NCERT Solutions Class 7 Mathematics Algebraic Expressions Ex 12.2

Class 6 - Mathematics
Chapter - Algebraic Expressions : Exercise 12.2

NCERT Solutions Class 7 Mathematics textbook
Top Block 1

Question : 1.Simplify combining like terms:

  1. 21b – 32 + 7b – 20b
  2. – z2 + 13z2 – 5z + 7z3 – 15z
  3. p – (p – q) – q – (q – p)
  4. 3a – 2b – ab – (a – b + ab) + 3ab + b – a
  5. 5𝓍2y – 5𝓍2 + 3y𝓍2 – 3y2 + 𝓍2 – y2 + 8𝓍y2 – 3y2
  6. (3y2 + 5y – 4) – (8y – y2 – 4)

Answer :
(i) 21b – 32 + 7b – 20b = 21b + 7b – 20b – 32

= 28b – 20b – 32 = 8b – 32

(ii) – z2 + 13z2 – 5z + 7z3 – 15z = 7z3 + ( – z2 + 13z2) – (5z + 15z)

= 7z3 + 12z2 – 20z

(iii) p – (p – q) – q – (q – p) = p – p + q – q – q + p

= p – p + p + q – q – q = p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a = 3a – 2b – ab – a + b – ab + 3ab + b – a

= 3a – a – a – 2b + b + b – ab – ab + 3ab

= (3a – a – a) – (2b – b – b) – (ab + ab – 3ab)

= a – 0 – ( – ab)

= a + ab

(v) 5𝓍2y – 5𝓍2 + 3y𝓍2 – 3y2 + 𝓍2 – y2 + 8𝓍y2 – 3y2

= 5𝓍2y + 3y𝓍2 + 8𝓍y2 – 5𝓍2 + 𝓍2 – 3y2 – y2 – 3y2

= (5𝓍2y + 3𝓍2y) + 8𝓍y2 – (5𝓍2 – 𝓍2) – (3y2 + y2 + 3y2)

= 8𝓍2y + 8𝓍y2 – 4𝓍2 – 7y2

(vi) (3y2 + 5y – 4) – (8y – y2 – 4) = 3y2 + 5y – 4 – 8y + y2 + 4

= (3y2 + y2) + (5y – 8y) – (4 – 4)

= 4y2 – 3y – 0 = 4y2 – 3y

Question : 2.Add:

  1. 3mn, – 5mn, 8mn – 4mn
  2. t – 8tz, 3tz – z, z – t
  3. – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
  4. a + b – 3,b – a + 3,a – b + 3
  5. 14𝓍 + 10y – 12𝓍y – 13, 18 – 7𝓍 – 10y + 8𝓍y, 4𝓍y
  6. 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
  7. 4𝓍2y, – 3𝓍y2, – 5𝓍y2, 5𝓍2y
  8. 3p2q2 – 4pq + 5, – 10p2q2, 15 + 9pq + 7p2q2
  9. ab – 4a, 4b – ab, 4a – 4b
  10. 𝓍2 – y2 – 1, y2 – 1 – 𝓍2, 1 – 𝓍2 – y2

Answer :
(i)
3mn, – 5mn,8mn, – 4mn = 3mn + ( – 5mn) + 8mn + ( – 4mn)

= (3 – 5 + 8 – 4)mn = 2mn

(ii)
t – 8tz,3tz – z,z – t = t – 8tz + 3tz – z + z – t

= t – t – 8tz + 3tz – z + z

= (1 – 1)t + ( – 8 + 3)tz + ( – 1 + 1)z

= 0 – 5tz + 0 = – 5tz

(iii)
– 7mn + 5,12mn + 2,9mn – 8, – 2mn – 3 = – 7mn + 5 + 12mn + 2 + 9mn – 8 + ( – 2mn) – 3

= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3 – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3

