To solve this equation, we need to consider the cases when the absolute value term is positive and when it is negative.
Case 1: If $(2x-y-3) \geq 0$ then $(2x-y-3) = 0$, which means $2x - y - 3 = 0$.
Substitute this into the second term of the equation:$(2x-y-3)+(x+3y-6)^2=0$
$2x - y - 3 + (x + 3y - 6)^2 = 0$
$2x - y - 3 + (x + 3y - 6)(x + 3y - 6) = 0$
$2x - y - 3 + x^2 + 6xy + 9y^2 - 6x - 18y - 6 = 0$
$x^2 + 6xy + 9y^2 - 4x - 19y + 3 = 0$
This is a quadratic equation in terms of x and y. We can attempt to solve it using the quadratic formula, but it may result in a complex solution.
Case 2: If $(2x-y-3) < 0$ then $-(2x - y - 3) = 0$, which means $2x - y - 3 = 0$.
Substitute this into the second term of the equation and follow the same steps as above to solve for x and y.
To solve this equation, we need to consider the cases when the absolute value term is positive and when it is negative.
Case 1: If $(2x-y-3) \geq 0$ then $(2x-y-3) = 0$, which means $2x - y - 3 = 0$.
Substitute this into the second term of the equation:
$(2x-y-3)+(x+3y-6)^2=0$
$2x - y - 3 + (x + 3y - 6)^2 = 0$
$2x - y - 3 + (x + 3y - 6)(x + 3y - 6) = 0$
$2x - y - 3 + x^2 + 6xy + 9y^2 - 6x - 18y - 6 = 0$
$x^2 + 6xy + 9y^2 - 4x - 19y + 3 = 0$
This is a quadratic equation in terms of x and y. We can attempt to solve it using the quadratic formula, but it may result in a complex solution.
Case 2: If $(2x-y-3) < 0$ then $-(2x - y - 3) = 0$, which means $2x - y - 3 = 0$.
Substitute this into the second term of the equation and follow the same steps as above to solve for x and y.