{"id":26932,"date":"2021-06-28T18:01:10","date_gmt":"2021-06-28T12:31:10","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26932"},"modified":"2021-06-28T18:01:10","modified_gmt":"2021-06-28T12:31:10","slug":"ncert-solutions-for-class-7-maths-chapter-12-ex-12-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-12-ex-12-2\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 12 Algebraic Expressions Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2<\/h2>\n

Question 1.
\nSimplify combining like terms:
\n(i) 21b – 32 + 7b – 20b
\n(ii) – z2<\/sup> + 13z2<\/sup> – 5z + 7z3<\/sup> – 15z
\n(iii) p – (p – q) – q – (q – p)
\n(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
\n(v) 5x2<\/sup>y – 5x2<\/sup> + 3yx2<\/sup> – 3y2<\/sup> + x2<\/sup> – y2<\/sup> + 8xy2<\/sup> – 3y2<\/sup>
\n(vi) (3y2<\/sup> + 5y – 4) – (8y – y2<\/sup> – 4)
\nAnswer:
\n(i) 21b – 32 + 76 – 20b
\nCombining the like terms, we have
\n(21b + 7b – 20b) + (- 32) = (21 + 7 – 20) b + (- 32)
\n= 8b – 32<\/p>\n

\"NCERT<\/p>\n

(ii) -z2<\/sup> + 13z2<\/sup> – 5z + 7z3<\/sup> – 15z
\nCombining like terms, we get
\n= 7z3<\/sup> + (-z2<\/sup> + 13z2<\/sup>) + (-5z – 15z)
\n= 7z3<\/sup> + (-1 + 13)z2<\/sup> + (- 5 – 15)z
\n= 7z3<\/sup> + 12z2<\/sup> – 20z<\/p>\n

(iii) p-(p-q)-q-(q-p) = p- p + q – q – q + p
\nCombining like terms, we get
\n= p – p + p + q – q – q
\n= (1 – 1 + 1) p + (1 – 1 – 1) q
\n= p + (-1)q
\n= p – q<\/p>\n

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
\n= 3a – 2b – ab – a + b – ab + 3ab + b -a
\nCombining like terms, we get
\n= 3a – a – a -2b + b + b – ab – ab + 3ab
\n= (3 – 1 – 1) a + (-2 + 1 + 1) b + (-1 -1 + 3)ab
\n= (3 – 2) a + (-2 + 2) b + (-2 + 3) ab
\n= (1) a + (0) b + (1) ab
\n= a + ab<\/p>\n

(v) 5x2<\/sup>y – 5x2<\/sup> + 3yx2<\/sup> – 3y2<\/sup> + x2<\/sup> – y2<\/sup> + 8xy2<\/sup> – 3y2<\/sup>
\nCombining the like terms, we get
\n(5x2<\/sup>y + 3yx2<\/sup>) + 8xy2<\/sup> + (-5x2<\/sup> + x2<\/sup> ) + (-3y2<\/sup> – y2<\/sup> – 3y2<\/sup> )
\n(5 + 3) x2<\/sup>y + 8xy2<\/sup> + (-5 + 1) x2<\/sup> + (-3 – 1 – 3)y2<\/sup>
\n8x2<\/sup>y + 8xy2<\/sup> + (- 4) x2<\/sup> + (-7) y2<\/sup>
\n8x2<\/sup>y + 8xy2<\/sup> – 4x2<\/sup> – 7y2<\/sup><\/p>\n

\"NCERT<\/p>\n

(vi) (3y2<\/sup> + 5y – 4)-(8y – y2<\/sup> – 4)
\n= 3y2<\/sup> + 5y – 4 – 8y + y2<\/sup> + 4
\nCombining the like terms, we get
\n= 3y2<\/sup> + y2<\/sup> + 5y – 8y – 4 + 4
\n= (3 + 1)y2<\/sup> + (5-8)y + (-4 + 4)
\n= 4y2<\/sup> – 3y<\/p>\n

Question 2.
\nAdd:
\n(i) 3mn, – 5mn, 8mn, – 4mn
\n(ii) t – 8tz, 3tz – z, z – t
\n(iii) – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
\n(iv) a + b-3, b-a + 3, a-b + 3
\n(v) 14x + lOy – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
\n(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
\n(vii) 4x2<\/sup>y, – 3xy2<\/sup>, – 5xy2<\/sup>, 5x2<\/sup>y
\n(viii) 3p2<\/sup>q2<\/sup> – 4pq + 5, – 10p2<\/sup>q2<\/sup>, 15 + 9pq + 7p2<\/sup>q2<\/sup>
\n(ix) ab – 4a, 4b – ab, 4a – 4b
\n(x) x2<\/sup> – y2<\/sup> – 1, y2<\/sup> -1 – x2<\/sup>, 1 – x2<\/sup> – y2<\/sup>
\nAnswer:
\n(i) 3mn, – 5mn, 8mn, – 4mn
\n3mn + (-5mn) + 8mn + (- 4mn)
\n= 3mn – 5mn + 8mn – 4mn
\n= (3 – 5 + 8 – 4) mn = 2mn<\/p>\n

