Question 81 (iv)
Add :
5x2−3xy+4y2−9,7y2+5xy−2x2+13
We have,
5x2−3xy+4y2−9,7y2+5xy−2x2+13
=5x2−3xy+4y2−9+7y2+5xy−2x2+13
=(5x2−2x2)+(−3xy+5xy)+(4y2+7y2)+(−9+13)
[grouping like terms]
=3x2+2xy+11y2+4