Question 81 (vii)
Add :
3a(2b+5c),3c(2a+2b)
We have, 3a(2b+5c)+3c(2a+2b) =(6ab+15ac)+(6ac+6bc)=6ab+15ac+6ac+6bc [grouping like terms] =6ab+21ac+6bc
Question 81 (i)
7a2bc,−3abc2,3a2bc,2abc2
Question 81 (vi)
Add:
3a(a - b + c) , 2b(a - b + c)
Question 81 (ii)
9ax+3by−cz,−5by+ax+3cz
Question 81 (iv)
5x2−3xy+4y2−9,7y2+5xy−2x2+13