Question 81 (ii)
Add :
9ax+3by−cz,−5by+ax+3cz
We have, (9ax+3by−cz)+(−5by+ax+3cz) =9ax+3by−cz−5by+ax+3cz =(9ax+ax)+(3by−5by)+(−cz+3cz) [grouping like terms]
=10ax−2by+2cz
Question 81 (iv)
5x2−3xy+4y2−9,7y2+5xy−2x2+13
Question 81 (i)
7a2bc,−3abc2,3a2bc,2abc2