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

These NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths 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)
(i) 21b – 32 + 76 – 20b
Combining the like terms, we have
(21b + 7b – 20b) + (- 32) = (21 + 7 – 20) b + (- 32)
= 8b – 32

(ii) -z2 + 13z2 – 5z + 7z3 – 15z
Combining like terms, we get
= 7z3 + (-z2 + 13z2) + (-5z – 15z)
= 7z3 + (-1 + 13)z2 + (- 5 – 15)z
= 7z3 + 12z2 – 20z

(iii) p-(p-q)-q-(q-p) = p- p + q – q – q + p
Combining like terms, we get
= p – p + p + q – q – q
= (1 – 1 + 1) p + (1 – 1 – 1) q
= p + (-1)q
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b -a
Combining like terms, we get
= 3a – a – a -2b + b + b – ab – ab + 3ab
= (3 – 1 – 1) a + (-2 + 1 + 1) b + (-1 -1 + 3)ab
= (3 – 2) a + (-2 + 2) b + (-2 + 3) ab
= (1) a + (0) b + (1) ab
= a + ab

(v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
Combining the like terms, we get
(5x2y + 3yx2) + 8xy2 + (-5x2 + x2 ) + (-3y2 – y2 – 3y2 )
(5 + 3) x2y + 8xy2 + (-5 + 1) x2 + (-3 – 1 – 3)y2
8x2y + 8xy2 + (- 4) x2 + (-7) y2
8x2y + 8xy2 – 4x2 – 7y2

(vi) (3y2 + 5y – 4)-(8y – y2 – 4)
= 3y2 + 5y – 4 – 8y + y2 + 4
Combining the like terms, we get
= 3y2 + y2 + 5y – 8y – 4 + 4
= (3 + 1)y2 + (5-8)y + (-4 + 4)
= 4y2 – 3y

Question 2.
(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 + lOy – 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
(i) 3mn, – 5mn, 8mn, – 4mn
3mn + (-5mn) + 8mn + (- 4mn)
= 3mn – 5mn + 8mn – 4mn
= (3 – 5 + 8 – 4) mn = 2mn

(ii) t – 8tz, 3tz – z; z -t
t – 8tz + 3tz + (-z) + z + (-t)
= t – 8tz + 3tz – z + z – t
Combining like terms, we get
= t – t-z + z – 8tz + 3tz
= (1 – 1) t + (- 1 + 1) z + (- 8 + 3) tz
= (0) t + (0) z + (-5) tz
= 0 + 0 – 5tz
= -5tz

(iii) -7mn + 5; 12mn + 2; 9mn – 8; – 2mn -3
-7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn – 3)
– 7mn + 5 + 12mn + 2 + 9mn – 8 -2mn – 3
Combining like terms, we get
– 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3
= (-7 + 12 + 9-2) mn + (5 + 2 – 8 -3)
= (21-9) mn + (7 – 11)
= (12) mn + (- 4)
= 12mn – 4

(iv) a + b-3;b-a + 3;a-b + 3
a + b – 3 + b – a + 3 + a – b + 3
Combining like terms, we get
= (a – a + a) + (b + b – b) + (-3 + 3 + 3)
= (1 – 1 + 1) a + (1 + 1 – 1)b + (6 – 3)
= (2 – 1) a + (2 -1) b + 3
= a + b + 3

(v) 14x + 10y – 12xy – 13; 18 – 7x – lOy + 8xy, 4xy
14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Combining like terms, we get
= (14x – 7x) + (10y – 10y) + (-12xy + 8xy + 4xy)
= (14 – 7)x + (10 – 10) y + (-12 + 8 + 4) xy+ (-13 + 18)
= 7x + (0)y + (0)xy + (+5)
= 7x + 5

(vi) 5m – 7n, 3n – 4m + 2; 2m – 3mn – 5
5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
Combining like terms, we get
= (5m – 4m + 2m) + (-7n + 3n) + (-3mn) + (2 – 5)
= (5 – 4 + 2) m + (-7 + 3) n – 3mn + (-3)
= (7 – 4) m + (- 4) n – 3mn -3
= 3m – 4n – 3mn – 3

(vii) 4x2y; – 3xy2, – 5xy2; 5x2y
4x2y + (-3xy2) + (-5xy2) + 5x2y
= 4x2y – 3xy2 – 5xy2 + 5x2y
Combining like terms, we get
= 4x2y + 5x2y – 5xy2 – 5xy2
= (4 + 5) x2y + (- 3 – 5) xy2
= 9x2y + (- 8) xy2
= 9x2y – 8xy2

