{"id":17000,"date":"2022-01-29T14:50:25","date_gmt":"2022-01-29T09:20:25","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=17000"},"modified":"2023-03-29T16:17:11","modified_gmt":"2023-03-29T10:47:11","slug":"ap-board-7th-class-maths-solutions-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-board-7th-class-maths-solutions-chapter-9-ex-9-3\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 9 Algebraic Expressions Ex 9.3"},"content":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf<\/a> Chapter 9 Algebraic Expressions Ex 9.3 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.3<\/h2>\n

Question 1.
\nWrite standard form and additive inverse of the following expressions.
\n(i) – 6a
\nAnswer:
\nAdditive inverse of – 6a = – (- 6a) = 6a<\/p>\n

\"AP<\/p>\n

(ii) 2 + 7c2<\/sup>
\nAnswer:
\nStandard form of 2 + 7c2<\/sup> = 7c2<\/sup> + 2
\nAdditive inverse of 7c2<\/sup> + 2
\n= – (7c2<\/sup> + 2)
\n= – 7c2<\/sup> – 2<\/p>\n

(iii) 6x2<\/sup> + 4x – 5
\nAnswer:
\nGiven expression is in standard form.
\nAdditive inverse of 6x2<\/sup> + 4x – 5
\n= – (6x2<\/sup> + 4x – 5)
\n= – 6x2<\/sup> – 4x + 5<\/p>\n

(iv) 3c + 7a – 9b
\nAnswer:
\nStandard form of 3c + 7a – 9b = 7a – 9b + 3c
\nAdditive inverse of 7a – 9b + 3c
\n= – (7a – 9b + 3c)
\n= – 7a + 9b – 3c<\/p>\n

Question 2.
\nWrite the following expressions in standard form:
\n(i) 6x + x2<\/sup> – 5
\nAnswer:
\nStandard form of 6x + x2<\/sup> – 5
\n= x2<\/sup> + 6x – 5<\/p>\n

(ii) 3 – 4a2<\/sup> – 5a
\nAnswer:
\nStandard form of 3 – 4a2<\/sup> – 5a
\n= – 4a2<\/sup> – 5a + 3<\/p>\n

(iii) – m + 6 + 3m2<\/sup>
\nAnswer:
\nStandard form of – m + 6 + 3m2<\/sup>
\n= 3m2<\/sup> – m + 6<\/p>\n

\"AP<\/p>\n

(iv) c3<\/sup> + 1 + c + 2c2<\/sup>
\nAnswer:
\nStandard form of
\nc3<\/sup> + 1 + c + 2c2<\/sup> = c3<\/sup> + 2c2<\/sup> + c + 1<\/p>\n

(v) 9 – p2<\/sup>
\nAnswer:
\nStandard form of 9 – p2<\/sup> = – p2<\/sup> + 9<\/p>\n

Question 3.
\nAdd the following algebraic expressions using both horizontal and vertical methods. Did you get the same answer with both the methods? Verify.
\n(i) 2x2<\/sup> – 6x +3; 4x2<\/sup> + 9x + 5
\nAnswer:
\n\"AP<\/p>\n

(ii) a2<\/sup> + 6ab + 8; – 3a2<\/sup> – ab – 2
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

(iii) – p2<\/sup> + 2p – 10; 4 – 5p – 2p2<\/sup>
\nAnswer:
\n\"AP<\/p>\n

Question 4.
\nSubtract the second expression from the first expression:
\n(i) 2x + y , x – y
\nAnswer:
\nLet A = 2x + y and B = x – y
\nA – B = A + (- B) Additive inverse of B is
\n-B = – (x – y) = – x + y
\n\u2234 A – B = A + (- B)
\n= 2x + y + (- x + y)
\n= 2x + y – x + y
\n= 2x – x + y + y
\n= (2 – 1)x + (1 + 1)y
\n\u2234 A – B = x + 2y<\/p>\n

\"AP<\/p>\n

(ii) a + 2b + c, – a – b – 3c
\nAnswer:
\nLet A = a + 2b + c and B = – a – b – 3c
\nA – B = A + (- B)
\nAdditive inverse of B is
\n– B = – (- a – b – 3c)
\n= a + b + 3c
\n\u2234 A – B = A + (- B)
\n= a + 2b + c + (a + b + 3c)
\n= a + 2b + c + a + b + 3c
\n= (a + a) + (2b + lb) + (c + 3c)
\n\u2234 A – B = 2a + 3b + 4c<\/p>\n

