Solve for x and y simultaneously:

X-2y=3

4xsquared-5xy+6y=3

X = 3 + 2y

4 (3+2y) ^2 - 5 (3+2y) y + 6y = 3

4 (4y^2 + 12y + 9) - 5 (3y + 2y^2) + 6y - 3 = 0

16y^2 + 48y + 36 - 15y - 10y^2 + 6y - 3 = 0

6y^2 + 39y + 33 = 0

6y^2 + 6y + 33y + 33 = 0

6y (y+1) + 33 (y+1) = 0

(y+1) (6y+33) = 0

y = - 1, - 11/2

x = 3 + 2y

x = 3 + 2 (-1)

x = 3 + 2 (-11/2)

x = 1, - 8

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