This is a system of three equations with two unknowns, x and y. To solve it, we can use the method of elimination. First, we can eliminate y by subtracting the first equation from the second equation:
x^2-2y=3 - (x-2y=3) => x^2-x=0
Then we can factor the resulting equation and find the values of x:
x^2-x=0 => x(x-1)=0 => x=0 or x=1
Next, we can substitute these values of x into any of the original equations and find the corresponding values of y. For example, using the first equation:
x-2y=3 => 0-2y=3 => y=-3/2 when x=0 x-2y=3 => 1-2y=3 => y=-1 when x=1
Finally, we can substitute these values of x and y into the expression x2y-2xy2 and evaluate it:
x2y-2xy2 => 02(-3/2)-2(0)(-3/2)2 => 0 when x=0 and y=-3/2 x2y-2xy2 => 12(-1)-2(1)(-1)2 => -3 when x=1 and y=-1
Therefore, the value of x2y-2xy2 can be either 0 or -3 depending on the values of x and y.