{"id":1611,"date":"2023-11-21T10:05:22","date_gmt":"2023-11-21T04:35:22","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=1611"},"modified":"2023-11-22T11:35:21","modified_gmt":"2023-11-22T06:05:21","slug":"samacheer-kalvi-7th-maths-term-1-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-7th-maths-term-1-chapter-3-ex-3-2\/","title":{"rendered":"Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2"},"content":{"rendered":"

Students can Download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2<\/h2>\n

Question 1.
\nFill in the blanks
\n(i) The addition of – 7b and 2b is _______
\n(ii) The subtraction of 5m from -3m is ______
\n(iii) The additive inverse of -37xyz is _____
\nSolution:
\n(i) -5b
\n(ii) -8m
\n(iii) 37xyz<\/p>\n

Question 2.
\nSay True or False
\n(i) The expressions 8x + 3y and 7x + 2y cannot be added
\n(ii) If x is a natural number, then x + 1 is its predecessor.
\nHint: x – 1 is its predecessor.
\n(iii) Sum of a – b + c and -a + b – c is zero
\nSolution:
\n(i) False
\n(ii) False
\n(iii) True<\/p>\n

Question 3.
\nAdd: (i) 8x, 3x
\n(ii) 7mn, 5mn
\n(iii) -9y, 11y, 2y
\nSolution:
\n(i) 8x + 3x = (8 + 3) x = 11x
\n(ii) 7mn + 5mn = (7 + 5)mn = 12mn
\n(iii) -9y + 11y + 2y =(-9 + 11 + 2 )y = (2 + 2)y = 4y<\/p>\n

Question 4.
\nSubtract:
\n(i) 4k from 12k
\n(ii) 15q from 25q
\n(iii) 7xyz from 17xyz
\nSolution:
\n(i) 4k from 12k
\n12k – 4k = (12 – 4) k = 8k
\n(ii) 15q from 25q
\n25q – 15q = (25 – 15)q = 10q
\n(iii) 7xyz from 17xyz
\n17xyz – 7xyz = (17 – 7)xyz = 10xyz<\/p>\n

\"SamacheerKalvi.Guru\"<\/p>\n

Question 5.
\nFind the sum of the following expressions
\n(i) 7p + 6q, 5p – q, q + 16p
\nSolution:
\n(7p + 6q) + (5p – q) + (q + 16p) = 7p + 6q + 5p – q + q + 16p
\n= (7p + 5p + 16p) + (6q – q + q)
\n= (7 + 5 + 16) p + (6 – 1 + 1) q
\n= (12 + 16) p + 6q = 28p + 6q<\/p>\n

(ii) a + 5b + 7c, 2a + 106 + 9c
\nSolution:
\n(a + 5b + 7c) + (2a + 10b + 9c) = a + 5b + 7c + 2a + 10b + 9c
\n= a + 2a + 5b + 10b + 7c + 9c
\n= (1 + 2)a + (5 + 10)b + (7 + 9)c
\n= 3a + 15b + 16c<\/p>\n

(iii) mn + t, 2mn – 2t, – 3t + 3mn
\nSolution:
\n(mn + t) + (2mn – 2t) + (-3t + 3mn)
\n= mn + t + 2mn – 2t + (-3t) + 3mn
\n= (mn + 2mn + 3mn) + (t – 2t – 3t)
\n= (1 + 2 + 3) mn + (1 – 2 – 3) t
\n= 6mn + (1 – 5)t
\n= 6mn + (- 4) t
\n= 6mn – 4t<\/p>\n

(iv) u + v, u – v, 2u + 5v, 2u – 5v
\nSolution:
\n(u + v) + (u – v) + (2u + 5v) + (2u – 5v)
\n= u + v + u – v + 2u + 5v + 2u – 5v
\n= u + u + 2u + 2u + v – v + 5v – 5v
\n= (1 + 1 + 2 + 2) u +(1 – 1 + 5 – 5)v = 6u + 0v
\n= 6u<\/p>\n

(v) 5xyz – 3xy, 3zxy – 5yx
\nSolution:
\n5xyz – 3xy + 3zxy – 5yx = 5xyz + 3xyz – 3xy – 5xy
\n= (5 + 3) xyz + [(-3) + (-5)] xy = 8xyz + (-8) xy
\n= 8xyz – 8xy<\/p>\n

