Quadratic Diophantine Problem in two variables

Question

I have a quick question in regards to solving a quadratic two-variable diophantine problem. The equation is $6x^2 - 2xy + 3y - 17x = 6$. My attempt thus far starts by making y the subject: $$y = \frac{6x^2 - 17x - 6}{2x - 3}$$ After this I state that $2x - 3$ divides $6x^2 - 17x - 6$ and therefore $6x^2 - 17x - 6 \equiv 0$ (mod $2x - 3$). I've been trying to use $2x - 3 \equiv 0$ (mod $2x - 3$) to somehow get something going but I simply can't work out how to solve this (most likely simple) problem from this point.

I'd greatly appreciate any input on whether this method is the way to go (and hence how to continue), or if there is a "better" approach.

Thanks in advance

Answer 1

$$6x^2-17x-6=3x(2x-3)-4(2x-3)-6-12\equiv-18\pmod{2x-3}$$

$\implies2x-3$ must divide $18$

Also, $2x-3\equiv1\pmod2$ i.e, $2x-3$ is odd

Answer 2

Your idea works: $\,\color{#c00}{2x\equiv 3}\,$ so $\,0\equiv 2f = \overbrace{3(\color{#c00}{2x})^2\!-17(\color{#c00}{2x})-12}^{\large 2\,(6\,x^2\ -\ 17\,x\ -\ 6)\quad }\!\equiv -36\equiv -2^2\cdot 9.\,$ Since $\,2x\!-\!3\,$ is odd we can cancel $\,2\,$ so $\,9\equiv 0,\,$ i.e. $\,2x\!-\!3\mid 9,\ $ so $\ 2x\!-\!3 = \pm\, 1,3,9\ $ so $\,x = \ldots$