= ( – 7 + 12 + 9 – 2)mn + 7 – 11

= 12mn – 4

(iv)
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 + b + 3

(v)
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

= 7𝓍 + 0y + 0𝓍y + 5 = 7𝓍 + 5

(vi)
5m – 7n,3n – 4m + 2,2m – 3mn – 5 = 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5

= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5

= (5 – 4 + 2)m + ( – 7 + 3)n – 3mn – 3

= 3m – 4n + 3mn – 3

(vii)
4𝓍2y, – 3𝓍y2, – 5𝓍y2,5𝓍2y = 4𝓍2y + ( – 3𝓍y2) + ( – 5𝓍y2) + 5𝓍2y

= 4𝓍2y + 5𝓍2y – 3𝓍y2 – 5𝓍y2

= 9𝓍2y – 8𝓍y2

(viii)
3p2q2 – 4pq + 5, – 10p2q2,15 + 9pq + 7p2q2

= 3p2q2 – 4pq + 5 + ( – 10p2q2) + 15 + 9pq + 7p2q2

= 3p2q2 – 10p2q2 + 7p2q2 + 4pq + 9pq + 5 + 15

= (3 – 10 + 7)p2q2 + ( – 4 + 9)pq + 20

= 0p2q2 + 5pq + 20 = 5pq + 20

(ix)
ab – 4a,4b – ab,4a – ab = ab – 4a + 4b – ab + 4a – ab

= – 4a + 4a + 4b – 4b + ab – ab

= 0 + 0 + 0 = 0

(x)
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

= (1 – 1 – 1)𝓍2 + ( – 1 + 1 – 1)y2 – 1 – 1 + 1

= – 𝓍2 – y2 – 1

Question : 3.Subtract:

  1. – 5y2 fromy2

  2. 6𝓍y from – 12𝓍y

  3. (a – b) from (a + b)

  4. a(b – 5) from b(5 – a)

  5. – m2 + 5mn from 4m2 – 3mn + 8

  6. – 𝓍2 + 10𝓍 – 5 from 5𝓍 – 10

  7. 5a2 – 7ab + 5b2 from ab – 2a2 – 2b2

  8. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq

Answer :
(i) y2 – ( – 5y2) = y2 + 5y2 = 6y2

(ii) – 12𝓍y – (6𝓍y) = – 12𝓍y – 6𝓍y = – 18𝓍y

(iii) (a + b) – (a – b) = a + b – a + b = a – a + b + b = 2b

(iv) b(5 – a) – a(b – 5) = 5b – ab – ab + 5a = 5b – 2ab + 5a = 5a + 5b – 2ab

(v) 4m2 – 3mn + 8 – ( – m2 + 5mn) = 4m2 – 3mn + 8 + m2 – 5mn

= 4m2 + m2 – 3mn – 5mn + 8

= 5m2 – 8mn + 8

(vi) 5𝓍 – 10 – ( – 𝓍2 + 10𝓍 – 5) = 5𝓍 – 10 + 𝓍2 – 10𝓍 + 5

= 𝓍2 + 5𝓍 – 10𝓍 – 10 + 5 = 𝓍2 – 5𝓍 – 5

(vii) 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2) = 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2

= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2

= 10ab – 7a2 – 7b2

= – 7a2 – 7b2 + 10ab

(viii) 5p2 + 3q2 – pq – (4pq – 5q2 – 3p2) = 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2

= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq

= 8p2 + 8q2 – 5pq

Mddle block 1

Question : 4.

(a) What should be added to 𝓍2 + 𝓍y + y2 to obtain 2𝓍2 + 3𝓍y?

(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16 ?

Answer :
(a) Let pp should be added. Then according to Question:

, 𝓍2 + 𝓍y + y2 + p = 2𝓍2 + 3𝓍y

⇒p = 2𝓍2 + 3𝓍y – (𝓍2 + 𝓍y + y2)

⇒p = 2𝓍2 + 3𝓍y – 𝓍2 – 𝓍y – y2

⇒p = 2𝓍2 – 𝓍2 – y2 + 3𝓍y – 𝓍y

⇒p = 𝓍2 – y2 + 2𝓍y

Hence, 𝓍2 – y2 + 2𝓍y should be added.

