{"id":15977,"date":"2020-09-10T16:58:33","date_gmt":"2020-09-10T11:28:33","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=15977"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-solutions-for-class-7-maths-chapter-12-ex-12-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-12-ex-12-2\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2"},"content":{"rendered":"

Students can Download Chapter 12 Algebraic Expressions Ex 12.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2<\/h2>\n

Question 1.
\nSimplify combining like terms :
\ni) 21b – 32 + 7b – 20b
\n(re-arranging the terms)
\n= 21b + 7b – 20b – 32
\n= 28b – 20b – 32
\n= 8b – 32<\/p>\n

ii) z2<\/sup> + 13z2<\/sup> – 5z + 7Z2<\/sup> – 15z
\n(re-arranging the terms)
\n= -z2<\/sup> + 13z2<\/sup> – 5z – 15z + 7z3<\/sup>
\n= 12z2<\/sup> – 20z + 7z3<\/sup><\/p>\n

iii) p – (p – q) – q – (q – p)
\n= p – p + q – q – q + p
\n(re-arranging the terms)
\n\"KSEEB
\n= p – q<\/p>\n

\"KSEEB<\/p>\n

iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
\n= 3a – 2b – ab – a + b – ab + 3ab + b – a
\n= 3a – a – a – 2b + b + b – ab – ab + 3ab
\n\"KSEEB
\n= a + ab<\/p>\n

v) 5x2<\/sup>y – 5x2<\/sup> + 3yx2<\/sup> – 3y2<\/sup> + x2<\/sup> – y2<\/sup> + 8xy2<\/sup> – 3y2<\/sup>
\n(re-arranging the terms)
\n= 5x2<\/sup>y + 3yx2<\/sup> – 5x2<\/sup> + x2<\/sup> – 3y2<\/sup> – y2<\/sup> – 3y2<\/sup> + y2<\/sup>
\n= 8x2<\/sup>y – 4x2<\/sup> – 7y2<\/sup> + 8xy2<\/sup><\/p>\n

vi) (3y2<\/sup> + 5y – 4) – (8y – y2<\/sup> – 4) .
\n3y2<\/sup> + 5y – 4 – 8y – y2<\/sup> – 4
\n(re-arranging the terms)
\n3y2<\/sup> + y2<\/sup> + 5y – 8y – 4 + 4
\n= 14y2<\/sup> – 3y<\/p>\n

Question 2.
\nAdd:
\ni) 3mn, -5mn, 8mn, -4mn
\n3mn + 8nm – 5mn – 4mn
\n(re-arranging the terms)
\n= 11mn – 9mn = 2mn<\/p>\n

ii) t – 8tz, 3tz – z, z – t
\n\"KSEEB
\n(re-arranging the terms)
\n= -8tz + 3tz = -5tz<\/p>\n

iii) -7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
\n-7mn + 12 mn + 9mn – 2mn + 5 + 2 – 8 – 3
\n(re-arranging the terms)
\n= -7mn – 2mn + 12mn + 9mn + 7 – 11
\n= -9mn + 21 mn -4
\n= 12mn – 4<\/p>\n

iv) a + b -3, b – a +3, a – b + 3
\n\"KSEEB
\n(re-arranging the terms)
\n= (2a – a + 2b – b + 6 – 3)
\n= a + b + 3<\/p>\n

v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
\n(re-arranging the terms)
\n\"KSEEB
\n= 17x + 51<\/p>\n

vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
\n(re-arranging the terms)
\n= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
\n= 7m – 4m – 7n + 3n – 3mn – 3
\n= 13m – 4n – 3mn – 3<\/p>\n

vii) 4x2<\/sup>y, – 3xy2<\/sup>, 5y2<\/sup>, 5x2<\/sup>y
\n(Re-arranging the terms)
\n4x2<\/sup>y + 5x2<\/sup>y – 3xy2<\/sup> – 5xy2<\/sup>
\n= 9x2<\/sup>y – 8xy2<\/sup><\/p>\n

viii) 3p2<\/sup>q2<\/sup> – 4pq + 5, 10p2<\/sup>q2<\/sup>, 15 + 9pq + 7p2<\/sup>q2<\/sup>
\n3p2<\/sup>q2<\/sup> – 10p2<\/sup>q2<\/sup> + 7p2<\/sup>q2<\/sup> – 4pq + 9pq + 5+ 15
\n(Re-arranging the terms)
\n= 3p2<\/sup>q2<\/sup> + 7p2<\/sup>q2<\/sup> – 10p2<\/sup>q2<\/sup> – 4pq + 9pq + 5 + 15
\n\"KSEEB
\n= 5pq + 20<\/p>\n

\"KSEEB<\/p>\n

ix) ab – 4a, 4b – ab, 4a – 4b
\n\"KSEEB
\n(Re-arranging the terms)
\n(All terms are cancelling)
\n= 0<\/p>\n

x) x2<\/sup> – y2<\/sup> – 1, y2<\/sup> – 1 – x2<\/sup>, 1 – x2<\/sup> – y2<\/sup>
\n(Re arranging the terms)
\n\"KSEEB
\n= -x2<\/sup> – y2<\/sup> – 1<\/p>\n

