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

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

Question 1.
Simplify by 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
= 21b + 7b – 20b – 32
= (21 + 7 – 20)b – 32
= 8b – 32

(ii) -z2 + 13z2 – 5z + 7z3 – 15z
= 7z3 + (-1 + 13) z2 + (-5 -15) z
= 7z3 + 12z2 – 20z

(iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3 ab + b – a
= (3 – 1 – 1)a + (-2 + 1 + 1)b + (-1 – 1 + 3)ab
= a + ab

(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
= (5 + 3) x2y + (-5 + 1) x2 + (- 3 – 1 – 3 )y2 + 8xy2
= 8x2y – 4x2 – 7y2 + 8xy2

(vi) (3y2 + 5y – 4) – (8y – y2 – 4)
= 3y2 + 5y – 4 – 8y + y2 + 4
= (3 + 1) y2 + (5 – 8) y + 4 – 4
= 4y2 – 3y

Question 2.
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz – z, z – 1
(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) 3 p2q2 – 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) 3 mn + (-5 mn) + 8 mn + (-4 mn)
= (3 – 5 + 8 – 4 )mn = 2 mn

(ii) (t – 8tz) + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= (1 – 1)t + (- 8 + 3)tz + (-1 + 1)z
= -5 tz

(iii) (-7mn + 5) + (12mn + 2) + (9mn – 8) + (-2mn – 3)
=-7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3
= (-7 + 12 + 9 – 2)mn + (5 + 2 – 8 – 3)
= 12mn – 4

(iv) (a + b – 3) + (b – a + 3) + (a – b + 3)
= a + b – 3 + b – a + 3 + a – b + 3
= (1 -1 + 1)a + (1 + 1 – 1)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
= (14 – 7)x + (10 – 10)y + (12 + 8 + 4 )xy + (-13 + 18)
= 7x + 5

(vi) (5m – 7n) + (3n – 4m + 2) + (2m – 3mn – 5)
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= (5 – 4 + 2)m + (- 7 + 3)n – 3mn + (2 – 5)
= 3m – 4n – 3mn – 3

(vii) (4x2y) + (-3xy2) + (-5xy2) + (5x2y)
= 4x2y – 3xy2 – 5xy2 + 5x2y
= (4 + 5) x2y + (-3 – 5) xy2
= 9x2y – 8xy2

(viii) (3p2q2 – 4pq + 5) + (-10p2q2) + (15 + 9pq +7p2q2)
= 3p2q2 – 4pq + 5 – 10p2q2 + 15 + 9pq + 7p2q2
= (3 – 10 + 7) p2q2 + (-4 + 9)pq + (5 + 15)
= 5pq + 20

(ix) (ab – 4a) +(4 b – ab) + (4a- 4b)
= ab – 4a + 4b – ab + 4a – 4b
= (1 – 1)ab + (-4 + 4 )a + (4 – 4)b = 0

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

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 + 5 mn from 4mi2 – 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) y2 – (-5y2) = y2 + 5y2 = 6y2
(ii) -12xy – (6xy) = -12xy – 6xy = -18xy
(iii) (a + b) – (a – b) = a + b – a + b = 2b

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

(v) (4m2 – 3mn + 8) – (- m2 + 5mn)
= 4m2 – 3mn + 8 + m2 – 5mn
= (4 + 1)m2 + (- 3 – 5 )mn + 8
= 5m2 – 8mn + 8

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

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

(viii) (5p2 + 3q2 – pq) – (4pq – 5a2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2
= (5 + 3 )p2 + (3 + 5 )q2 + (-1 – 4 )pq
= 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) Let a be the required term.
∴ a + (x2 + y2 + xy) = 2x2 + 3xy
⇒ a = 2x2 + 3xy – (x2 + y2 + xy)
= 2x2 + 3xy – x2 – y2 – xy
= (2 – 1) x2 – y2 + (3 – 1)xy
= x2 – y2 + 2xy

(b) Let p be the required term.
∴ (2a + 8b + 10) -p = -3a + 7b + 16
⇒ p = 2a + 8b + 10 – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= (2 + 3)a + (8 – 7)b + (10 – 16)
= 5a + b – 6

Question 5.
What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20 ?
Solution:
Required term
= (3x2 – 4y2 + 5xy + 20) – (-x2 – y2 + 6xy + 20)
= 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20
= (3 + 1)x2 + (- 4 + 1) y2 + (5 – 6)xy + (20 – 20)
= 4x2 – 3y2 – xy

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 + 2x2, subtract the sum of 3x2 – 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 + (- 1 – 1) y + (11 – 11)
= 3x – 2y
Now, required difference
= (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= (3 – 3)x + (- 2 + 1) y + 11
= -y + 11

(b) Sum of 4 + 3x and 5 – 4x + 2x2
= (4 + 3x) + (5 – 4x + 2x2)
= 4 + 3x + 5 – 4x + 2x2
= (3 – 4)x + 2x2 + 4 + 5
= – x + 2x2 + 9
Now, sum of 3x2 – 5x and -x2 + 2x + 5
= (3x2 – 5x) + (-x2 + 2x + 5)
= 3x2 – 5x – x2 + 2x + 5
= (3 – 1) x2 + (- 5 + 2)x + 5
= 2x2 – 3x + 5
Required difference
= (- x + 2x2 + 9) – (2x2 – 3x + 5)
= -x + 2x2 + 9 – 2x2 + 3x – 5
= (-1 + 3)x + (2 – 2) x2 + (9 – 5)
= 2x + 4