(iii) 2l2<\/sup> – 3lm + 5m2<\/sup>, 3l2<\/sup> – 4lm + 6m2<\/sup>
\nAnswer:
\nLet A = 2l2<\/sup> – 3lm + 5m2<\/sup> and
\nB = 3l2<\/sup> – 4lm + 6m2<\/sup>
\nA – B = A + (- B)
\nAdditive inverse of B is
\n– B = – (3t2<\/sup> – 4lm + 6m2<\/sup>)
\n= – 3l2<\/sup> + 4lm – 6m2<\/sup>
\n\u2234 A – B = A + (- B)
\n= (2l2<\/sup> – 3lm + 5m2<\/sup>) + (- 3l2<\/sup> + 4lm – 6m2<\/sup>)
\n= 2l2<\/sup> – 3lm + 5m2<\/sup> – 3l2<\/sup> + 4lm – 6m2<\/sup>
\n= 2l2<\/sup> – 3l2<\/sup> – 3lm + 4lm + 5m2<\/sup> – 6m2<\/sup>
\n= (2 – 3)l2<\/sup> + (- 3 + 4)lm + (5 – 6)m2<\/sup>
\n= (- 1) l2<\/sup> + 1 lm + (- 1)m2<\/sup>
\n\u2234 A – B = – l2<\/sup> + lm – m2<\/sup><\/p>\n

(iv) 7 – x – 3x2<\/sup>, 2x2<\/sup> – 5x – 3
\nAnswer:
\nLet A = 7 – x – 3x2<\/sup> and B = 2x2<\/sup> – 5x – 3
\nWrite the given expressions in standard form.
\n\u2234 A = – 3x2<\/sup> – x + 7 and B = 2x2<\/sup> – 5x – 3
\nA – B = A + (- B)
\nAdditive inverse of B is
\n– B = – (2x2<\/sup> – 5x – 3)
\n= – 2x2<\/sup> + 5x + 3
\n\u2234 A – B = A + (- B)
\n= (- 3x2<\/sup> – x + 7) + (- 2x2<\/sup> + 5x + 3)
\n= – 3x2<\/sup> – x + 7 – 2x2<\/sup> + 5x + 3
\n= (- 3x2<\/sup> – 2x2<\/sup>) + (- x + 5x) + (7 + 3)
\n= (- 3 – 2)x2<\/sup> + (- 1 + 5)x + 10
\n\u2234 A – B = – 5x2<\/sup> + 4x + 10<\/p>\n

\"AP<\/p>\n

(v) 6m3<\/sup> + 4m2<\/sup> + 7m – 3, 2m3<\/sup> + 4
\nAnswer:
\nLet A = 6m3<\/sup> + 4m2<\/sup> + 7m – 3 and
\nB = 2 m2<\/sup> + 4
\nA – B = A + (- B)
\nAdditive inverse of B is
\n– B = – (2m3<\/sup> + 4)
\n= – 2m3<\/sup> – 4
\n\u2234 A – B = A + (- B)
\n= (6m3<\/sup> + 4m2<\/sup> + 7m – 3) + (- 2m3<\/sup> – 4)
\n= 6m3<\/sup> + 4m2<\/sup> + 7m – 3 – 2m3<\/sup> – 4
\n= (6m3<\/sup> – 2m3<\/sup>) + 4m2<\/sup> + 7m + (- 3 – 4)
\n= (6 – 2)m3<\/sup> + 4m2<\/sup> + 7m + (- 7)
\n\u2234 A – B = 4m3<\/sup> + 4m2<\/sup> + 7m – 7<\/p>\n

Question 5.
\nFind the perimeter of the beside rect\u00acangle whose length is 6x + y and breadth is 3x – 2y.
\n\"AP
\nAnswer:
\nGiven length of rectangle l = 6x + y
\nbreadth b = 3x – 2y
\nPerimeter of Rectangle = 2 (l + b)
\n= 2[(6x + y) + (3x – 2y)]
\n= 2[6x + y + 3x – 2y]
\n= 2[(6 + 3)x + (1 – 2)y]
\n= 2[9x + (- 1) y]
\n= 2[9x – 1y]
\n= 2 \u00d7 (9x) – 2 \u00d7 (1y)
\n\u2234 Perimeter of rectangle = (18x – 2y) units.<\/p>\n

\"AP<\/p>\n

Question 6.
\nFind the perimeter of triangle whose sides are a + 3b, a – b and 2a – b.
\n\"AP
\nAnswer:
\nLet the sides of triangle are
\nx = a + 3b, y = a – b and z = 2a – b
\nPerimeter of triangle = x + y + z
\n= (a + 3b) + (a – b) + (2a – b)
\n= a + 3b + a – b + 2a – b
\n= (a + a + 2a) + (3b – b – b)
\n= (1 + 1 + 2)a + (3 – 1 – 1)b
\nPerimeter of triangle.
\n= (4a + b) units.<\/p>\n

