Solve for x and y simultaneously:
x-2y=3
4xsquared-5xy+6y=3

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