Find $x+y$, if integers $x$ and $y$ satisfy the equation $y+1/x=25/3$

Question

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?

Answer 1

If

$$ \frac{x y +1}{x} = \frac{25}{3} \Rightarrow \left\{\begin{array}{rcl}x y +1 & = & 25 k\\ x & = & 3 k\end{array}\right. $$

hence

$25k-3ky = 1\Rightarrow k(25-3y)=1\Rightarrow y = 8$ etc.

Answer 2

Hint:

Then $$y = {25x-3\over 3x}\implies 3x\mid 25x-3\implies x\mid 25x-3$$ $$\implies x\mid -3\implies x\in\{\pm 1,\pm3\}$$

Since $3\mid 25x-3 \implies 3\mid 25x\implies 3\mid x$, so $x=\pm 3$.

Answer 3

We have: $y = 8+\dfrac{x-3}{3x}\implies \dfrac{x-3}{3x}\in \mathbb{Z}\implies x-3=3nx\implies x-3nx=3\implies(3n-1)x=-3\implies x\mid 3 \implies x = \pm 1, \pm 3$. Since $y$ is an integer, $x = 3$. Thus $y = 8$, and $x+y = 3+8 = 11$.

Answer 4

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

Answer 5

Correct me if wrong .

Pedestrian approach.

$y+1/x=25/3= 8+1/3$

Note: $y, x \not = 0$

1) Let $x,y$ be positive integers.

$x>0$, since

$y+1/x = (8+1/3$) $\not \in \mathbb{Z^+}$,

hence $1/x <1$; $y \in \mathbb{Z^+}$ ,

$\rightarrow$ $y=8$, and $x=3$.

2) $x,y <0$, ruled out .

3) $y <0, x >0$ ,

ruled out , recall $1/x <1$.

4) Remains: $y>0$ , $x <0$, i, e.

$y-1/|x| = 8 +1/3.$

Recall: $1/|x| <1$.

Check:

$y= 9$, ruled out, $y >9$, ruled out.

Remains option 1).

Answer 6

Given: $y+1/x=25/3$, multiply both sides by $3$: $$3y+\frac3x=25 \Rightarrow x=\{\pm1;\pm3\}.$$ Hence: $$y=8+\frac13-\frac1x \Rightarrow (x,y)=(3,8).$$