Last updated at Oct. 25, 2021 by Teachoo

Transcript

Ex 12.2, 3 Subtract: (i) –5y2 from y2 y2 − (−5y2) = y2 + 5y2 = y2 (1 + 5) = y2 (6) = 6y2 Ex 12.2, 3 (Method 1) Subtract: (viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq (5p2 + 3q2 – pq) − (4pq – 5q2 – 3p2) = 5p2 + 3q2 – pq − 4pq + 5q2 + 3p2 = 5p2 + 3p2 + 3q2 + 5q2 – pq − 4pq = p2 (5 + 3) + q2 (3 + 5) + pq (−1 − 4) = p2 (8) + q2 (8) + pq (−5) = 8p2 + 8q2 − 5pq

Ex 12.2

Ex 12.2, 1 (i)

Ex 12.2, 1 (ii)

Ex 12.2, 1 (iii)

Ex 12.2, 1 (iv)

Ex 12.2, 1 (v)

Ex 12.2, 1 (vi)

Ex 12.2, 2 (i)

Ex 12.2, 2 (ii)

Ex 12.2, 2 (iii)

Ex 12.2, 2 (iv)

Ex 12.2, 2 (v)

Ex 12.2, 2 (vi)

Ex 12.2, 2 (vii)

Ex 12.2, 2 (viii)

Ex 12.2, 2 (ix)

Ex 12.2, 2 (x)

Ex 12.2, 3 (i) Important You are here

Ex 12.2, 3 (ii)

Ex 12.2, 3 (iii)

Ex 12.2, 3 (iv)

Ex 12.2, 3 (v)

Ex 12.2, 3 (vi)

Ex 12.2, 3 (vii)

Ex 12.2, 3 (viii)

Ex 12.2, 4 (a) Important

Ex 12.2, 4 (b)

Ex 12.2, 5 Important

Ex 12.2, 6 (a)

Ex 12.2, 6 (b)

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.