How to solve the system $\frac{35-12b}{a-b}= \frac bk,\frac{12a-35}{a-b}= \frac {a}{1-k}$

Question

I am trying to solve the system of diophantine equations: $$ \begin{align*} \frac{35-12b}{a-b} &= \frac bk \\[6pt] \frac{12a-35}{a-b} &= \frac {a}{1-k} \end{align*} $$ Where $a-b\ne 0,$ and $ k\,(1-k) \ne 0$ . I know for fact that the solutions are $(7\,(1-k),5k)$. I just cannot figure how to get those results.

Answer 1

We have: $a-b = \dfrac{k(35-12b)}{b} = \dfrac{(1-k)(12-35b)}{a}$. Thus: $35ak - 12akb =k(35a - 12ab) = (1-k)(12b - 35b^2) = 12b - 35b^2 - 12kb + 35kb^2$. From this rewrite the equation as a quadratic equation in variable $b$, and set its $\triangle = n^2$. That's a preliminary thought of mine. You can explore this idea to see if it works.