How can I find the integers $x,y$? $$ax+by=c$$
$$ax^2+by^2=d$$
where $ab\neq 0$ and $x,y$ are coprime?
 A: You can solve this by talking the first equation and solving for y. Then substituting y in the second equation, as so:
$$ax+by=c$$
$$ax-c=-by$$
$$(ax-c)/(-b)=y$$
Now substitute into second equation:
$$ax^{2}+\frac{b\left(ax-c\right)^{2}}{b^{2}}=d$$
Expand:
$$ax^{2}+\frac{a^{2}x^{2}-2acx+c^{2}}{b}=d$$
Simplify:
$$\frac{bax^{2}}{b}+\frac{a^{2}x^{2}-2acx+c^{2}}{b}=d$$
$$\frac{bax^{2}+a^{2}x^{2}-2acx+c^{2}}{b}=d$$
Subtract d from both sides:
$$\frac{bax^{2}+a^{2}x^{2}-2acx+c^{2}-bd}{b}=0$$
Simplify more...
$$\frac{\left(a^{2}+ba\right)}{b}x^{2}-\frac{2ac}{b}x+\frac{\left(c^{2}-bd\right)}{b}=0$$
This is just a quadratic. You can solve it with the quadratic equation:
$$x=\frac{-g+\sqrt{g^{2}-4fh}}{2f}$$
And
$$x=\frac{-g-\sqrt{g^{2}-4fh}}{2f}$$
Where $$f=\frac{\left(a^{2}+ba\right)}{b}$$
$$g=-\frac{2ac}{b}$$
$$h=\frac{\left(c^{2}-bd\right)}{b}$$
Try it out: https://www.desmos.com/calculator/dw9xe3ppxe
A: Above simultaneous equation shown below:
$ax^2+by^2=d$
$ax+by=c$
Taking, $(a,b,c,d)=(4,5,22,56)$ we get:
$(x,y)=(3,2)$ & $(x,y)=(17/9,26/9)$
A: Easiest Possible solution. Its a simple solution using trignometry. I use this to solve many problems
