# HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.2 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 12 Algebraic Expressions Exercise 12.2

Question 1.
Simplify combining like terms :
(i) 21b – 32 + 7b – 20b
(ii) – z2, + 13z2 -5z + 7z3 – 15z
(iii) p-(p-q)-q- (q – p)
(iv) 3a – 2b – ab – (a -b + ab) + 3ab + b-a
(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
Solution:
(i) 21b – 32 + 7b – 20b
Rearranging terms:
= 21b + 7b – 20b – 32
= 28b – 20b – 32 = 8b – 32.

(ii) -z2 + 13z2 – 5z + 7z3 – 15z
Rearranging terms:
= 7z3 – z2 + 13z2 – 5z – 15z
= 7z3 + 12z2 – 20z

(iii) p – (p – q) – q – (q – p)
= P – P + q – q – q + p
Rearranging terms:
= 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
Rearranging terms:
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= 3a – 2a – 2b – 2ab + 3ab
= a + ab.

(v) 5x2y — 5x2 + 3yx2 — 3y2 + x2 — y2 + 8xy2 – 3y2.
Rearranging terms:
= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2 + 8xy2.
= 8x2y – 4x2 – 7y2 + 8xy2

(vi) (3y2 + 5y – 4) – (8y -y2 – 4).
= 3y2 + 5y – 4 – 8y + y2 + 4
Rearranging terms:
= 3y2 + y2 + 5y – 8y – 4 + 4
= 4y2 + 3y.

Question 2.
(i) 3mn, – 5mn, 8 mn, – 4 mn.
(ii) t – 8tz, 3tz -z, z- t.
(iii) – 7mn + 5, 12mn +2, 9 mn -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) + 8 mn + (- 4mn)
= 3mn + 8mn – 5mn – 4mn
[Rearranging Terms]
= 11mn – 9mn – 2mn.

(ii) t – 8tz + 3tz -z + z -t
= t – t – z + z – 8tz + 3tz
[Rearranging Terms]
= – 5 tz.

(iii) – 7mn + 5 + 12mn + 2 + 9mn – 8 + (- 2 mn) – 3
= 12mn + 9mn – 7mn – 2mn + 5 + 2 – 8 – 3
[Rearranging Terms]
= 12mn – 4.

(iv) a + b – 3 + b – a + 3 + a – b + 3
= a- a + a + b + b – b + 3 + 3 – 3
[Rearranging Terms]
= a + b + 3.

(v) 14x + 10y – 12rxy – 13 + 18 – 7x – 10y + 8xy + 4xy
= 14x – 7x – 12xy + 8xy + 4xy + 10y – 10y + 18 – 13
[Rearranging Terms]
= 7x + 5.

(vi) 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= 5m- 4m + 2m -7n + 3n – 3mn -5 + 2
[Rearranging Terms]
= 3 m – 4n – 3 mn – 3.

(vii) 4x2y + (- 3xy2) + (- 5xy2) + 5x2y
= 4x2y + 5xy – 5xy2 + 3xy2
[Rearranging Terms]
= 9x2y – 8xy2.

(viii) 3p2q2 – 4pq + 5 + (- 10 p2q2) + 15 + 9pq + 7p2q2
= 3p2q2 -10p2q2 + 7p2q2 + 9pq – 4pq + 15 + 5 [Rearranging Terms]
= 5pq + 20.

(ix) ab – 4a + 46 – ab + 4a – 46
= ab – ab – 4a + 4a + 46 – 46
[Rearranging Terms]
= 0.

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

Question 3.
Subtract:
(i) – 5y2 from y2.
(ii) 6xy from – 12xy
(iii) (a – b) from (a + b)
(iii) 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 = – 4y2.

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

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

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

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

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

(vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2
= 3ab – 2a2 – 2b2 – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2
= (- 2a2 – 5a2) + (-2b2 – 5b2) + (3b + 7ab)
= -7a2 – 7b2 + 10ab

(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.

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 ?
Solution:
(a) (2x2 + 3xy) – (x2 + xy + y2)
= 2x2 + 3xy – x2 – xy – y2
= 2x2 – x2 + 3xy – xy – y2
= x2 + 2xy – y2.

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

Question 5.
(a) What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain – x2 – y2 + 6xy + 20?
Solution:
(a) (3x2 – 4y2 + 5xy + 20) – (- x2 – y2 + 6xy + 20)
= 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20
= 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20
= 4x2 – 3y2 – xy.

Question 6.
(a) From the sum of 3x – y + 11 and-y – 11, subtract 3x – y – 11.
(6) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2– 5x and -x2 + 2x + 5.
Solution:
(a) (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
Then, subtract 3x -y – 11 from obtained expression:
(3ac – 2y) – (3x-y- 11)
= 3x — 2y — 3x+y + 11
= 3x – 3x – 2y + y + 11
= -y + 11

(b) (4 + 3x) + (5 – 4x + 2x2 )
= 4 + 3x + 5 – 4x + 2x2
= 3x – 4x + 4 + 5 + 2x2
= – x + 9 + 2x2
Then, (3x2 – 5x) + (- x2 + 2x + 5)
= 3x2 – 5x – x2 + 2x + 5
= 3x2 – x2 – 5x + 2x + 5
= 2x2 – 3x + 5
Then according to question :
= (-x + 9 + 2x2) – (2x2 – 3x + 5)
= – x + 9 + 2x2 – 2x2 + 3x – 5
= – x + 3x + 2x2 – 2x2 + 9 – 5
= + 2x + 4 = 2x + 4.