NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

• Class 7 Maths Algebraic Expressions Exercise 12.1
• Class 7 Maths Algebraic Expressions Exercise 12.2
• Class 7 Maths Algebraic Expressions Exercise 12.3
• Class 7 Maths Algebraic Expressions Exercise 12.4

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.2
Ex 12.2 Class 7 Maths Question 1.
Simplify combining like terms:
(i) 21b -32 + 7b- 206
(ii) -z ² + 13z2 -5z + 7z ³ – 15 ²
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a
(v) 5x ² y – 5x ² + 3yx ² – 3y ² + x ² – y ² + 8xy ² -3y ²
(vi) (3y ² + 5y – 4) – (8y – y ² – 4)
Solution:
(i) 21b – 32 + 7b – 206
Re-arranging the like terms, we get
216 + 7b – 206 – 32
= (21 + 7 – 20)b – 32
= 8b – 32 which is required.

(ii) -z ² + 13z ² – 5z – 15z
Re-arranging the like terms, we get
7z ³ – z ² + 13z ² – 5z + 5z – 15z
= 7z ³ + (-1 + 13)z ² + (-5 – 15)z
= 7z ³ + 12z ² – 20z which is required.

(iii) p – (p – q) – q – (q – p)
=p – p + q – q – q + p
Re-arranging the like terms, we get

= p – q which is required.

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
Re-arranging the like terms, we get
= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a + ab which is required.

(v) 5x ² y – 5x ² + 3yx ² – 3y ² + x ² – y ² + 8xy ² – 3y ²
Re-arranging the like terms, we get
5x ² y + 3x ² y + 8xy ² – 5x ² + x ² – 3y ² – y ² – 3y ²
= 8x ² y + 8xy ² – 4x ² – 7y ² which is required.

(vi) (3y ² + 5y – 4) – (8y – y ² – 4)
= 3y ² + 5y – 4 – 8y + y ² + 4 (Solving the brackets)
Re-arranging the like terms, we get
= 3y ² + y ² + 5y – 8y – 4 + 4
= 4y ² – 3y which is required.

Ex 12.2 Class 7 Maths Question 2.
Add:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz, -z, z – t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x ² y, -3xy ² , -5xy ² , 5x ² y
(viii) 3p ² q ² – 4pq + 5, -10p ² q ² , 15 + 9pq + 7p ² q ²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x ² – y ² – 1, y ² – 1 – x ² , 1 – x ² – y ²
Solution:
(i) 3mn, -5mn, 8mn, -4mn
= (3 mn) + (-5 mn) + (8 mn) + (- 4 mn)
= (3 – 5 + 8 – 4)mn
= 2mn which is required.

(ii) t – 8tz, 3tz – z, z – t
t – 8tz + 3tz – z + z – t
Re-arranging the like terms, we get
t – t – 8tz + 3tz – z + z
⇒ 0 – 5 tz + 0
⇒ -5tz which is required.

(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
= -7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn) – 3
Re-arranging the like terms, we get
-7mn + 12 mn + 9mn – 2 mn + 5 + 2 – 8 – 3

= 12 mn – 4 which is required.

(iv) a + b – 3, b – a + 3, a – b + 3
⇒ a + b – 3 + b – a + 3 + a – b + 3
Re-arranging the like terms, we get
a – a + a + b + b – b – 3 + 3 + 3
⇒ a + b + 3 which is required.

(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
∴ 14x + 14y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Re-arranging the like terms, we get
-12xy + 8xy + 4xy + 14x – 7x + 10y – 10y – 13 + 18

= 0 + 7x + 0 + 5
= 7x + 5 which is required

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5 5m -In + 3n — 4m + 2 + 2m – 3mn – 5
Re-arranging the like terms, we get
5m – 4m + 2m – 7n + 3n- 3mn + 2 – 5
= 3m – 4n – 3mn – 3 which is required.

(vii) 4x ² y, -3xy ² , -5xy ² , 5x ² y
Re-arranging the like terms and adding, we get
4x ² y – 5xy ² – 3xy ² + 5x ² y
=9x ² y — 8xy ² which is required.

(viii) 3p ² q ² – 4pq + 5, -10p ² q ² , 15 + 9pq + 7p ² q ²
= (3p ² q ² – 4pq + 5) + (-10p ² q ² ) + (15 + 9pq + 7p ² q ² )
= 3p ² q ² – 4pq + 5 – 10p ² q ² + (15 + 9pq + 7p ² q ² )
= 3p ² q ² + 7p ² q ² – 10p ² q ² – 4pq + 9pq + 5 + 15
= 10p ² q ² – 10p ² q ² + 5pq + 20
= 0 + 5pq + 20
= 5pq + 20 which is required.

