How many ordered pairs of integers satisfy the equation $\frac{7}{x}+\frac{3}{y}=\frac{1}{4}$? I tried doing it by trial and error method but i found it very lengthy. Is there any other approach to solve questions like this?
 A: Your equation is equivalent to $12x + 28y = xy$, with $x,y \ne 0$.


*

*Since $4(3x + 7y) = xy$, one possibility is that $y \mid 4$, whence it follows that $y \in \{ \pm 1, \pm 2, \pm 4 \}$. Plugging each of these values back in the equation it is easy to check whether they lead to integer values for $x$ or not.

*If $y \nmid 4$, then $y \mid 3x + 7y$, whence it follows that $y \mid 3x$. Again, there are two posibilities.
a. If $y \mid 3$, then $y \in \{ \pm 1, \pm 3 \}$ and proceeding as above test them to see whether they lead to integer values of $x$.
b. If $y \nmid 3$, then $y \mid x$, so there exist $k \in \Bbb Z$ with $x = ky$, so your equation becomes (after dividing it by $y$) $12k + 28 = ky$, whence $k \mid 28$, whence it follows that $k \in \{ \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28 \}$. As above, plug each value of $k$ in the simplified equation and see if you get integer values for $y$; if you do, don't forget to compute the $x$ corresponding to each such $k$.
The subfields of number theory, in general, require a fair bit of explicit computations. Diophantine equations is such a domain. Be happy, though, there are fields of mathematics that are highly non-computable!
A: Hint : The given equation is equivalent to $$(x-28)(y-12)=336$$ , if $x$ and $y$ are non-zero.
A: As Alex M. wrote
your equation is equivalent to
$12x + 28y = xy$.
Consider the more general equation
$xy = ax+by$.
Write it as
$xy - ax-by=0$.
Since
$(x-b)(y-a)
=xy-ax-by+ab
$,
this means that
$(x-b)(y-a)
=ab
$.
Each solution to this
corresponds to
a factorization of $ab$,
of which there are always 2:
$1\cdot ab$ and
$a \cdot b$.
For every factorization
$ab = uv$,
a solution is gotten
by setting
$x-b = u$
and
$y-a = v$
so that
$x = b+u$,
$y = a+v$.
In your case,
$ab = 12\cdot 28
=336
=2^4\cdot 3 \cdot 7
$.
Asking Google for
the factors of $336$ gives
$1,2,3,4,6,7,8,12,14,16,21,24,28,42,48,56,84,112,168,336$,
which is the list
of possible values of $x-b$.
The complementary factor,
$y-a$,
is just this list in reverse order.
A: Multiply the sistem with $4xy$ and you get;
$$28y+12x=xy$$ now let's get rid of one variable;
$$x=y(x-28)\rightarrow y=\frac{x}{x-28}$$ so $x-28 \mid x$
$$\frac{(x-28)+28}{x-28}=1+\frac{28}{x-28}$$ now we can conclude $x-28 \mid  28$ 
$$x=0, \quad x=7, \quad x=14, \quad x=27, \quad x=29, \quad x=56, \quad x=35, \quad x=42$$ $y$ values could be found after we plug $x$'s in the equation, I didn't try to write all the $x$ values, so I might have missed one or two,
Nice Studies:)
