Question

-1/2 (2/5y - 2) (1/10y-4)

Answer 1

`-(1)/(2)((2)/(5)y-2)((1)/(10)y-4)`

we multiply the first parenthesis by its coefficient

`\begin{gathered} ((-(1)/(2)*(2)/(5)y)+(-(1)/(2)*-2))((1)/(10)y-4) \\ \\ (-(2)/(10)y+(2)/(2))((1)/(10)y-4) \\ \\ (-(1)/(5)y+1)((1)/(10)y-4) \end{gathered}`

now multiply each value and add the solutions

`\begin{gathered} (-(1)/(5)y*(1)/(10)y)+(-(1)/(5)y*-4)+(1*(1)/(10)y)+(1*-4) \\ \\ (-(1)/(50)y^2)+((4)/(5)y)+((1)/(10)y)+(-4) \\ \\ -(1)/(50)y^2+((4)/(5)y+(1)/(10)y)-4 \\ \\ -(1)/(50)y^2+(9)/(10)y-4 \end{gathered}`