# RS Aggarwal Class 8 Math Sixth Chapter Operation on Algebric Expressions Exercise 6A Solution

## EXERCISE 6A

### Add:

**(1) 8ab, – 5ab, 3ab, – ab**

= 8ab + (- 5ab) + 3ab + (- ab)

= 8ab – 5ab + 3ab – ab

= 11ab – 6ab = 5ab

**(2) 7x, – 3x, 5x, – x, – 2x**

= 7x + (- 3x) + 5x + (- x) + (- 2x)

= 7x – 3x + 5x – x – 2x

= 12x – 6x = 6x

**(3) 3a – 4b + 4c, 2a + 3b – 8c, a – 6b + c**

= 3a – 4b + 4c + 2a + 3b – 8c + a – 6b + c

= 6a – 7b – 3c

**(4) 5x – 8y + 2z, 3z – 4y – 2x, 6y – z – x and 3x – 2z – 3y **

= 5x – 8y + 2z + 3z – 4y – 2x + 6y – z – x + 3x – 2z – 3y

= 5x – 9y + 2z

**(5) 6ax – 2by + 3cz, 6by – 11ax – cz and 10cz – 2ax – 3by**

= 6ax – 2by + 3cz + 6by – 11ax – cz + 10cz – 2ax – 3by

= – 7ax + by + 12cz

**(6) 2x ^{3} – 9x^{2} + 8, 3x^{2} – 6x – 5, 7x^{3} – 10x + 1 and 3 + 2x – 5x^{2} – 4x^{3}**

= 2x^{3 }– 9x^{2} + 8 + 3x^{2} – 6x – 5 + 7x^{3} – 10x + 1 + 3 + 2x – 5x^{2} – 4x^{3}

= 5x^{3} – 11x^{2} – 14x + 12

**(7) 6p + 4q – r + 3, 2r – 5p – 6, 11q – 7p + 2r – 1 and 2q – 3r + 4**

= 6p + 4q – r + 3 + 2r – 5p – 6 + 11q – 7p + 2r – 1 + 2q – 3r + 4

= – 6p + 17q

**(8) 4x ^{2} – 7xy + 4y^{2} – 3, 5 + 6y^{2} – 8xy + x^{2} and 6 – 2xy + 2x^{2} – 5y^{2.}**

= 4x^{2} – 7xy + 4y^{2} – 3 + 5 + 6y^{2} – 8xy + x^{2} + 6 – 2xy + 2x^{2 }– 5y^{2}

= 7x^{2} – 17xy + 5y^{2} + 8

### Subtract:

**(9) 3a ^{2}b from – 5a^{2}b**

= – 5a^{2}b – 3a^{2}b

= – 8a^{2}b

**(10) – 8pq from 6pq**

= 6pq – (- 8pq)

= 6pq + 8pq

= 14pq

**(11) – 2abc from – 8abc**

= – 8abc – (- 2abc)

= – 8abc + 2abc

= – 6abc

**(12) – 16p from – 11p**

= -11p – (- 16p)

= – 11p + 16p

= 5p

**(13) 2a – 5b + 2c – 9 from 3a – 4b – c + 6**

= 3a – 4b – c + 6 – (2a – 5b + 2c – 9)

= 3a – 4b – c + 6 – 2a + 5b – 2c + 9

= a + b – 3c + 15

**(14) – 6p + q + 3r + 8 from p – 2q – 5r – 8**

= p – 2q – 5r – 8 – (- 6p + q + 3r + 8)

= p – 2q – 5r – 8 + 6p – q – 3r – 8

= 7p – 3q – 8r – 16

**(15) x ^{3} + 3x^{2} – 5x + 4 from 3x^{3} – x^{2} + 2x – 4**

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

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

= 2x^{3} – 4x^{2} + 7x – 8

**(16) 5y ^{4} – 3y^{3} + 2y^{2} + y – 1 from 4y^{4} – 2y^{3} – 6y^{2} – y + 5**

= 4y^{4} – 2y^{3} – 6y^{2} – y + 5 – (5y^{4} – 3y^{3} + 2y^{2} + y – 1)

= 4y^{4} – 2y^{3} – 6y^{2} – y + 5 – 5y^{4} + 3y^{3} – 2y^{2} – y + 1

= – y^{4} + y^{3} – 8y^{2} – 2y + 6

**(17) 4p ^{2} + 5q^{2} – 6r^{2} + 7 from 3p^{2} – 4q^{2} – 5r^{2} – 6**

= 3p^{2} – 4q^{2} – 5r^{2} – 6 – (4p^{2} + 5q^{2} – 6r^{2} + 7)

= 3p^{2} – 4q^{2} – 5r^{2} – 6 – 4p^{2} – 5q^{2} + 6r^{2} – 7

= – p^{2} – 9q^{2} + r^{2 }– 13

**(18) What must be subtracted from 3a ^{2} – 6ab – 3b^{2} – 1 to get 4a^{2} – 7ab – 4b^{2} + 1?**

Solution: 3a^{2} – 6ab – 3b^{2} – 1 – 4a^{2} + 7ab + 4b^{2} – 1

= – a^{2} + ab + b^{2 }– 2

**(19) The two adjacent sides of a rectangle are 5x ^{2} – 3y^{2} and x^{2} + 2xy. Find the perimeter.**

Solution: Perimeter of the rectangle = 2 [(5x^{2} – 3y^{2}) + (x^{2} + 2xy)]

= 2 (5x^{2} – 3y^{2} + x^{2} + 2xy)

= 2 (6x^{2} – 3y^{2} + 2xy)

= 12x^{2} – 6y^{2} + 4xy

**(20) The perimeter of a triangle is 6p ^{2} – 4p + 9 and two of its sides are p^{2} – 2p + 1 and 3p^{2} – 5p + 3. Find the third side of a triangle.**

Solution: 6p^{2} – 4p + 9 – [(p^{2} – 2p + 1) + (3p^{2} – 5p + 3)]

= 6p^{2} – 4p + 9 – (p^{2} – 2p + 1 + 3p^{2} – 5p + 3)

= 6p^{2} – 4p + 9 – (4p^{2} – 7p + 4)

= 6p^{2} – 4p + 9 – 4p^{2} + 7p – 4

= 2p^{2} + 3p + 5

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