Class 6 - Mathematics
Chapter - Algebraic Expressions : Exercise 12.2
Top Block 1
Question : 1.Simplify combining like terms:
- 21b – 32 + 7b – 20b
- – z2 + 13z2 – 5z + 7z3 – 15z
- p – (p – q) – q – (q – p)
- 3a – 2b – ab – (a – b + ab) + 3ab + b – a
- 5𝓍2y – 5𝓍2 + 3y𝓍2 – 3y2 + 𝓍2 – y2 + 8𝓍y2 – 3y2
- (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:
- 3mn, – 5mn, 8mn – 4mn
- t – 8tz, 3tz – z, z – t
- – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
- a + b – 3,b – a + 3,a – b + 3
- 14𝓍 + 10y – 12𝓍y – 13, 18 – 7𝓍 – 10y + 8𝓍y, 4𝓍y
- 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
- 4𝓍2y, – 3𝓍y2, – 5𝓍y2, 5𝓍2y
- 3p2q2 – 4pq + 5, – 10p2q2, 15 + 9pq + 7p2q2
- ab – 4a, 4b – ab, 4a – 4b
- 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:
- – 5y2 fromy2
- 6𝓍y from – 12𝓍y
- (a – b) from (a + b)
- a(b – 5) from b(5 – a)
- – m2 + 5mn from 4m2 – 3mn + 8
- – 𝓍2 + 10𝓍 – 5 from 5𝓍 – 10
- 5a2 – 7ab + 5b2 from ab – 2a2 – 2b2
- 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