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

= 0 + 0 + 0 = 0 which is required.

(x) x ² – y ² – 1, y ² – 1 – x ² , 1 – x ² – y ²
= x ² – y ² – 1 + y ² – 1 – x ² + 1 – x ² – y ²

= -x ² – y ² – 1
= -(x ² + y ² + 1) which is required.

Ex 12.2 Class 7 Maths Question 3.
Subtract:
(i) -5y ² from y ²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b(5 – a)
(v) -m ² + 5mn from 4m ² – 3mn + 8
(vi) -x ² + 10x – 5 from 5x – 10
(vii) 5a ² – 7ab + 5b ² from 3ab – 2a ² – 2b ²
(viii) 4pq – 5q ² – 3p ² from 5p ² + 3q ² – pq
Solution:
(i) -5y ² from y ² = y ² – (-5y ² )
= y ² + 5y ² = 6y ²

(ii) 6ry from -12ry = -12xy – 6xy = -18xy which is required.
(iii) (a – b) from (a + b)
= (a + b) – (a – b)
= a + b – a + b = 2b which is required

(iv) a(b – 5) from b(5 – a)
= b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a – 2ab + 5b
= 5a + 5b – 2ab which is required.

(v) -m ² + 5mn from 4m ² – 3mn + 8
= (4m ² – 3mn + 8) – (-m ² + 5mn)
= 4m ² – 3mn + 8 + m ² – 5mn
= 4m ² + m ² – 3mn – 5mn + 8
= 5m ² – 8mn + 8 which is required.

(vi) -x ² + 10x – 5 from 5x – 10
= (5x – 10) – (-x ² + 10x – 5)
= 5x – 10 + x ² – 10x + 5
= x ² + 5x – 10x – 10 + 5
= x ² – 5x – 5 which is required.

(vii) 5a ² – 7ab + 5b ² from 3ab – 2a ² – 2b ²
= (3ab – 2a ² – 2b ² ) – (5a ² – 7ab + 5b ² )
= 3ab – 2a ² – 2b ² – 5a ² + 7ab – 5b ²
= 3ab + 7ab – 2a ² – 5a ² – 2b ² – 5b ²
= 10ab – 7a ² – 7b ²
which is required.

(viii) 4pq – 5q ² – 3p ² from 5p ² + 3q ² – pq
= (5p ² + 3q ² – pq) – (4pq – 5q ² – 3p ² )
= 5p ² + 3q ² – pq – 4pq + 5q ² + 3p ²
= 5p ² + 3p ² + 3q ² + 5q ² – pq – 4pq
= 8p ² + 8q ² – 5pq
which is required.

Ex 12.2 Class 7 Maths Question 4.
(a) What should be added to x ² + xy + y ² to obtain 2x ² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16?
Solution:
(a) (2x ² + 3xy) – (x ² + xy + y ² )
= 2x ² + 3xy – x ² – xy – y ²
= 2x ² – x ² + 3xy – xy – y ²
= x ² + 2xy – y ² is required expression.

(b) (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
= 5a + b – 6 is required expression.

Ex 12.2 Class 7 Maths Question 5.
What should be taken away from 3x ² – 4y ² + 5xy + 20 to obtain -x ² – y ² + 6xy + 20?
Solution:
Let A be taken away.
∴ (3x ² – 4y ² + 5 xy + 20)-A
= -x ² – y ² + 6xy + 20
⇒ A = (3x ² – 4y ² + 5xy + 20) – (-x ² 2 – y ² + 6xy + 20)
= 3x ² – 4y ² + 5xy + 20 + x ² + y ² – 6xy – 20
= 3x ² + x ² – 4y ² + y ² + 5xy – 6xy + 20 – 20
= 4x ² – 3y ² – xy is required expression.

Ex 12.2 Class 7 Maths Question 6.
(a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x ² , subtract the sum of 3x ² – 5x and -x ² + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and -y – 11
= (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
∴ 3x – 2y – (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11 is required solution.

(b) Sum of (4 + 3x) and (5 – 4x + 2x ² )
= 4 + 3x + 5 – 4x + 2x ²
= 2x ² – 4x + 3x + 9 = 2x ² – x + 9
Sum of (3x ² – 5x) and (-x ² + 2x + 5)
= (3x ² – 5x) + (-x ² + 2x + 5)
= 3x ² – 5x – x ² + 2x + 5 = 2x ² – 3x + 5
Now (2x ² – x + 9) – (2x ² – 3x + 5)
= 2x ² – x + 9 – 2x ² + 3x – 5
= 2x ² – 2x ² + 3x – x + 4
= 2x + 4 is required expression.

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