Question 3 (vii) - Adding and Subtracting Like terms - Worksheet - Chapter 10 Class 7 Algebraic Expressions

Last updated at Dec. 16, 2024 by

Ex 12.2, 3 - Chapter 12 Class 7 Algebraic Expressions - Part 7

Ex 12.2, 3 - Chapter 12 Class 7 Algebraic Expressions - Part 8

Go Ad-free

Transcript

Question 3 (Method 1) Subtract: (vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2 (3ab – 2a2 – 2b2) − (5a2 – 7ab + 5b2 ) = 3ab – 2a2 – 2b2 − 5a2 + 7ab − 5b2 = −2a2 − 5a2 − 2b2 − 5b2 + 3ab + 7ab = a2 (−2 − 5) + b2 (−2 − 5) + ab (3 + 7) = a2 (−7) + b2 (−7) + ab (10) = −7a2 − 7b2 + 10ab Question 3 (Method 2) Subtract: (vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2 We need to subtract ((5a2 – 7ab + 5b2 ) from (3ab – 2a2 – 2b2) 5a2 − 7ab + 5b2 from 3ab − 2a2 − 2b2 Answer is 10ab – 7a2 – 7b2

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo