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

(iii) p – (p – q) – q – (q – p)

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

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

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

Solution:

(i) 21b – 32 + 7b – 20b

= 21b + 7b – 20b – 32

= (21 + 7 – 20)b – 32

= 8b – 32

(ii) -z^{2} + 13z^{2} – 5z + 7z^{3} – 15z

= 7z^{3} + (-1 + 13) z^{2} + (-5 -15) z

= 7z^{3} + 12z^{2} – 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) 5x^{2}y – 5x^{2} + 3yx^{2} – 3y^{2} + x^{2} – y^{2} + 8xy^{2} – 3y^{2}

= (5 + 3) x^{2}y + (-5 + 1) x^{2} + (- 3 – 1 – 3 )y^{2} + 8xy^{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

= (3 + 1) y^{2} + (5 – 8) y + 4 – 4

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

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

(ix) ab – 4a, 4b – ab, 4a – 4b

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

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) (4x^{2}y) + (-3xy^{2}) + (-5xy^{2}) + (5x^{2}y)

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

= (4 + 5) x^{2}y + (-3 – 5) xy^{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} – 4pq + 5 – 10p^{2}q^{2} + 15 + 9pq + 7p^{2}q^{2}

= (3 – 10 + 7) p^{2}q^{2} + (-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) (x^{2} – y^{2} – 1) + (y^{2} – 1 – x^{2}) + (1 – x^{2} – y^{2})

= x^{2} – y^{2} – 1 + y^{2} – 1 – x^{2} + 1 – x^{2} – y^{2}

= (1 – 1 – 1)x^{2} + (-1 + 1 – 1)y^{2} + (-1 – 1 + 1)

= -x^{2} – y^{2} – 1

Question 3.

Subtract:

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

(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) -x^{2} + 10x – 5 from 5x – 10

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

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

Solution:

(i) y^{2} – (-5y^{2}) = y^{2} + 5y^{2} = 6y^{2}

(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) (4m^{2} – 3mn + 8) – (- m^{2} + 5mn)

= 4m^{2} – 3mn + 8 + m^{2} – 5mn

= (4 + 1)m^{2} + (- 3 – 5 )mn + 8

= 5m^{2} – 8mn + 8

(vi) (5x – 10) – (-x^{2} + 10x – 5)

= 5x – 10 + x^{2} – 10x + 5

= x^{2} + (5 – 10)x + (-10 + 5)

= x^{2} – 5x – 5

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

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

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

= 10ab – 7a^{2} – 7b^{2}

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

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

= (5 + 3 )p^{2} + (3 + 5 )q^{2} + (-1 – 4 )pq

= 8p^{2} + 8q^{2} – 5pq

Question 4.

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

(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16?

Solution:

(a) Let a be the required term.

∴ a + (x^{2} + y^{2} + xy) = 2x^{2} + 3xy

⇒ a = 2x^{2} + 3xy – (x^{2} + y^{2} + xy)

= 2x^{2} + 3xy – x^{2} – y^{2} – xy

= (2 – 1) x^{2} – y^{2} + (3 – 1)xy

= x^{2} – y^{2} + 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 3x^{2} – 4y^{2} + 5xy + 20 to obtain -x^{2} – y^{2} + 6xy + 20 ?

Solution:

Required term

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

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

= (3 + 1)x^{2} + (- 4 + 1) y^{2} + (5 – 6)xy + (20 – 20)

= 4x^{2} – 3y^{2} – 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 + 2x^{2}, subtract the sum of 3x^{2} – 5x and -x^{2} + 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 + 2x^{2}

= (4 + 3x) + (5 – 4x + 2x^{2})

= 4 + 3x + 5 – 4x + 2x^{2}

= (3 – 4)x + 2x^{2} + 4 + 5

= – x + 2x^{2} + 9

Now, sum of 3x^{2} – 5x and -x^{2} + 2x + 5

= (3x^{2} – 5x) + (-x^{2} + 2x + 5)

= 3x^{2} – 5x – x^{2} + 2x + 5

= (3 – 1) x^{2} + (- 5 + 2)x + 5

= 2x^{2} – 3x + 5

Required difference

= (- x + 2x^{2} + 9) – (2x^{2} – 3x + 5)

= -x + 2x^{2} + 9 – 2x^{2} + 3x – 5

= (-1 + 3)x + (2 – 2) x^{2} + (9 – 5)

= 2x + 4