, 3y+10) = (5, -5), (-1, 25), (-25, 1)$ fitting the required form.   

Thus, the only solutions are $(6, -5), (4,5), (-4, -3)$.  
A: $$3xy + 10x - 13y - 35=0$$
$$x(3y+10)  = 13y + 35 $$
$$x = \frac{13y+35}{3y+10}$$
Since $x$ is an integer , $(3y+10)$ must divide $(13y + 35)$.
Also $$3(13y+35) = 13(3y+10) -25$$ 
So $3y+10$ must divide $-25$.
Looking at the possible factors of $-25$ , 
$$\begin{cases} 3y+10 = \pm1  \\3y+10 = \pm5 \\ 3y+10 = \pm25\end{cases}$$
Which yields $y=-3$ and $y = \pm5$
Inputting the values in $x$ gives the following the results : 
$$(x,y) = \,(-4,-3) \,,(4,5) \,,(6,-5)$$
A: To have $x=\dfrac{13y+35}{3y+10}$ be an integer, note the following:
$4<x<5$ when $y<-8$
$x$ is not an integer when $y=-6$ or $-7$
$x$ is an integer $(6)$ when $y=-5$
$x$ is not an integer when $y=-4$
$x$ is an integer $(-4)$ when $y=-3$
$x$ is not an integer when $y\in\{-2,-1,0,1,2,3,4\}$
$x$ is an integer ($4$) when $y=5$
$4<x<5$ when $y>5$