(ii) t – 8tz, 3tz – z; z -t
\nt – 8tz + 3tz + (-z) + z + (-t)
\n= t – 8tz + 3tz – z + z – t
\nCombining like terms, we get
\n= t – t-z + z – 8tz + 3tz
\n= (1 – 1) t + (- 1 + 1) z + (- 8 + 3) tz
\n= (0) t + (0) z + (-5) tz
\n= 0 + 0 – 5tz
\n= -5tz<\/p>\n

\"NCERT<\/p>\n

(iii) -7mn + 5; 12mn + 2; 9mn – 8; – 2mn -3
\n-7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn – 3)
\n– 7mn + 5 + 12mn + 2 + 9mn – 8 -2mn – 3
\nCombining like terms, we get
\n– 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
\n= (-7 + 12 + 9-2) mn + (5 + 2 – 8 -3)
\n= (21-9) mn + (7 – 11)
\n= (12) mn + (- 4)
\n= 12mn – 4<\/p>\n

(iv) a + b-3;b-a + 3;a-b + 3
\na + b – 3 + b – a + 3 + a – b + 3
\nCombining like terms, we get
\n= (a – a + a) + (b + b – b) + (-3 + 3 + 3)
\n= (1 – 1 + 1) a + (1 + 1 – 1)b + (6 – 3)
\n= (2 – 1) a + (2 -1) b + 3
\n= a + b + 3<\/p>\n

(v) 14x + 10y – 12xy – 13; 18 – 7x – lOy + 8xy, 4xy
\n14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
\nCombining like terms, we get
\n= (14x – 7x) + (10y – 10y) + (-12xy + 8xy + 4xy)
\n= (14 – 7)x + (10 – 10) y + (-12 + 8 + 4) xy+ (-13 + 18)
\n= 7x + (0)y + (0)xy + (+5)
\n= 7x + 5<\/p>\n

(vi) 5m – 7n, 3n – 4m + 2; 2m – 3mn – 5
\n5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
\nCombining like terms, we get
\n= (5m – 4m + 2m) + (-7n + 3n) + (-3mn) + (2 – 5)
\n= (5 – 4 + 2) m + (-7 + 3) n – 3mn + (-3)
\n= (7 – 4) m + (- 4) n – 3mn -3
\n= 3m – 4n – 3mn – 3<\/p>\n

(vii) 4x2<\/sup>y; – 3xy2<\/sup>, – 5xy2<\/sup>; 5x2<\/sup>y
\n4x2<\/sup>y + (-3xy2<\/sup>) + (-5xy2<\/sup>) + 5x2<\/sup>y
\n= 4x2<\/sup>y – 3xy2<\/sup> – 5xy2<\/sup> + 5x2<\/sup>y
\nCombining like terms, we get
\n= 4x2<\/sup>y + 5x2<\/sup>y – 5xy2<\/sup> – 5xy2<\/sup>
\n= (4 + 5) x2<\/sup>y + (- 3 – 5) xy2<\/sup>
\n= 9x2<\/sup>y + (- 8) xy2<\/sup>
\n= 9x2<\/sup>y – 8xy2<\/sup><\/p>\n

\"NCERT<\/p>\n

(viii) 3p2<\/sup>q2<\/sup> – 4pq + 5; – 10p2<\/sup>q2<\/sup>; 15 + 9pq + 7p2<\/sup>q2<\/sup>
\n3p2<\/sup>q2<\/sup> + (- 4pq) + 5 + (-10p2<\/sup>q2<\/sup>) + 15 + 9pq + 7p2<\/sup>q2<\/sup>
\n= 3p2<\/sup>q2<\/sup> – 4pq + 5 – 10p2<\/sup>q2<\/sup> + 15 + 9pq + 7p2<\/sup>q2<\/sup>
\nCombining like terms, we get
\n= 3p2<\/sup>q2<\/sup> – 10p2<\/sup>q2<\/sup> + 7p2<\/sup>q2<\/sup> – 4pq + 9pq + 5 + 15
\n= (3 – 10 + 7) p2<\/sup>q2<\/sup> + (- 4 + 9) pq + (5 + 15)
\n= (10 – 10) p2<\/sup>q2<\/sup> + (5) pq + 20
\n= (0) p2<\/sup>q2<\/sup> + 5pq + 20
\n= 5pq + 20<\/p>\n

