RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.1: Here we provided the RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.1 which focuses on operations with algebraic expressions.
By working through these solutions students can gain a clearer understanding of how to handle algebraic expressions and perform these fundamental operations effectively. These solutions are designed to help students build a strong foundation in algebra and improve their problem-solving skills, which will be beneficial for their overall mathematical proficiency.RS Aggarwal Solutions for Class 8 Maths Chapter 6 Exercise 6.1 PDF
Add:
(Question 1) 8ab, – 5ab, 3ab, – ab
Solution:
= 8ab + (- 5ab) + 3ab + (- ab) = 8ab – 5ab + 3ab – ab = 11ab – 6ab = 5ab(Question 2) 7x, – 3x, 5x, – x, – 2x
Solution:
= 7x + (- 3x) + 5x + (- x) + (- 2x) = 7x – 3x + 5x – x – 2x = 12x – 6x = 6x(Question 3) 3a – 4b + 4c, 2a + 3b – 8c, a – 6b + c
Solution:
= 3a – 4b + 4c + 2a + 3b – 8c + a – 6b + c = 6a – 7b – 3c(Question 4) 5x – 8y + 2z, 3z – 4y – 2x, 6y – z – x and 3x – 2z – 3y
Solution:
= 5x – 8y + 2z + 3z – 4y – 2x + 6y – z – x + 3x – 2z – 3y = 5x – 9y + 2z(Question 5) 6ax – 2by + 3cz, 6by – 11ax – cz and 10cz – 2ax – 3by
Solution:
= 6ax – 2by + 3cz + 6by – 11ax – cz + 10cz – 2ax – 3by = – 7ax + by + 12cz(Question 6) 2x 3 – 9x 2 + 8, 3x 2 – 6x – 5, 7x 3 – 10x + 1 and 3 + 2x – 5x 2 – 4x 3
Solution:
= 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(Question 7) 6p + 4q – r + 3, 2r – 5p – 6, 11q – 7p + 2r – 1 and 2q – 3r + 4
Solution:
= 6p + 4q – r + 3 + 2r – 5p – 6 + 11q – 7p + 2r – 1 + 2q – 3r + 4 = – 6p + 17q(Question 8) 4x 2 – 7xy + 4y 2 – 3, 5 + 6y 2 – 8xy + x 2 and 6 – 2xy + 2x 2 – 5y 2.
Solution:
= 4x 2 – 7xy + 4y 2 – 3 + 5 + 6y 2 – 8xy + x 2 + 6 – 2xy + 2x 2 – 5y 2 = 7x 2 – 17xy + 5y 2 + 8Subtract:
(Question 9) 3a 2 b from – 5a 2 b
Solution:
= – 5a 2 b – 3a 2 b = – 8a 2 b(Question 10) – 8pq from 6pq
Solution:
= 6pq – (- 8pq) = 6pq + 8pq = 14pq(Question 11) – 2abc from – 8abc
Solution:
= – 8abc – (- 2abc) = – 8abc + 2abc = – 6abc(Question 12) – 16p from – 11p
Solution:
= -11p – (- 16p) = – 11p + 16p = 5p(Question 13) 2a – 5b + 2c – 9 from 3a – 4b – c + 6
Solution:
= 3a – 4b – c + 6 – (2a – 5b + 2c – 9) = 3a – 4b – c + 6 – 2a + 5b – 2c + 9 = a + b – 3c + 15(Question 14) – 6p + q + 3r + 8 from p – 2q – 5r – 8
Solution:
= p – 2q – 5r – 8 – (- 6p + q + 3r + 8) = p – 2q – 5r – 8 + 6p – q – 3r – 8 = 7p – 3q – 8r – 16(Question 15) x 3 + 3x 2 – 5x + 4 from 3x 3 – x 2 + 2x – 4
Solution:
= 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(Question 16) 5y 4 – 3y 3 + 2y 2 + y – 1 from 4y 4 – 2y 3 – 6y 2 – y + 5
Solution:
= 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(Question 17) 4p 2 + 5q 2 – 6r 2 + 7 from 3p 2 – 4q 2 – 5r 2 – 6
Solution:
= 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(Question 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(Question 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(Question 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