How many ordered pairs of integers satisfy the equation $\frac{7}{x}+\frac{3}{y}=\frac{1}{4}$?

Question

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?

Answer 1

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.

Answer 2

Your equation is equivalent to $12x + 28y = xy$, with $x,y \ne 0$.

1. 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.

2. 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!

Answer 3

Hint : The given equation is equivalent to $$(x-28)(y-12)=336$$ , if $x$ and $y$ are non-zero.

Answer 4

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:)