(ix) ab – 4a; 4b – ab; 4a – 4b
\nab – 4a + 4b – ab + 4a – 4b = (ab – ab) + (- 4a + 4a) + (4b – 4b)
\n= (1 – 1) ab + (- 4 + 4) a + (4 – 4) b
\n= (0) ab + (0) a + (0) b
\n= 0 + 0 + 0
\n= 0<\/p>\n

(x) x2<\/sup> – y2<\/sup> – 1; y2<\/sup> – 1 – x2<\/sup>; 1 – x2<\/sup> – y2<\/sup>
\nx2<\/sup>– y2<\/sup> – 1 + y2<\/sup> – 1 – x2<\/sup> + 1 – x2<\/sup> – y2<\/sup>
\n= (x2<\/sup> – x2<\/sup> – x2<\/sup>) + (- y2<\/sup> + y2<\/sup> – y2<\/sup>) + (-1 – 1 + 1)
\n= (1 – 1 – 1)x2<\/sup> + (-2 + 1)y2<\/sup> + (-1)
\n= (1 – 2) x2<\/sup> + (-2 + 1)y2<\/sup> + (-1)
\n= (- 1) x2<\/sup> + (-1) y2<\/sup> +(-1)
\n= – x2<\/sup> – y2<\/sup> – 1<\/p>\n

Question 3.
\nSubtract:
\n(i) – 5y2<\/sup> from y2<\/sup>
\n(ii) 6xy from -12xy
\n(iii) (a – b) from (a + b)
\n(iv) a (b – 5) from b (5 – a)
\n(v) -m2<\/sup> + 5 mn from 4m2<\/sup> – 3mn + 8
\n(vi) – x2<\/sup> + 10x – 5 from 5x – 10
\n(vii) 5a2<\/sup> – 7ab + 5b2<\/sup> from 3ab – 2a2<\/sup> – 2b2<\/sup>
\n(viii) 4pq – 5q2<\/sup> – 3p2<\/sup> from 5p2<\/sup> + 3q2<\/sup> – pq
\nAnswer:
\n(i) Subtract – 5y2<\/sup> from y2<\/sup>
\ny2<\/sup> – (-5y2<\/sup> ) = y2<\/sup> + 5y2<\/sup> – 12xy – (6xy) = 6y2<\/sup><\/p>\n

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

\"NCERT<\/p>\n

(iii) Subtract (a – b) from (a + b)
\n(a + b)-(a-b) = a + b- a + b =a-a+b+b
\n= (1 – 1)a + (1 + 1)b
\n= (0)a + 2b
\n= 0 + 2b
\n= 2b<\/p>\n

(iv) Subtract a (b – 5) from b (5 – a) b (5 – a) – a (b – 5)
\n= 5b – ab – ab + 5a
\n= 5b + 5a – ab – ab
\n= 5a + 5b + (-1 -1) ab
\n= 5a + 5b – 2ab<\/p>\n

(v) Subtract -m2<\/sup> + 5mn from 4m2<\/sup> – 3mn + 8
\n4m2<\/sup> – 3mn + 8 – (-m2<\/sup> + 5mn)
\n= 4m2<\/sup> – 3mn + 8 + m2<\/sup> – 5mn
\n= 4m2<\/sup> + m2<\/sup> – 3mn – 5mn + 8
\n= (4 + 1) m2<\/sup> + (-3 – 5) mn + 8 = 5m2<\/sup> + (- 8) mn + 8
\n= 5m2<\/sup> – 8mn + 8<\/p>\n

(vi) Subtract -x2<\/sup> + 10x – 5 from 5x – 10
\n5x – 10 – (-x2<\/sup> + 10x – 5)
\n= 5x – 10 + x2<\/sup> – 10x + 5
\n= x2<\/sup> + 5x – 10x – 10 + 5
\n= x2<\/sup> + (5 – 10) x + (-10 + 5)
\n= x2<\/sup> + (-5) x + (-5)
\n= x2<\/sup> – 5x – 5<\/p>\n

\"NCERT<\/p>\n

(vii) Subtract 5a2<\/sup> – 7ab + 5b2<\/sup> from 3ab – 2a2<\/sup> – 2b2<\/sup>
\n3ab – 2a2<\/sup> – 2b2<\/sup> – (5a2<\/sup> – 7ab + 5b2<\/sup>)
\n= 3ab – 2a2<\/sup> – 2b2<\/sup> – 5a2<\/sup> + 7ab – 5b2<\/sup>
\n= -2a2<\/sup> – 5a2<\/sup> – 2b2<\/sup> – 5b2<\/sup> + 3ab + 7ab
\n= (-2 -5) a2<\/sup> + (-2 – 5) b2<\/sup> + (3 + 7) ab
\n= -7a2<\/sup> – 7b2<\/sup> + 10ab (or) 10ab – 7a2<\/sup> – 7b2<\/sup><\/p>\n

