RS Aggarwal Class 6 Solutions Chapter 8 Algebraic Expressions Ex 8D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D.

Other Exercises

Simplify :

Question 1.
Solution:
We have : a – (b – 2a)
= a – b + 2a
= a + 2a – b
= (1 + 2) a – b
= 3a – b.

Question 2.
Solution:
We have : 4x – (3y – x + 2z)
= 4x – 3y + x – 2z
= 4x + x – 2y – 2z
= 5x – 3y – 2z

Question 3.
Solution:
We have :
(a2 + b2 + 2ab) – (a2 + b2 – 2ab)
= a2 + b2 + 2ab – a2 – b2 + 2ab
= a2 – a2 + b2 – b2 + 2ab + 2ab
= 0 + 0 + (2 + 2) ab
= 4 ab

Question 4.
Solution:
We have :
– 3 (a + b) + 4 (2a – 3b) – (2a – b)
= – 3a – 3b + 8a – 12b – 2a + b
= – 3a + 8a – 2a – 3b – 12b + b
= ( – 3 + 8 – 2) a + ( – 3 – 12 + 1) b
= 3a – 14 b.

Question 5.
Solution:
We have :
– 4x2 + {(2x2 – 3) – (4 – 3x2)}
= – 4x2 + {2x2 – 3 – 4 + 3x2}
[removing grouping symbol]
= – 4x2 + {5x2 – 7)
= – 4x2 + 5x2 – 7
(removing grouping symbol {})
= x2 – 7

Question 6.
Solution:
We have :
– 2 (x2 – y2 + xy) – 3 (x2 + y2 – xy)
= – 2x2 + 2y2 – 2xy – 3x2 – 3y2 + 3xy
= – 2x2 – 3x2 + 2y2 – 3y2 – 2xy + 3xy
= ( – 2 – 3)x2 + (2 – 3) y2 + ( – 2 + 3)xy
= – 5x2 – y2 + xy

Question 7.
Solution:
a – [2b – {3a – (2b – 3c)}]
= a – [2b – {3a – 2b + 3c}]
[removing grouping symbol( )]
= a – [2b – 3a + 2b – 3c]
(removing grouping symbol {})
= a – [4b – 3a – 3c]
= a – 4b + 3a + 3c
(removing grouping symbol [ ])
= 4a – 4b + 3c

Question 8.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
– x + [5y – {x – (5y – 2x)}]
= – x + [5y – {x – 5y + 2x}]
= – x + [5y – {3x – 5y}]
= – x + [5y – 3x + 5y]
= – x + [ 10y – 3x]
= – x + 10y – 3x
= – x – 3x + 10y
= – 4x + 10y

Question 9.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
86 – [15x – 7 (6x – 9) – 2 {10x – 5(2 – 3x)}]
= 86 – [15x – 42x + 63 – 2 {10x – 10 + 15x}
= 86 – [ 15x – 42x + 63 – 2 {25x – 10}]
= 86 – [15x – 42x + 63 – 50x + 20]
= 86 – 15x + 42x – 63 + 50x – 20
= (86 – 63 – 20) – 15x + 42x + 50x
= (86 – 83) + (- 15 + 42 + 50) x
= 3 + 77x

Question 10.
Solution:
Removing the innermost grouping ‘ symbol () first, then { } and then [ ], we have :
12x – [3x3 + 5x2 – {7x2 – (4 – 3x – x3) + 6x3} – 3x]
= 12x – [3x3 – 5x2 – {7x2 – 4 + 3x + x3 + 6x3} – 3x]
= 12x – [3x3 + 5x2 – {7x2 – 4 + 3x + 7x3} – 3x]
= 12x – [3x3 + 5x2 – 7x2 + 4 – 3x – 7x3 – 3x]
= 12x – [3x3 – 7x3 + 5x2 – 7x2 + 4 – 3x – 3x]
= 12x – [ – 4x3 + 2x2 + 4 – 6x]
= 12x + 4x3 + 2x2 – 4 + 6x
= 12x + 6x + 4x3 + 2x2 – 4
= 18x + 4x3 + 2x2 – 4
= 4x3 + 2x2 + 18x – 4

