Question 359858
{{{2y^2 - x^2 = 1}}}
{{{y+1 = x}}}.  substitute the bottom equation into the top equation:
{{{2y^2 -(y+1)^2 = 1}}},
{{{2y^2 -y^2 - 2y-1 = 1}}},
{{{y^2 -2y-2 = 0}}},
{{{y = (2 +- sqrt( 2^2-4*1*(-2) ))/(2*1)}}},
{{{y = (2 +- sqrt( 12 ))/2}}},
{{{y = (2+-2sqrt(3))/2}}},
{{{y = 1+-sqrt(3)}}}.  Then {{{x = 2+-sqrt(3)}}}.