X - 2y = 3 <span>4x^2 - 5xy + 6y = 3 lets solve for x the first and substitute in the second: x = 3 + 2y 4(</span>3 + 2y)^2 - 5(3 + 2y)y + 6y = 3 4(9 + 12y + 4y^2) - 15y - 10y^2 = 3 36 + 48y +16y^2<span> - 15y - </span><span>10y^2 = 3 6y^2 + 33y + 33 = 0 we can solve using the general quadratic formula: y = (-33 +- </span>√(33^2 - 4*6*33)<span>)/12 </span>y = (-33 +- √(297)<span>)/12 </span>so there are 2 solutions for y: y1 = (-33 + √(297)<span>)/12 </span>y2 = (-33 - √(297)<span>)/12 </span>pick one and then substitute the y value in the first equation to find x