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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) z^{2} + 13z^{2} – 5z + 7Z^{2} – 15z

(re-arranging the terms)

= -z^{2} + 13z^{2} – 5z – 15z + 7z^{3}

= 12z^{2} – 20z + 7z^{3}

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) 5x^{2}y – 5x^{2} + 3yx^{2} – 3y^{2} + x^{2} – y^{2} + 8xy^{2} – 3y^{2}

(re-arranging the terms)

= 5x^{2}y + 3yx^{2} – 5x^{2} + x^{2} – 3y^{2} – y^{2} – 3y^{2} + y^{2}

= 8x^{2}y – 4x^{2} – 7y^{2} + 8xy^{2}

vi) (3y^{2} + 5y – 4) – (8y – y^{2} – 4) .

3y^{2} + 5y – 4 – 8y – y^{2} – 4

(re-arranging the terms)

3y^{2} + y^{2} + 5y – 8y – 4 + 4

= 14y^{2} – 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) 4x^{2}y, – 3xy^{2}, 5y^{2}, 5x^{2}y

(Re-arranging the terms)

4x^{2}y + 5x^{2}y – 3xy^{2} – 5xy^{2}

= 9x^{2}y – 8xy^{2}

viii) 3p^{2}q^{2} – 4pq + 5, 10p^{2}q^{2}, 15 + 9pq + 7p^{2}q^{2}

3p^{2}q^{2} – 10p^{2}q^{2} + 7p^{2}q^{2} – 4pq + 9pq + 5+ 15

(Re-arranging the terms)

= 3p^{2}q^{2} + 7p^{2}q^{2} – 10p^{2}q^{2} – 4pq + 9pq + 5 + 15

= 5pq + 20

ix) ab – 4a, 4b – ab, 4a – 4b

(Re-arranging the terms)

(All terms are cancelling)

= 0

x) x^{2} – y^{2} – 1, y^{2} – 1 – x^{2}, 1 – x^{2} – y^{2}

(Re arranging the terms)

= -x^{2} – y^{2} – 1

Question 3.

Subtract :

i) -5y^{2} from y^{2}

= y^{2} – (-5y^{2})

= y^{2} + 5y^{2} = 6y^{2}

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) -m^{2} + 5mn from 4m^{2} – 3mn + 8

= (4m^{2} – 3mn + 8) – (-m^{2} + 5mn)

= 4m^{2} – 3mn + 8 + m^{2} – 5mn

= 4m^{2} + m^{2} – 3mn – 5mn + 8

= 5m^{2} – 8mn + 8

vi) -x^{2} + 10x – 5 from 5x – 10

= (5x – 10) – (-x^{2} + 10x – 5)

= 5x – 10 + x^{2} – 10x + 5

= 5x – 10x + x^{2} – 10 + 5

= x^{2} – 5x – 5

vii) 5a^{2} – 7 ab + 5b^{2} from 3ab – 2a^{2} – 2b^{2}

= (3ab – 2a^{2} – 2b^{2}) – (5a^{2} – 7ab + 5b^{2})

= 3ab – 2a^{2} – 2b^{2} – 5a^{2} + 7ab – 5b

= -2a^{2} – 5a^{2} – 2b^{2} – 5b^{2} + 7ab + 3ab

= -7a^{2} – 7b^{2} + 10ab

viii) 4pq – 5q^{2} – 3p^{2} from 5p^{2} + 3q^{2} – pq

= (5p^{2} + 3q^{2} – pq) – (4pq – 5q^{2} – 3p^{2})

= 5p^{2} + 3q^{2} – pq^{2} – 4pq + 5q^{2} + 3p^{2}

= 5p^{2} + 3p^{2} + 3q^{2} + 5q^{2} – 4pq – pq

= 8p^{2} + 8q^{2} – 5pq

Question 4.

a) What should be added to x^{2} + xy + y^{2} to obtain 2x^{2} + 3xy ?

Solution:

= 2x^{2} + 3xy – (x^{2} + xy + y^{2})

= 2x^{2} + 3xy – x^{2} – xy – y^{2} – 2x^{2} – x^{2} + 3xy – xy – y^{2}

(re-arranging the terms)

∴ x^{2} + 2xy – y^{2} should be added to x^{2} + xy + y^{2} to get 2x^{2} + 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 – 4y^{2} + 5xy + 20 to obtain -x^{2} – y^{2} + 6xy + 20?

Solution:

= (3x^{2} – 4y^{2} + 5xy + 20) – (-x^{2} – y^{2} + 6xy + 20)

= 3x^{2} – 4y^{2} + 5xy + 20 + x^{2} + y^{2} – 6xy – 20

= 3x^{2} + x^{2} – 4y^{2} + y^{2} + 5xy – 6xy + 20 – 20

(re – arranging the terms)

= 4x^{2} – 3y – xy

∴ 4x^{2} + 3y^{2} – xy should be taken away from 3x^{2} – 4y^{2} + 5xy + 20 to obtain -x^{2} – y^{2} + 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 + 2x^{2}, subtract the sum of 3x^{2} – 5x and -x^{2} + 2x + 5.

Solution:

or

[(4 + 3x) + (5 – 4x + 2x^{2})] – [(3x^{2} – 5x) + (x^{2} + 2x + 5)]

= 4 + 3x + 5 – 4x + 2x^{2} – 3x^{2} + 5x + x^{2} – 2x – 5 = 4 + 3x – 4x + 2x^{2} + 5 – 3x^{2} + x^{2} + 5x – 2x – 5

(re-arranging the terms)

= 2x + 4