Question 11.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
5a – [a2 – {2a (1 – a + 4a2) – 3a (a2 – 5a – 3)}] – 8a
= 5a – [a2 – {2a – 2a2 + 8a3 – 3a3 + 15a2 + 9a}] – 8a
= 5a – [a2 – {2a + 9a – 2a2 + 15a2 + 8a3 – 3a3}] – 8a
= 5a – [a2 – {11a + 13a2 + 5a3}] – 8a
= 5a – [a2 – 11a – 13a2 – 5a3] – 8a
= 5a – a2 + 11a + 13a2 + 5a3 – 8a
= 5a + 11a – 8a – a2 + 13a2 + 5a3
= 8a + 12a2 + 5a3
= 5a3 + 12a2 + 8a.

Question 12.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
3 – [x – {2y – (5x + y – 3) + 2x2} – (x2 – 3y)]
= 3 – [x – {2y – 5x – y + 3 + 2x2} – x2 + 3y]
= 3 – [x – {y – 5x + 3 + 2x2} – x2 + 3y]
= 3 – [x – y + 5x – 3 – 2x2 – x2 + 3y]
= 3 – [6x + 2y – 3 – 3x2]
= 3 – 6x – 2y + 3 + 3x2
= 6 – 6x – 2y + 3x2

Question 13.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
xy – [yz – zx – {yx – (3y – xz} – (xy – zy)}]
= xy – [yz – zx – {yx – 3y + xz – xy + zy}]
= xv – [yz – zx – {- 3y + xz + zy}]
= xy – [yz – zx + 3y – xz – zy]
= xy – [ – 2xz + 3y]
= xy + 2xz – 3y

Question 14.
Solution:
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
2a – 3b – [3a – 2b – {a – c – (a – 2b)}]
= 2a – 3b – [3a – 2b – {a – c – a + 2b}]
= 2a – 3b – [3a – 2b – { – c + 2b}]
= 2a – 3b – [3a – 2b + c – 2b]
= 2a – 3b – 3a + 2b – c + 2b
= 2a – 3a – 3b + 2b + 2b – c
= – a + b – c

Question 15.
Solution:
Removing the innermost grouping symbol () first, then { } and ten [ ], we have:
– a – [a + {a + b – 2a – (a – 2b)} – b]
= – a – [a + {a + b – 2a – a + 2b} – b]
= – a – [a + { – 2a + 3b} – b]
= – a – [a – 2a + 3b – b]
= – a – a + 2a – 3b + b
= – 2a + 2a – 2b
= – 2 b

Question 16.
Solution:
Removing the innermost grouping symbol ‘—’ first, then ( ), then { } and then [ ], we have
2a – [4b – {4a – (3b – \(\overline { 2a+2b } \))}]
= 2a – [4b – {4a – (3b – 2a – 2b)}]
= 2a – [4b – {4a – (b – 2a)}]
= 2a – [4b – {4a – b + 2a}]
= 2a – [4b – {6a – b}]
= 2a – [4b – 6a + b]
= 2a – [5b – 6a]
= 2a – 5b + 6a
= 8a – 5b.

Question 17.
Solution:
Removing the innermost grouping < symbol ( ) first, then { } and then [ ], we have :
5x – [4y – {7x – (3z – 2y) + 4z – 3(x + 3y – 2z)}]
= 5x – [4y – {7x – 3z + 2y + 4z – 3x – 9y + 6z}]
= 5x – [4y – {4x + 7z – 7y}]
= 5x – [4y – 4x – 7z + 7y]
= 5x – [11y – 4x – 7z]
= 5x – 11y + 4x + 7z
= 9x – 11y + 7z

Hope given RS Aggarwal Solutions Class 6 Chapter 8 Algebraic Expressions Ex 8D are helpful to complete your math homework.

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