Question 6.
\nSubtract
\n(i) 13x + 12y – 5 from 27x + 5y – 43
\nSolution:
\n27x + 5y – 43 – (13x + 12y – 5) = 27z + 5y – 43 + (-13x – 12y + 5)
\n= 27x + 5y – 43 – 13x – 12y + 5
\n= (27 – 13) x + (5 – 12)y + (- 43) + 5
\n= 14x + (- 7) y + (- 38) = 14x – 7y – 38<\/p>\n

(ii) 3p + 5 from p – 2q + 7
\nSolution:
\np – 2q + 7 – (3p + 5) = p – 2q + 7 + (- 3p – 5)
\n= p – 2q + 7 – 3p – 5 = p – 3p – 2q + 7 – 5
\n= (1 – 3)p – 2q + 2 = -2p – 2q + 2<\/p>\n

(iii) m + n from 3m – 7n
\nSolution:
\n3m – 7n – (m + n) = 3m – 7n + (-m – n)
\n= 3m – 7n – m – n = (3m – m) + (-7n – n)
\n= (3 – 1 )m + (-7 – 1) n = 2m + (-8) n
\n= 2m – 8n<\/p>\n

(iv) 2y + z from 6z – 5y
\nSolution:
\n6z – 5y – (2y + z) = 6z – 5y + (-2y – z)
\n= 6z – 5y – 2y – z = 6z – z – 5y – 2y
\n= (6 – 1) z + (-5 -2) y = 5z + (-7) y
\n= 5z – 7y = -7y + 5z<\/p>\n

Question 7.
\nSimplify
\n(i) (x + y – z) + (3x – 5y + 7z) – (14x + 7y – 6z)
\nSolution:
\n(x + y – z) + (3x – 5y + 7z) – (14x – 7y – 6z)
\n= (x + y – z) + (3x – 5y + 7z) + (-14x – 7y + 6z)
\n= (x + 3x – 14x) + (y – 5y – 7y) + (-z + 7z + 6z)
\n= (1 + 3 – 14) x + (1 – 5 – 7)y + (-1 + 7 + 6) z
\n= – 10x – 11y + 12z<\/p>\n

(ii) p + p + 2 + p + 3 + p – 4 – p – 5 + p + 10
\nSolution:
\np + p + 2 + 3 – p – 4 – p – 5 + p + 10 = (p + p + p – p – p + p) + (2 + 3 – 4 – 5 + 10)
\n= (1 + 1 + 1 – 1 – 1 + 1) p + 6 = 2p + 6<\/p>\n

(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
\nSolution:
\nn + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
\n= n + m + 1 + n + 2 + m + 3 + n + 4 + m + 5
\n= n + n + n + m + m + m + 1 + 2 + 3 + 4 + 5
\n= (1 + 1 + 1)n + (1 + 1 + 1)m + 15
\n= 3n + 3m + 15 = 3m + 3n + 15<\/p>\n

Objective Type Questions<\/span><\/p>\n

Question 8.
\nThe addition of 3mn, -5mn, 8mn and – 4mn is
\n(i) mn
\n(ii) – mn
\n(iii) 2mn
\n(iv) 3mn
\nSolution:
\n(iii) 2mn
\nHint: = 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn = 2 mn<\/p>\n

\"SamacheerKalvi.Guru\"<\/p>\n

Question 9.
\nWhen we subtract ‘a’ from \u2018-a\u2019, we get ______
\n(i) a
\n(ii) 2a
\n(iii) -2a
\n(iv) -a
\nSolution:
\n(iii) -2a
\nHint: – a – a = – 2a<\/p>\n

Question 10.
\nIn an expression, we can add or subtract only _____
\n(i) like terms
\n(ii) unlike terms
\n(iii) all terms
\n(iv) None of the above
\nSolution:
\n(i) like terms<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2 Question 1. […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[5],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1611"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=1611"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1611\/revisions"}],"predecessor-version":[{"id":55089,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1611\/revisions\/55089"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=1611"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=1611"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=1611"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}