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.
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.
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.
Hint
Solve this equation with respect to $x$.
Think about conditions you should put on denominator in order to get $x\in\mathbb{Z}$ and $x>0$.