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.
Add:
(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.