# Solving $11x-2=kx+15$ for a positive integral solution for $x$ given that $k$ is an integer

My question is- Find the integer value of $k$ such that the equation $11x-2=kx+15$ has positive integral solution for $x$. Find that solution.

Any guidance to solve this question would be helpful.

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I guess you meant an integer solution? – data Jun 1 '12 at 12:24
@m.woj yes an integer solution – mgh Jun 1 '12 at 12:26
There are two integers $k$ for which the equation has a positive integer solution for $x$. – André Nicolas Jun 1 '12 at 13:56

It is clear that $x=\frac{17}{11-k}$. Now for $x$ to be an integer $11-k$ should divide 17, or in other words there is some integer, say $n$ such that $17=n(11-k)$. We seem to have gone from one problem to another problem of equal difficulty. But no! Since 17 is a prime number, it can be factorized only as $17\times 1$ or as $1\times 17$. So either $n=17,11-k=1$ or $n=1,11-k=17$. It follows that $k=10$ or $k=-6$. You can now easily find the two possible values of $x$ by substitution.

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thanks helped a lot! – mgh Jun 1 '12 at 14:12

Hint

1. Solve this equation with respect to $x$.

2. Think about conditions you should put on denominator in order to get $x\in\mathbb{Z}$ and $x>0$.

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ya i got u. Well i have come up to the step : x=17/11-k but i am unable to go further. – mgh Jun 1 '12 at 12:30