Question 7.
\nSubtract the sum of x2<\/sup> – 5xy + 2y2<\/sup> and y2<\/sup> – 2xy – 3x2<\/sup> from the sum of 6x2<\/sup> – 8xy – y2<\/sup> and 2xy – 2y2<\/sup> – x2<\/sup>.
\nAnswer:
\nGiven expressions are
\nx2<\/sup> – 5xy + 2y2<\/sup> and y2<\/sup> – 2xy – 3x2<\/sup> and 6x2<\/sup> – 8xy – y2<\/sup> and 2xy – 2y2<\/sup> – x2<\/sup>
\nWrite the given expressions in the standard form
\nx2<\/sup> – 5xy + 2y2<\/sup> and – 3x2<\/sup> – 2xy + y2<\/sup> and 6x2<\/sup> – 8xy – y2<\/sup> and – x2<\/sup> + 2xy – 2y2<\/sup>
\nLet the Sum
\nA = (x2<\/sup> – 5xy + 2y2<\/sup>) + (- 3x2<\/sup> – 2xy + y2<\/sup>)
\n= x2<\/sup> – 5xy + 2y2<\/sup> – 3x2<\/sup> – 2xy + y2<\/sup>
\n= x2<\/sup> – 3x2<\/sup> – 5xy – 2xy + 2y2<\/sup> + y2<\/sup>
\n= (1 – 3)x2<\/sup> + (- 5 – 2)xy + (2 + 1)y2<\/sup>
\nA = – 2x2<\/sup> – 7xy + 3y2<\/sup><\/p>\n

Let the Sum
\nB = (6x2<\/sup> – 8xy – y2<\/sup>) + (- x2<\/sup> + 2xy – 2y2<\/sup>)
\n= 6x2<\/sup> – 8xy – y2<\/sup> – x2<\/sup> + 2xy – 2y2<\/sup>
\n= (6 – 1 )x2<\/sup> + (- 8 + 2<\/sup>)xy + (- 1 – 2)y2<\/sup>
\nB = 5x2<\/sup> – 6xy – 3y2<\/sup>
\nB – A = B + (- A)
\nAdditive inverse of A is
\n– A = – (A)
\n= – (- 2x2<\/sup> – 7xy + 3y2<\/sup>)
\n\u2234 – A = 2x2<\/sup> + 7xy – 3y2<\/sup>
\nB – A = B + (- A)
\n= (5x2<\/sup> – 6xy – 3y2<\/sup>) + (2x2<\/sup> + 7xy – 3y2<\/sup>)
\n= (5 + 2)x2<\/sup> + (- 6 + 7)xy + (- 3 – 3)y2<\/sup>
\n\u2234 B – A = 7x2<\/sup> + xy – 6y2<\/sup><\/p>\n

\"AP<\/p>\n

Question 8.
\nWhat should be added to 1 + 2<\/sup>p – 3P2<\/sup> to get p2<\/sup> – p – 1 ?
\nAnswer:
\nGiven expressions are
\n1 + 2p – 3p2<\/sup> and p2<\/sup> – p – 1
\nWrite the given expressions in the standard form.
\n– 3p2<\/sup> + 2<\/sup>p + 1 and p2<\/sup> – p – 1
\nLet A should be added to B to get C.
\ni.e. A + B = C
\n\u2234 A = C – B
\nLet B = – 3p2<\/sup> + 2p + 1 and
\nC = p2<\/sup> – p – 1
\nA = C + (- B )
\nAdditive inverse B is
\n– B = – (- 3p2<\/sup> + 2p + 1) .
\n– B = 3p2<\/sup> – 2p – 1 .
\nA = (p2<\/sup> – p – 1) + (3p2<\/sup> – 2p – 1)
\n= p2<\/sup> – p – 1 + 3p2<\/sup> – 2p – 1
\n= (1 + 3)p2<\/sup> + (- 1 – 2)p + (- 1 – 1)
\n\u2234 A = 4p2<\/sup> – 3p – 2<\/sup>
\n\u2234 4p2<\/sup> – 3p – 2<\/sup> is added to 1 + 2p – 3p2<\/sup> to get p2<\/sup> – p – 1.<\/p>\n