Question 3.
\nSubtract :
\ni) -5y2<\/sup> from y2<\/sup>
\n= y2<\/sup> – (-5y2<\/sup>)
\n= y2<\/sup> + 5y2<\/sup> = 6y2<\/sup><\/p>\n

ii) 6xy from – 12xy
\n= -12xy – (6xy)
\n= -12xy – 6xy = -18xy<\/p>\n

iii) (a – b) from (a + b)
\n= a + b – (a – b)
\n\"KSEEB
\n= 2b<\/p>\n

iv) a(b – 5) from b(5 – a)
\n= ab – 5a from 5b – ab
\n= (5b – ab) – (ab – 5a)
\n= 5b – ab – ab + 5a
\n= 5a – 5b – 2ab<\/p>\n

v) -m2<\/sup> + 5mn from 4m2<\/sup> – 3mn + 8
\n= (4m2<\/sup> – 3mn + 8) – (-m2<\/sup> + 5mn)
\n= 4m2<\/sup> – 3mn + 8 + m2<\/sup> – 5mn
\n= 4m2<\/sup> + m2<\/sup> – 3mn – 5mn + 8
\n= 5m2<\/sup> – 8mn + 8<\/p>\n

vi) -x2<\/sup> + 10x – 5 from 5x – 10
\n= (5x – 10) – (-x2<\/sup> + 10x – 5)
\n= 5x – 10 + x2<\/sup> – 10x + 5
\n= 5x – 10x + x2<\/sup> – 10 + 5
\n= x2<\/sup> – 5x – 5<\/p>\n

vii) 5a2<\/sup> – 7 ab + 5b2<\/sup> from 3ab – 2a2<\/sup> – 2b2<\/sup>
\n= (3ab – 2a2<\/sup> – 2b2<\/sup>) – (5a2<\/sup> – 7ab + 5b2<\/sup>)
\n= 3ab – 2a2<\/sup> – 2b2<\/sup> – 5a2<\/sup> + 7ab – 5b
\n= -2a2<\/sup> – 5a2<\/sup> – 2b2<\/sup> – 5b2<\/sup> + 7ab + 3ab
\n= -7a2<\/sup> – 7b2<\/sup> + 10ab<\/p>\n

\"KSEEB<\/p>\n

viii) 4pq – 5q2<\/sup> – 3p2<\/sup> from 5p2<\/sup> + 3q2<\/sup> – pq
\n= (5p2<\/sup> + 3q2<\/sup> – pq) – (4pq – 5q2<\/sup> – 3p2<\/sup>)
\n= 5p2<\/sup> + 3q2<\/sup> – pq2<\/sup> – 4pq + 5q2<\/sup> + 3p2<\/sup>
\n= 5p2<\/sup> + 3p2<\/sup> + 3q2<\/sup> + 5q2<\/sup> – 4pq – pq
\n= 8p2<\/sup> + 8q2<\/sup> – 5pq<\/p>\n

Question 4.
\na) What should be added to x2<\/sup> + xy + y2<\/sup> to obtain 2x2<\/sup> + 3xy ?
\nSolution:
\n= 2x2<\/sup> + 3xy – (x2<\/sup> + xy + y2<\/sup>)
\n= 2x2<\/sup> + 3xy – x2<\/sup> – xy – y2<\/sup> – 2x2<\/sup> – x2<\/sup> + 3xy – xy – y2<\/sup>
\n(re-arranging the terms)
\n\u2234 x2<\/sup> + 2xy – y2<\/sup> should be added to x2<\/sup> + xy + y2<\/sup> to get 2x2<\/sup> + 3xy<\/p>\n

b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16 ?
\nSolution:
\n= (2a + 8b + 10) – (-3a + 7b + 16)
\n= 2a + 8b+ 10 + 3a – 7b – 16
\n= 2a + 3a + 8b – 7b + 10 – 16
\n(re-arranging the terms)
\n= 5a + b – 6
\n\u2234 (5a + b – 6) should be subtracted from 2a + 8b +10 to get -3a + 7b + 16.<\/p>\n

Question 5.
\nWhat should be taken away from – 4y2<\/sup> + 5xy + 20 to obtain -x2<\/sup> – y2<\/sup> + 6xy + 20?
\nSolution:
\n= (3x2<\/sup> – 4y2<\/sup> + 5xy + 20) – (-x2<\/sup> – y2<\/sup> + 6xy + 20)
\n= 3x2<\/sup> – 4y2<\/sup> + 5xy + 20 + x2<\/sup> + y2<\/sup> – 6xy – 20
\n= 3x2<\/sup> + x2<\/sup> – 4y2<\/sup> + y2<\/sup> + 5xy – 6xy + 20 – 20
\n(re – arranging the terms)
\n= 4x2<\/sup> – 3y – xy
\n\u2234 4x2<\/sup> + 3y2<\/sup> – xy should be taken away from 3x2<\/sup> – 4y2<\/sup> + 5xy + 20 to obtain -x2<\/sup> – y2<\/sup> + 6xy + 20<\/p>\n

Question 6.
\nFrom the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
\nSolution:
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

b) From the sum of 4 + 3x and 5 – 4x + 2x2<\/sup>, subtract the sum of 3x2<\/sup> – 5x and -x2<\/sup> + 2x + 5.
\nSolution:
\n\"KSEEB
\nor
\n[(4 + 3x) + (5 – 4x + 2x2<\/sup>)] – [(3x2<\/sup> – 5x) + (x2<\/sup> + 2x + 5)]
\n= 4 + 3x + 5 – 4x + 2x2<\/sup> – 3x2<\/sup> + 5x + x2<\/sup> – 2x – 5 = 4 + 3x – 4x + 2x2<\/sup> + 5 – 3x2<\/sup> + x2<\/sup> + 5x – 2x – 5
\n(re-arranging the terms)
\n\"KSEEB
\n= 2x + 4<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 12 Algebraic Expressions Ex 12.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Question 1. Simplify combining like terms …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15977"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=15977"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15977\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=15977"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=15977"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=15977"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}