(b) Let q should be subtracted.

Then according to Question ,

2a + 8b + 10 – q = – 3a + 7b + 16

⇒ – q = – 3a + 7b + 16 – (2a + 8b + 10)

⇒ – q = – 3a + 7b + 16 – 2a – 8b – 10

⇒ – q = – 3a – 2a + 7b – 8b + 16 – 10

⇒ – q = – 5a – b + 6

⇒ q = – ( – 5a – b + 6)

⇒ q = 5a + b – 6

Question : 5.What should be taken away from 3𝓍2 – 4y2 + 5𝓍y + 203𝓍2 – 4y2 + 5𝓍y + 20 to obtain – 𝓍2 – y2 + 6𝓍y + 20 – 𝓍2 – y2 + 6𝓍y + 20 ?

Answer :
Let qq should be subtracted.

Then according to Question ,

3𝓍2 – 4y2 + 5𝓍y + 20 – q = – 𝓍2 – y2 + 6𝓍y + 20

⇒q = 3𝓍2 – 4y2 + 5𝓍y + 20 – ( – 𝓍2 – y2 + 6𝓍y + 20)q

= 3𝓍2 – 4y2 + 5𝓍y + 20 – ( – 𝓍2 – y2 + 6𝓍y + 20)

⇒q = 3𝓍2 – 4y2 + 5𝓍y + 20 + 𝓍2 + y2 – 6𝓍y – 20q

= 3𝓍2 – 4y2 + 5𝓍y + 20 + 𝓍2 + y2 – 6𝓍y – 20

⇒q = 3𝓍2 + 𝓍2 – 4y2 + y2 + 5𝓍y – 6𝓍y + 20 – 20q

= 3𝓍2 + 𝓍2 – 4y2 + y2 + 5𝓍y – 6𝓍y + 20 – 20

⇒q = 4𝓍2 – 3y2 – 𝓍y + 0q = 4𝓍2 – 3y2 – 𝓍y + 0

Hence, 4𝓍2 – 3y2 – 𝓍y4𝓍2 – 3y2 – 𝓍y should be subtracted.

Question : 6.(a) From the sum of 3x – y + 113x – y + 11 and – y – 11, – y – 11, subtract the sum of 3𝓍2 – 5𝓍3𝓍2 – 5𝓍and – 𝓍2 + 2𝓍 + 5. – 𝓍2 + 2𝓍 + 5.

Answer :


(a) According to Question ,

(3𝓍 – y + 11) + ( – y – 11) – (3𝓍 – y – 11)

= 3𝓍 – y + 11 – y – 11 – 3𝓍 + y + 11

= 3𝓍 – 3𝓍 – y – y + y + 11 – 11 + 11

= (3 – 3)𝓍 – (1 + 1 – 1)y + 11 + 11 – 11

= 0𝓍 – y + 110 = – y + 11

(b) According to Question,

[(4 + 3𝓍) + (5 – 4𝓍 + 2𝓍2)] – [(3𝓍2 – 5𝓍) + ( – 𝓍2 + 2𝓍 + 5)]

= [4 + 3𝓍 + 5 – 4𝓍 + 2𝓍2] – [3𝓍2 – 5𝓍 – 𝓍2 + 2𝓍 + 5]

= [2𝓍2 + 3𝓍 – 4𝓍 + 5 + 4] – [3𝓍2 – 𝓍2 + 2𝓍 – 5𝓍 + 5]

= [2𝓍2 – 𝓍 + 9] – [2𝓍2 – 3𝓍 + 5]

= 2𝓍2 – 𝓍 + 9 – 2𝓍2 + 3𝓍 – 5

= 2𝓍2 – 2𝓍2 – 𝓍 + 3𝓍 + 9 – 5

= 2𝓍 + 4

Bottom Block 3
Share with your friends

Leave a Reply