Question 9.
\nWhat should be taken away from 3a2<\/sup> – 4b2<\/sup> + 5ab + 20 to get – a2<\/sup> – b2<\/sup> + 6ab + 3 ?
\nAnswer:
\nGiven expressions are
\n3a2<\/sup> – 4b2<\/sup> + 5ab + 20 and – a2<\/sup> – b2<\/sup> + 6ab +3
\nLet A be taken away from B to get C. that is A = B – C = B + (- C)
\nLet B = 3a2<\/sup> – 4b2<\/sup> + 5ab + 20 and C = – a2<\/sup> – b2<\/sup> + 6ab + 3 Additive inverse of C is
\n(- C) = – (- a2<\/sup> – b2<\/sup> + 6ab + 3)
\n= a2<\/sup> + b2<\/sup> – 6ab – 3
\nA = B + (- C)
\n= (3a2<\/sup> – 4b2<\/sup> + 5ab + 20) + (a2<\/sup> + b2<\/sup> – 6ab – 3)
\n= 3a2<\/sup> – 4b2<\/sup> + 5ab + 20 + a2<\/sup> + b2<\/sup> – 6ab – 3
\n= 3a2<\/sup> + a2<\/sup> – 4b2<\/sup> + b2<\/sup> + 5ab – 6ab + 20 – 3
\n= (3 + 1)a2<\/sup> + (- 4 + 1)b2<\/sup> + (5 – 6) ab + (20 – 3)
\nA = 4a2<\/sup> – 3b2<\/sup> – 1 ab + 17
\nSo, 4a2<\/sup> – 3b2<\/sup> – 1 ab + 17 is taken away from 3a2<\/sup> – 4b2<\/sup> + 5ab + 20 to get – a2<\/sup> – b2<\/sup> + 6ab + 3.<\/p>\n

\"AP<\/p>\n

Question 10.
\nIf A = 4x2<\/sup> + y2<\/sup> – 6xy;
\nB = 3y2<\/sup> + 12x2<\/sup> + 8xy;
\nC = 6x2<\/sup> + 8y2<\/sup> + 6xy then,
\nfind(i) A + B + C (ii) (A – B) – C
\nAnswer:
\nGiven A = 4x2<\/sup> + y2<\/sup> – 6xy
\nB = 3y2<\/sup> + 12x2<\/sup> + 8xy
\nC = 6x2<\/sup> + 8y2<\/sup> + 6xy
\nWrite the given expressions in standard form.
\nA = 4x2<\/sup> – 6xy + y2<\/sup>
\nB = 12x2<\/sup> + 8xy + 3y2<\/sup>
\nC = 6x2<\/sup> + 6xy + 8y2<\/sup><\/p>\n

(i) A + B + C = (4x2<\/sup> – 6xy + y2<\/sup>) + (12x2<\/sup> + 8xy + 3y2<\/sup>) + (6x2<\/sup> + 6xy + 8y2<\/sup>)
\n= 4x2<\/sup> – 6xy + y2<\/sup> + 12x2<\/sup> + 8xy + 3y2<\/sup> + 6x2<\/sup> + 6xy + 8y2<\/sup>
\n= (4x2<\/sup> + 12x2<\/sup> + 6x2<\/sup>) + (- 6xy + 8xy + 6xy) + (y2<\/sup> + 3y2<\/sup> + 8y2<\/sup>)
\n= (4 + 12 + 6) x2<\/sup> + (- 6 + 8 + 6) xy + (1 + 3 + 8)y2<\/sup>
\n\u2234 A + B + C = 22x2<\/sup> + 8xy + 12y2<\/sup><\/p>\n

(ii) (A – B) – C
\nA + (- B) + (- C)
\nAdditive inverse of B is
\n– B = – (12x2<\/sup> + 8xy + 3y2<\/sup>)
\n\u2234 – B = – 12x2<\/sup> – 8xy – 3y2<\/sup>
\nAdditive inverse of C is
\n– C = -(6x2<\/sup> + 6xy + 8y2<\/sup>)
\n\u2234 – C = – 6x2<\/sup> – 6xy – 8y2<\/sup>
\nA + (- B) + (- C)
\n= (4x2<\/sup> – 6xy + y2<\/sup>) + (- 12x2<\/sup> – 8xy – 3y2<\/sup>) + (- 6x2<\/sup> – 6xy – 8y2<\/sup>)
\n= 4x2<\/sup> – 6xy + y2<\/sup> – 12x2<\/sup> – 8xy – 3y2<\/sup> – 6x2<\/sup> – 6xy – 8y2<\/sup>
\n= (4x2<\/sup> – 12x2<\/sup> – 6x2<\/sup>) + (- 6xy – 8xy – 6xy) + (y2<\/sup> – 3y2<\/sup> – 8y2<\/sup>)
\n= (4 – 12 – 6)x2<\/sup> + (- 6 – 8 – 6)xy + (1 – 3 – 8)y2<\/sup>
\n\u2234 (A – B) – C = – 14x2<\/sup> – 20xy – 10y2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions Ex 9.3 Textbook Exercise Questions and Answers. AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions Ex 9.3 Question 1. Write standard form and additive inverse of the following expressions. (i) – 6a Answer: Additive inverse of – 6a = – (- 6a) … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":37565,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/01\/Chapter-9-Algebraic-Expressions-Ex-9.3.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/17000"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=17000"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/17000\/revisions"}],"predecessor-version":[{"id":17001,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/17000\/revisions\/17001"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/37565"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=17000"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=17000"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=17000"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}