3xy + 14x + 17y + 71 = 0 need some advice $$3xy + 14x + 17y + 71 = 0$$
Need to find both $x$ and $y$. If there was only one variable then this is easy problem.
Have tried:
$$\begin{align}3xy &= -14x - 17y - 71 \\
x &= \frac{-14x - 17y - 71}{3y}\end{align}$$
Then tried to put this expression everywhere instead of $x$ but it tooks forever to find both $x$ and $y$.
I don't even know how to get on right track.
Please give any advice. Thanks.
 A: First, I will assume that you want to solve for $x$ in terms of $y$ and vice-versa. To do this, you first need to keep all the terms with $x$ on one side of the equation and move all the others to the other side:
$$3xy + 14x = -17y - 71$$
Next, factor out the $x$:
$$x(3y + 14) = -17y - 71$$
Now divide by $3y + 14$:
$$x = \frac{-17y - 71}{3y + 14}$$
The steps are similar to solve for $y$ in terms of $x$. I will leave that as an exercise for the reader.
A: This is another way to find the integer solutions.
Define $d=y-x$ and the equation becomes $3x(x+d)+14x+17(x+d)+71=0$
which boils down to $$3x^2+(3d+31)x+17d+71=0.$$
The quadratic formula gives the roots:$$6x=-(3d+31)\pm\sqrt{9d^2-18d+109},$$
and we want the discriminant, $9(d-1)^2+100 = (3(d-1))^2+10^2$ to be a square.
$d=1$ is obviously one candidate, leading to final values $x=-4, y=-3$.  
Otherwise we are just looking for Pythagorean triples. A quick look at a list of Pythagorean triples shows that there is just one triple with $10$ as one of the lower values: (10, 24, 26).
Thus $9(d-1)^2=24^2$, giving $d=9$ or $d=-7$, from which the appropriate values of $x$ and $y$ can be found.
A: There will be infinitely many pairs $(x,y)$ of real numbers that satisfy the equation.  If you are looking for integer solutions, that is another matter. 
I would rewrite the equation as $9xy+42x+51y+213=0$, and then as
$$(3x+17)(3y+14)-(17)(14)+213=0,$$
which turns into the attractive
$$(3x+17)(3y+14)=25.$$
Note that we did an analogue of completing the square.  We get a hyperbola.
Now for integer solutions the analysis becomes quite simple.  We take all ordered pairs $(s,t)$ of integers (both positive or both negative) such that $st=25$. There are only $6$ such pairs. 
For some but not all of these pairs, it turns out that $x$ and $y$ are integers.  Let's start. Look at $s=1, t=25$. No good, there is no integer $x$ such that $3x+17=1$.  Look at $s=-1,t=-25$. That gives $3x+17=-1$, $3y+14=-25$, which has the integer solution $x=-6, y=-13$. Continue. There is not far to go!
A: If $x$ and $y$ can be real numbers, with one equation in two unknowns you will have one dimension of freedom.  Solving for $x$, for example, $x=- \frac {17y+71}{3y+14}$.  You can substitute in any value for $y$ you want except $\frac {-14}3$ and find $x$.  If $x$ and $y$ are integers you can use divisibility testing to restrict the options. 
