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
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