# Ex 12.2, 2 - Chapter 12 Class 7 Algebraic Expressions

Last updated at Dec. 21, 2018 by Teachoo

Last updated at Dec. 21, 2018 by Teachoo

Transcript

Ex 12.2, 2 Add: (i) 3mn, – 5mn, 8mn, – 4mn 3mn + (–5mn) + 8mn + (–4mn) = 3mn – 5mn + 8mn – 4mn = mn (3 – 5 + 8 – 4) = mn (3 + 8 – 5 – 4) = mn (11 – (5 + 4)) = mn (11 – 9) = mn × 2 = 2mn Ex 12.2, 2 Add: (ii) t – 8tz, 3tz – z, z – t (t – 8tz) + (3tz – z) + (z – t) = t – 8tz + 3tz – z + z – t = t – t – 8tz + 3tz – z + z = t (1 – 1) + tz (–8 + 3) + z (–1 + 1) = t (0) + tz (–5) + z (0) = –5tz Ex 12.2, 2 Add: (iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3 (–7mn + 5) + (12mn + 2) + (9mn – 8) + (–2mn – 3) = –7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3 = –7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3 = mn (–7 + 12 + 9 – 2) + (5 + 2 – 8 – 3) = mn (12 + 9 – 7 – 2) + (5 + 2 – 8 – 3) = mn (21 – 9) + (7 – 11) = mn (12) + (– 4) = 12mn – 4 Ex 12.2, 2 Add: (iv) a + b – 3, b – a + 3, a – b + 3 (a + b – 3) + (b – a + 3) + (a – b + 3) = a + b – 3 + b – a + 3 + a – b + 3 = a – a + a + b + b – b –3 + 3 + 3 = a (1 – 1 + 1) + b (1 + 1 – 1) + (–3 + 3 + 3) = a (0 + 1) + b (2 – 1) + (–3 + 6) = a (1) + b(1) + (6 – 3) = a + b + 3 Ex 12.2, 2 Add: (v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + (4xy) = 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy = 14x – 7x + 10y – 10y – 12xy + 8xy + 4xy – 13 + 18 = x (14 – 7) + y (10 – 10) + xy (–12 + 8 + 4) + (18 – 13) = x (7) + y (0) + xy (–12 + 12) + (5) = 7x + xy (0) + 5 = 7x + 5 Ex 12.2, 2 Add: (vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5 (5m – 7n) + (3n – 4m + 2) + (2m – 3mn – 5) = 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5 = 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5 = m (5 – 4 + 2) + n (–7 + 3) –3mn + (2 – 5) = m(1 + 2) + n (–4) – 3mn + (–3) = m (3) – 4n –3mn – 3 = 3m – 4n – 3mn – 3 1 Ex 12.2, 2 Add: (vii) 4x2y, – 3xy2, –5xy2, 5x2y 4x2y + (– 3xy2) + (– 5xy2) + 5x2y = 4x2y – 3xy2 – 5xy2 + 5x2y = 4x2y + 5x2y – 3xy2 – 5xy2 = x2y (4 + 5) + xy2 (–3 – 5) = x2y (9) + xy2 (–8) = 9x2y – 8xy2 Ex 12.2, 2 Add: (viii) 3p2q2 – 4pq + 5, – 10 p2q2, 15 + 9pq + 7p2q2 (3p2q2 – 4pq + 5) + (– 10 p2q2) + (15 + 9pq + 7p2q2) = 3p2q2 – 4pq + 5 – 10 p2q2 + 15 + 9pq + 7p2q2 = 3p2q2 – 10 p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15 = p2q2 (3 – 10 + 7) + pq (–4 + 9) + (5 + 15) = pq2 (–7 + 7) + pq (5) + (20) = pq2 (0) + 5pq + 20 = 5pq + 20 Ex 12.2, 2 Add: (ix) ab – 4a, 4b – ab, 4a – 4b (ab − 4a) + (4b − ab) + (4a − 4b) = ab − 4a + 4b − ab + 4a − 4b = −4a + 4a + 4b − 4b + ab − ab = a (−4 + 4) + b (4 − 4) + ab (1 − 1) = a (0) + b (0) + ab (0) = 0 + 0 + 0 = 0 Ex 12.2, 2 Add: (x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2 (x2 – y2 – 1) + (y2 – 1 – x2 ) + (1 – x2 – y2) = x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2 = x2 – x2 – x2 – y2 + y2 – y2 – 1 – 1 + 1 = x2 (1 – 1 – 1) + y2 (–1 + 1 – 1) + (–1 – 1 + 1) = x2 (0 – 1) + y2 (0 – 1) + (–2 + 1) = x2 (–1) + y2 (–1) + (–1) = –x2 – y2 – 1

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.