Verified Answer :

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 - 4633))/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

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

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