If x = -2 and y = 1, then what is the value of $(\frac{4}{x}+y)(\frac{16}{x^2}+y^2 -\frac{4y}{x})$?

-5 7 -7 5

-7

x = -2 y = 1 Then what is the value of $(\frac{4}{x}+y)(\frac{16}{x^2}+y^2 -\frac{4y}{x})$ = $(\frac{4}{-2}+1)(\frac{16}{-2^2}+1^2 -\frac{4×1}{-2})$ = (-1)(4+1+2) = -7