Question 81 (i)
Add :
7a2bc,−3abc2,3a2bc,2abc2
We have, 7a2bc+(−3abc2)+3a2bc+2abc2=7a2bc−3abc2+3a2bc+2abc2 =(7a2bc+3a2bc)+(−3abc2+2abc2) [grouping like terms] =10a2bc+(−abc2) =10a2bc−abc2
Question 81 (vii)
3a(2b+5c),3c(2a+2b)
Question 81 (ii)
9ax+3by−cz,−5by+ax+3cz
Question 81 (iv)
5x2−3xy+4y2−9,7y2+5xy−2x2+13
Question 81 (iii)
xy2z2+3x2y2z−4x2yz2,−9x2y2z+3xy2z2+x2yz2