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