The restricted values in (x^2 - 3x - 2)/-4y * (2x - 2)/(-xy^2 + 2y^2) are x ≠2 and y ≠ 0
How to determine the restricted values for the function?
The complete question is added as an attachment
The equation is given as:
(x^2 - 3x - 2)/-4y * (2x - 2)/(-xy^2 + 2y^2)
Set the products of the denominators to 0
-4y * (-xy^2 + 2y^2) = 0
Split the equation
-4y = 0 or (-xy^2 + 2y^2) = 0
Remove the bracket
-4y = 0 or -xy^2 + 2y^2 = 0
Divide both sides by -4 in -4y = 0
y = 0
Factor out y^2 in -xy^2 + 2y^2 = 0
y^2(-x + 2) = 0
This means that y^2 = 0 or -x + 2= 0
This gives y = 0 or -x + 2 = 0
Solve for x in -x + 2 = 0
x = 2
Hence, the restricted values in (x^2 - 3x - 2)/-4y * (2x - 2)/(-xy^2 + 2y^2) are x ≠2 and y ≠ 0
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