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

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) 2x3 – 9x2 + 8, 3x2 – 6x – 5, 7x3 – 10x + 1 and 3 + 2x – 5x2 – 4x3

= 2x3 – 9x2 + 8 + 3x2 – 6x – 5 + 7x3 – 10x + 1 + 3 + 2x – 5x2 – 4x3

= 5x3 – 11x2 – 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) 4x2 – 7xy + 4y2 – 3, 5 + 6y2 – 8xy + x2 and 6 – 2xy + 2x2 – 5y2.

= 4x2 – 7xy + 4y2 – 3 + 5 + 6y2 – 8xy + x2 + 6 – 2xy + 2x2 – 5y2

= 7x2 – 17xy + 5y2 + 8

Subtract:

(9) 3a2b from – 5a2b

= – 5a2b – 3a2b

= – 8a2b

(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) x3 + 3x2 – 5x + 4 from 3x3 – x2 + 2x – 4

= 3x3 – x2 + 2x – 4 – (x3 + 3x2 – 5x + 4)

= 3x3 – x2 + 2x – 4 – x3 – 3x2+ 5x – 4

= 2x3 – 4x2 + 7x – 8

(16) 5y4 – 3y3 + 2y2 + y – 1 from 4y4 – 2y3 – 6y2 – y + 5

= 4y4 – 2y3 – 6y2 – y + 5 – (5y4 – 3y3 + 2y2 + y – 1)

= 4y4 – 2y3 – 6y2 – y + 5 – 5y4 + 3y3 – 2y2 – y + 1

= – y4 + y3 – 8y2 – 2y + 6

(17) 4p2 + 5q2 – 6r2 + 7 from 3p2 – 4q2 – 5r2 – 6

= 3p2 – 4q2 – 5r2 – 6 – (4p2 + 5q2 – 6r2 + 7)

= 3p2 – 4q2 – 5r2 – 6 – 4p2 – 5q2 + 6r2 – 7

= – p2 – 9q2 + r2 – 13

(18) What must be subtracted from 3a2 – 6ab – 3b2 – 1 to get 4a2 – 7ab – 4b2 + 1?

Solution: 3a2 – 6ab – 3b2 – 1 – 4a2 + 7ab + 4b2 – 1

= – a2 + ab + b2 – 2

(19) The two adjacent sides of a rectangle are 5x2 – 3y2 and x2 + 2xy. Find the perimeter.

Solution: Perimeter of the rectangle = 2 [(5x2 – 3y2) + (x2 + 2xy)]

= 2 (5x2 – 3y2 + x2 + 2xy)

= 2 (6x2 – 3y2 + 2xy)

= 12x2 – 6y2 + 4xy

(20) The perimeter of a triangle is 6p2 – 4p + 9 and two of its sides are p2 – 2p + 1 and 3p2 – 5p + 3. Find the third side of a triangle.

Solution: 6p2 – 4p + 9 – [(p2 – 2p + 1) + (3p2 – 5p + 3)]

= 6p2 – 4p + 9 – (p2 – 2p + 1 + 3p2 – 5p + 3)

= 6p2 – 4p + 9 – (4p2 – 7p + 4)

= 6p2 – 4p + 9 – 4p2 + 7p – 4

= 2p2 + 3p + 5


1 Comment

Add a Comment
  1. it’s so helpful and awesome

Leave a Reply

Your email address will not be published. Required fields are marked *

17 − 12 =