$$x^2 - 3y^2 = 1$$ $$5y = 3x - 1$$
I've already tried substitution from the second equation to the first, but that ended up too complicated for me to do by hand. Any method other than the use of a calculator of any kind is allowed. Thanks in advance!
$$x^2 - 3 \left( \frac{3x-1}{5}\right)^2 = 1$$
$$25x^2-3(3x-1)^2 = 25$$
$$25x^2-3(9x^2-6x+1)=25$$
$$-2x^2+18x-28=0$$
$$-x^2+9x-14=0$$
$$x^2-9x+14=0$$
$$(x-2)(x-7)=0$$
I believe the rest should be manageable.