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

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.
Add:
(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

MP Board Class 7th Maths Solutions

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