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

**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) -z2 + 13z2 -5z + 7z3 – 152

(iii) p – (p – q) – q – (q – p)

(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a

(v) 5x2y – 5×2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 -3y2

(vi) (3y2 + 5y – 4) – (8y – y2 – 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) -z2 + 13z2 – 5z – 15z

Re-arranging the like terms, we get

7z3 – z2 + 13z2 – 5z + 5z – 15z

= 7z3 + (-1 + 13)z2 + (-5 – 15)z

= 7z3 + 12z2 – 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) 5x2y – 5×2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2

Re-arranging the like terms, we get

5x2y + 3x2y + 8xy2 – 5×2 + x2 – 3y2 – y2 – 3y2

= 8x2y + 8xy2 – 4×2 – 7y2 which is required.

(vi) (3y2 + 5y – 4) – (8y – y2 – 4)

= 3y2 + 5y – 4 – 8y + y2 + 4 (Solving the brackets)

Re-arranging the like terms, we get

= 3y2 + y2 + 5y – 8y – 4 + 4

= 4y2 – 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) 4x2y, -3xy2, -5xy2, 5x2y

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

(ix) ab – 4a, 4b – ab, 4a – 4b

(x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2

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) 4x2y, -3xy2, -5xy2, 5x2y

Re-arranging the like terms and adding, we get

4x2y – 5xy2 – 3xy2 + 5x2y

=9x2y — 8xy2 which is required.

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

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

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

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

= 10p2q2 – 10p2q2 + 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) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2

= x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2

= -x2 – y2 – 1

= -(x2 + y2 + 1) which is required.

Ex 12.2 Class 7 Maths Question 3.

Subtract:

(i) -5y2 from y2

(ii) 6xy from -12xy

(iii) (a – b) from (a + b)

(iv) a(b – 5) from b(5 – a)

(v) -m2 + 5mn from 4m2 – 3mn + 8

(vi) -x2 + 10x – 5 from 5x – 10

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

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

Solution:

(i) -5y2 from y2 = y2 – (-5y2)

= y2 + 5y2 = 6y2

(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) -m2 + 5mn from 4m2 – 3mn + 8

= (4m2 – 3mn + 8) – (-m2 + 5mn)

= 4m2 – 3mn + 8 + m2 – 5mn

= 4m2 + m2 – 3mn – 5mn + 8

= 5m2 – 8mn + 8 which is required.

(vi) -x2 + 10x – 5 from 5x – 10

= (5x – 10) – (-x2 + 10x – 5)

= 5x – 10 + x2 – 10x + 5

= x2 + 5x – 10x – 10 + 5

= x2 – 5x – 5 which is required.

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

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

= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2

= 3ab + 7ab – 2a2 – 5a2 – 2b2 – 5b2

= 10ab – 7a2 – 7b2

which is required.

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

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

= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2

= 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq

= 8p2 + 8q2 – 5pq

which is required.

Ex 12.2 Class 7 Maths Question 4.

(a) What should be added to x2 + xy + y2to obtain 2×2 + 3xy?

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

Solution:

(a) (2×2 + 3xy) – (x2 + xy + y2)

= 2×2 + 3xy – x2 – xy – y2

= 2×2 – x2 + 3xy – xy – y2

= x2 + 2xy – y2 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 3×2 – 4y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20?

Solution:

Let A be taken away.

∴ (3×2 – 4y2 + 5 xy + 20)-A

= -x2 – y2 + 6xy + 20

⇒ A = (3×2 – 4y2 + 5xy + 20) – (-x22 – y2 + 6xy + 20)

= 3×2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20

= 3×2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20

= 4×2 – 3y2 – 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 + 2×2, subtract the sum of 3×2 – 5x and -x2 + 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 + 2×2)

= 4 + 3x + 5 – 4x + 2×2

= 2×2 – 4x + 3x + 9 = 2×2 – x + 9

Sum of (3×2 – 5x) and (-x2 + 2x + 5)

= (3×2 – 5x) + (-x2 + 2x + 5)

= 3×2 – 5x – x2 + 2x + 5 = 2×2 – 3x + 5

Now (2×2 – x + 9) – (2×2 – 3x + 5)

= 2×2 – x + 9 – 2×2 + 3x – 5

= 2×2 – 2×2 + 3x – x + 4

= 2x + 4 is required expression.

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