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

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Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms :
i) 21b – 32 + 7b – 20b
(re-arranging the terms)
= 21b + 7b – 20b – 32
= 28b – 20b – 32
= 8b – 32

ii) z² + 13z² – 5z + 7Z² – 15z
(re-arranging the terms)
= -z² + 13z² – 5z – 15z + 7z³
= 12z² – 20z + 7z³

iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
(re-arranging the terms)

= p – q

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

= a + ab

v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(re-arranging the terms)
= 5x²y + 3yx² – 5x² + x² – 3y² – y² – 3y² + y²
= 8x²y – 4x² – 7y² + 8xy²

vi) (3y² + 5y – 4) – (8y – y² – 4) .
3y² + 5y – 4 – 8y – y² – 4
(re-arranging the terms)
3y² + y² + 5y – 8y – 4 + 4
= 14y² – 3y

Question 2.
Add:
i) 3mn, -5mn, 8mn, -4mn
3mn + 8nm – 5mn – 4mn
(re-arranging the terms)
= 11mn – 9mn = 2mn

ii) t – 8tz, 3tz – z, z – t

(re-arranging the terms)
= -8tz + 3tz = -5tz

iii) -7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
-7mn + 12 mn + 9mn – 2mn + 5 + 2 – 8 – 3
(re-arranging the terms)
= -7mn – 2mn + 12mn + 9mn + 7 – 11
= -9mn + 21 mn -4
= 12mn – 4

iv) a + b -3, b – a +3, a – b + 3

(re-arranging the terms)
= (2a – a + 2b – b + 6 – 3)
= a + b + 3

v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(re-arranging the terms)

= 17x + 51

vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(re-arranging the terms)
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= 7m – 4m – 7n + 3n – 3mn – 3
= 13m – 4n – 3mn – 3

vii) 4x²y, – 3xy², 5y², 5x²y
(Re-arranging the terms)
4x²y + 5x²y – 3xy² – 5xy²
= 9x²y – 8xy²

viii) 3p²q² – 4pq + 5, 10p²q², 15 + 9pq + 7p²q²
3p²q² – 10p²q² + 7p²q² – 4pq + 9pq + 5+ 15
(Re-arranging the terms)
= 3p²q² + 7p²q² – 10p²q² – 4pq + 9pq + 5 + 15

= 5pq + 20

ix) ab – 4a, 4b – ab, 4a – 4b

(Re-arranging the terms)
(All terms are cancelling)
= 0

x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
(Re arranging the terms)

= -x² – y² – 1

Question 3.
Subtract :
i) -5y² from y²
= y² – (-5y²)
= y² + 5y² = 6y²

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

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

= 2b

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

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

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

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

viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
= (5p² + 3q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq² – 4pq + 5q² + 3p²
= 5p² + 3p² + 3q² + 5q² – 4pq – pq
= 8p² + 8q² – 5pq

Question 4.
a) What should be added to x² + xy + y² to obtain 2x² + 3xy ?
Solution:
= 2x² + 3xy – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y² – 2x² – x² + 3xy – xy – y²
(re-arranging the terms)
∴ x² + 2xy – y² should be added to x² + xy + y² to get 2x² + 3xy

b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16 ?
Solution:
= (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b+ 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
(re-arranging the terms)
= 5a + b – 6
∴ (5a + b – 6) should be subtracted from 2a + 8b +10 to get -3a + 7b + 16.

Question 5.
What should be taken away from – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:
= (3x² – 4y² + 5xy + 20) – (-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20
(re – arranging the terms)
= 4x² – 3y – xy
∴ 4x² + 3y² – xy should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20

Question 6.
From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
Solution:

b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:

or
[(4 + 3x) + (5 – 4x + 2x²)] – [(3x² – 5x) + (x² + 2x + 5)]
= 4 + 3x + 5 – 4x + 2x² – 3x² + 5x + x² – 2x – 5 = 4 + 3x – 4x + 2x² + 5 – 3x² + x² + 5x – 2x – 5
(re-arranging the terms)

= 2x + 4