Given 20 – [5xy + 3[x2 – (xy – y) – (x – y)]]
First we have to remove the parentheses, then remove braces, and then the square brackets.
Then we get,
= 20 – [5xy + 3[x2 – (xy – y) – (x – y)]]
= 20 – [5xy + 3[x2 – xy + y – x + y]]
= 20 – [5xy + 3[x2 – xy + 2y – x]]
= 20 – [5xy + 3x2 – 3xy + 6y – 3x]
= 20 – [2xy + 3x2 + 6y – 3x]
On simplifying we get
= 20 – 2xy – 3x2 – 6y + 3x
= – 3x2 – 2xy – 6y + 3x + 20