(viii) Subtract 4pq – 5q2<\/sup> – 3p2<\/sup> from 5p2<\/sup> + 3q2<\/sup> – pq
\n5p2<\/sup> + 3q2<\/sup> – pq – (4pq – 5q2<\/sup> – 3p2<\/sup>)
\n= 5p2<\/sup> + 3q2<\/sup> – pq – 4pq + 5p2<\/sup> + 3p2<\/sup>
\n= (5 + 3) p2<\/sup> + (3 + 5)q2<\/sup> – pq – 4pq
\n= 8p2<\/sup> + 8q2<\/sup> -5pq<\/p>\n

Question 4.
\n(a) What should be added to x2<\/sup> + xy + y2<\/sup> to obtain 2x2<\/sup> + 3xy?
\n(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?
\nAnswer:
\n(a) The required expression is
\n2x2<\/sup> + 3xy – (x2<\/sup> + xy + y2<\/sup>)
\n= 2x2<\/sup> + 3xy – x2<\/sup> – xy – y2<\/sup>
\n= (2x2<\/sup> – x2<\/sup>) – y2<\/sup> + 3xy – xy
\n= (2 – 1) x2<\/sup> – y2<\/sup> + (3 – 1) xy
\n= (1) x2<\/sup> – y2<\/sup> + (2) xy
\n= x2<\/sup> + 2xy – y2<\/sup><\/p>\n

(b) The required expression is
\n(2a + 8b + 10) – (- 3a + 7b + 16)
\n= 2a + 8b + 10 + 3a – 7b – 16
\n= (2a + 3a) + (8b – 7b) + 10- 16
\n= (2 + 3) a + (8 – 7) b + (10 – 16)
\n= (5)a + (l)b + (-6)
\n= 5a + b – 6<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nWhat should be taken away from 3x2<\/sup> – 4y2<\/sup> + 5xy + 20 to obtain -x2<\/sup> – y2<\/sup> + 6xy + 20?
\nAnswer:
\nThe required expression is
\n3x2<\/sup> – 4y2<\/sup> + 5xy + 20 –
\n(-x2<\/sup> – y2<\/sup> + 6xy + 20)
\n= 3x2<\/sup> – 4y2<\/sup> + 5xy + 20 – (-x2 -y2 + 6xy + 20)
\n= 3x2<\/sup> – 4y2<\/sup> + 5xy + 20 + x2<\/sup> + y2<\/sup> – 6xy – 20)
\n= (3x2<\/sup> + x2<\/sup>) + (- 4y2<\/sup> + y2<\/sup>) +
\n(+ 5xy – 6xy) + (20 – 20) = (3 + 1) x2 + (- 4 + 1) y2 +
\n(+ 5 – 6) xy + (0)
\n= (4) x2<\/sup> + (-3) y2<\/sup> + (-1xy)
\n= 4x2<\/sup> – 3y2<\/sup> – xy<\/p>\n

Question 6.
\n(a) From the sum of 3x – y + 11 and – y -11, subtract 3x – y – 11.
\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.
\nAnswer:
\n(a) The required expression is (3x-y+ 11) + (- y-11) –
\n(3x – y -11)
\n= 3x – y + 11 – y – 11 – 3x + y + 11
\n= (3x – 3x) + (-y – y + y) + (11 – 11 + 11)
\n= (3 – 3)x + (-1 -1 + 1) y + (11 – 11 + 11)
\n= (3 – 3)x + (-2 + 1)y + (22 – 11)
\n– (0) x + (-1) y + (11)
\n= 0 – y + 11 = -y + 11<\/p>\n

\"NCERT<\/p>\n

(b) The required expression is
\n(4 + 3x) + (5 – 4x + 2x2<\/sup>) –
\n[(3x2<\/sup> – 5x + (-x2<\/sup> + 2x + 5)]
\n= 4 + 3x + 5 – 4x + 2x2<\/sup> – [3x2<\/sup> – 5x – x2 + 2x + 5]
\n= 4 + 3x + 5 – 4x + 2x2<\/sup> – 3x2<\/sup> +5x + x2 – 2x – 5
\n= (2x2<\/sup> – 3x2<\/sup> + x2<\/sup>) + (3x -4x + 5x- 2x) + (4 + 5 – 5)
\n= (2 – 3 + 1) x2<\/sup> + (3 – 4 + 5 – 2) x + (9-5)
\n= (3 – 3) x2<\/sup> + (8 – 6) x + 4
\n= (0)x2<\/sup> + (2) x + 4
\n= 0 + 2x + 4
\n= 2x + 4<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2 Question 1. Simplify combining like terms: (i) 21b – 32 + 7b – 20b (ii) – z2 + 13z2 …<\/p>\n

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NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2 Question 1. 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