(viii) 3p2q2 – 4pq + 5; – 10p2q2; 15 + 9pq + 7p2q2
3p2q2 + (- 4pq) + 5 + (-10p2q2) + 15 + 9pq + 7p2q2
= 3p2q2 – 4pq + 5 – 10p2q2 + 15 + 9pq + 7p2q2
Combining like terms, we get
= 3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5 + 15
= (3 – 10 + 7) p2q2 + (- 4 + 9) pq + (5 + 15)
= (10 – 10) p2q2 + (5) pq + 20
= (0) p2q2 + 5pq + 20
= 5pq + 20

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

(x) x2 – y2 – 1; y2 – 1 – x2; 1 – x2 – y2
x2– y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2
= (x2 – x2 – x2) + (- y2 + y2 – y2) + (-1 – 1 + 1)
= (1 – 1 – 1)x2 + (-2 + 1)y2 + (-1)
= (1 – 2) x2 + (-2 + 1)y2 + (-1)
= (- 1) x2 + (-1) y2 +(-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 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
(i) Subtract – 5y2 from y2
y2 – (-5y2 ) = y2 + 5y2 – 12xy – (6xy) = 6y2

(ii) Subtract 6xy from – 12xy = – 12xy – 6xy
= (-12 – 6)xy = -18xy

(iii) Subtract (a – b) from (a + b)
(a + b)-(a-b) = a + b- a + b =a-a+b+b
= (1 – 1)a + (1 + 1)b
= (0)a + 2b
= 0 + 2b
= 2b

(iv) Subtract a (b – 5) from b (5 – a) b (5 – a) – a (b – 5)
= 5b – ab – ab + 5a
= 5b + 5a – ab – ab
= 5a + 5b + (-1 -1) ab
= 5a + 5b – 2ab

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

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

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

(viii) Subtract 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq
5p2 + 3q2 – pq – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq – 4pq + 5p2 + 3p2
= (5 + 3) p2 + (3 + 5)q2 – 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?
(a) The required expression is
2x2 + 3xy – (x2 + xy + y2)
= 2x2 + 3xy – x2 – xy – y2
= (2x2 – x2) – y2 + 3xy – xy
= (2 – 1) x2 – y2 + (3 – 1) xy
= (1) x2 – y2 + (2) xy
= x2 + 2xy – y2

(b) The required expression is
(2a + 8b + 10) – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= (2a + 3a) + (8b – 7b) + 10- 16
= (2 + 3) a + (8 – 7) b + (10 – 16)
= (5)a + (l)b + (-6)
= 5a + b – 6

Question 5.
What should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20?
The required expression is
3x2 – 4y2 + 5xy + 20 –
(-x2 – y2 + 6xy + 20)
= 3x2 – 4y2 + 5xy + 20 – (-x2 -y2 + 6xy + 20)
= 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20)
= (3x2 + x2) + (- 4y2 + y2) +
(+ 5xy – 6xy) + (20 – 20) = (3 + 1) x2 + (- 4 + 1) y2 +
(+ 5 – 6) xy + (0)
= (4) x2 + (-3) y2 + (-1xy)
= 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.
(a) The required expression is (3x-y+ 11) + (- y-11) –
(3x – y -11)
= 3x – y + 11 – y – 11 – 3x + y + 11
= (3x – 3x) + (-y – y + y) + (11 – 11 + 11)
= (3 – 3)x + (-1 -1 + 1) y + (11 – 11 + 11)
= (3 – 3)x + (-2 + 1)y + (22 – 11)
– (0) x + (-1) y + (11)
= 0 – y + 11 = -y + 11

(b) The required expression is
(4 + 3x) + (5 – 4x + 2x2) –
[(3x2 – 5x + (-x2 + 2x + 5)]
= 4 + 3x + 5 – 4x + 2x2 – [3x2 – 5x – x2 + 2x + 5]
= 4 + 3x + 5 – 4x + 2x2 – 3x2 +5x + x2 – 2x – 5
= (2x2 – 3x2 + x2) + (3x -4x + 5x- 2x) + (4 + 5 – 5)
= (2 – 3 + 1) x2 + (3 – 4 + 5 – 2) x + (9-5)
= (3 – 3) x2 + (8 – 6) x + 4
= (0)x2 + (2) x + 4
= 0 + 2x + 4
= 2x + 4

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