1
$\begingroup$

I need to know how to solve this equation where x and y are both variables Find integer Solutions. $$ \frac{1}{x} + \frac{1}{y} = \frac{1}{2} $$ from what I know I need at least 2 equations to solve for 2 variables. So I need a Hint

enter image description here

$\endgroup$
1

5 Answers 5

4
$\begingroup$

$$y=\frac{2x}{x-2}=2\times \left(1+\frac{2}{x-2}\right)=2+\frac{4}{x-2}$$

Then $-2 \leq x < 2$ or $2<x\leq 6$. Since $y>0$, $2<x\leq 6$.

Therefore, There are three positive duplets $(x,y)$, namely, $(3,6),(4,4)$ and $(6,3)$.

$\endgroup$
2
$\begingroup$

The OP seeks the pairs of integer values $(x,y)$ that are solutions of the equation $y = 2x/(x-2)$.

We observe that for large $x$ (positive or negative) the value for $y$ converges to $2$. So beyond a certain maximum value for $x$ there can not be any integer solutions.

For positive $x$ the limiting value for $y$ is approached from above. Now the nearest possible value for $y$ greater then $2$ is $3$. The accompanying value for $x$ is found to be $x = 6$.

For negative $x$ the limiting value for $y$ is approached from below. The nearest possible value for $y$ smaller than $2$ is $1$. The accompanying value for $x$ is $x = -2$.

Hence only integer values in the range $-2 \le x \le 6$ can result in an integer pair $(x,y)$. Let us check the list of candidates. We find that $(-2, 1)$, $(0,0)$, $(1, -2)$, $(3,6)$, $(4,4)$, $(6,3)$ are the only solutions.

$\endgroup$
1
$\begingroup$

This is a really easy question how you need to do it is : $\frac{1}{x}+\frac{1}{y}$ =$\frac{1}{2}$ then $${\frac{1}{y} =\frac{1}{2}-\frac{1}{x}}$$ then manipulate it to get $${y=\frac{2x}{x-2}}$$ then put whatever values of x you want you will get the corresponding value of y.

If we want to get integer solutions of this equation then think like this : Let $${2x=k(x-2)}$$

  • Put $x =-2$ you get $k=1$.
  • Put $x=4$ you get $k=1$
  • Put $x=3$ we get $k=6$
  • Put $x=6$ you will get a soln.

After that $x-2$ doesn't remain a factor of $2x$ .

$\endgroup$
3
  • 1
    $\begingroup$ The title says "I need Positive Integer Solution to this Equation" $\endgroup$
    – user99914
    Jun 21, 2015 at 15:22
  • 1
    $\begingroup$ in order to ensure that both values are positive, all we should need is $x \ge 3$. $\endgroup$ Jun 21, 2015 at 16:09
  • $\begingroup$ @PaddlingGhost There are two words in "positive integer". $\endgroup$
    – Erick Wong
    Jun 21, 2015 at 16:10
1
$\begingroup$

Denote $S=x+y$, $P=xy$. As the problem is symmetric in $x$ and $y$, we may as well suppose $x\ge y$. $$\frac1x+\frac1y=\frac12\iff P=2S.$$ We'll examine the different possibilities according to the prime factorisation of $y$.

  • First note $x$ and $y$ have the same odd prime factors, if any. Indeed, if an odd prime $p$ divides one of them, it divides $P=2S$, hence it divides $S$ so that it divides the other.

  • If $y$ has an odd prime factor $p$, say $y=p^iy'$, then $$\frac12=\biggl\lvert \frac1x+\frac1y\biggr\rvert\le \frac2{p^i}$$ by the triangle inequality and the hypothesis $x\ge y$. This implies $\,p^i\le 4$, so $\,p=3,\, i=1$. For the same reason, $3$ is the only odd prime of $x$, not repeated.

  • In addition, $y$ cannot be even, otherwise $\,\biggl\lvert \dfrac1x+\dfrac1y\biggr\rvert\le \dfrac 26<\dfrac12$. Finally $y=\pm3$, and $$\frac1x=\frac12\mp\frac13= \frac16\enspace\text{or}\enspace\frac56.$$ The only compatible solution is $\,x=6$.

  • If $x$ and $y$ have no odd prime factor,we write $\,\lvert x\rvert=2^k,\, \lvert y\rvert=2^l\ (k\ge l)$. We have $$\frac12=\biggl\lvert \frac1x+\frac1y\biggr\rvert\le \frac2{2^l}$$ which implies $2^l\le 4$, i. e. $\,l\le 2$.

  • Now if $x$ and $y$ have the same sign, $P>0$ and the relation $P=2S$ shows this sign is positive. It is easy to check the only solution is $x=y=4$.

  • Else $y<0<x$. Then $\,\dfrac12=\dfrac1x+\dfrac1y<\dfrac1x$, so $\,x<2$, thus $x=1$, and $y=-2$.

Summary: Taking into account we've supposed $x\ge y$, there are $5$ solutions: $$\bigl\{(1,-2), (-2,1), (4,4), (6,3), (3,6)\bigr\}.$$

$\endgroup$
0
$\begingroup$

Because you have more vairables then equations, i.e. an undetermined system , so you can have either infinitely many solutions or no solutions.

Provided $xy \neq 0$, one obtains $$2x+2y = xy.$$ Then express $x$ in terms of $y$, i.e. $$x = \frac{2y}{y-2}$$

Take a value for $y$ without violating $xy \neq 0$, substitute into the above euqation. If the value of $x$ also does not violate the assumption, then you have obtained one solution.

Since you are looking for positive integer solutions, observe that $y>2$ is required. Besides, $\frac{2y}{y-2}\in \mathbb{N}^+$, which can be found by exhaustion.

So any pair $(\frac{2y}{y-2},y)$ with integer valued $y>2$ is a solution.

$\endgroup$
1
  • $\begingroup$ So when are $y, \frac{2y}{y-2}$ both integers? $\endgroup$
    – user99914
    Jun 21, 2015 at 15:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.