## 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 :

(a^{2} + b^{2} + 2ab) – (a^{2} + b^{2} – 2ab)

= a^{2} + b^{2} + 2ab – a^{2} – b^{2} + 2ab

= a^{2} – a^{2} + b^{2} – b^{2} + 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 :

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

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

[removing grouping symbol]

= – 4x^{2} + {5x^{2} – 7)

= – 4x^{2} + 5x^{2} – 7

(removing grouping symbol {})

= x^{2} – 7

**Question 6.****Solution:**

We have :

– 2 (x^{2} – y^{2 }+ xy) – 3 (x^{2} + y^{2} – xy)

= – 2x^{2} + 2y^{2} – 2xy – 3x^{2} – 3y^{2} + 3xy

= – 2x^{2} – 3x^{2} + 2y^{2} – 3y^{2} – 2xy + 3xy

= ( – 2 – 3)x^{2} + (2 – 3) y^{2} + ( – 2 + 3)xy

= – 5x^{2} – y^{2} + 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 – [3x^{3} + 5x^{2} – {7x^{2} – (4 – 3x – x^{3}) + 6x^{3}} – 3x]

= 12x – [3x^{3} – 5x^{2} – {7x^{2} – 4 + 3x + x^{3} + 6x^{3}} – 3x]

= 12x – [3x^{3} + 5x^{2} – {7x^{2} – 4 + 3x + 7x^{3}} – 3x]

= 12x – [3x^{3} + 5x^{2} – 7x^{2} + 4 – 3x – 7x^{3} – 3x]

= 12x – [3x^{3} – 7x^{3} + 5x^{2} – 7x^{2} + 4 – 3x – 3x]

= 12x – [ – 4x^{3} + 2x^{2} + 4 – 6x]

= 12x + 4x^{3} + 2x^{2} – 4 + 6x

= 12x + 6x + 4x^{3} + 2x^{2} – 4

= 18x + 4x^{3} + 2x^{2} – 4

= 4x^{3} + 2x^{2} + 18x – 4

**Question 11.****Solution:**

Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have

5a – [a^{2} – {2a (1 – a + 4a^{2}) – 3a (a^{2} – 5a – 3)}] – 8a

= 5a – [a^{2} – {2a – 2a^{2} + 8a^{3} – 3a^{3} + 15a^{2} + 9a}] – 8a

= 5a – [a^{2} – {2a + 9a – 2a^{2} + 15a^{2} + 8a^{3} – 3a^{3}}] – 8a

= 5a – [a^{2} – {11a + 13a^{2} + 5a^{3}}] – 8a

= 5a – [a^{2} – 11a – 13a^{2} – 5a^{3}] – 8a

= 5a – a^{2} + 11a + 13a^{2} + 5a^{3} – 8a

= 5a + 11a – 8a – a^{2} + 13a^{2} + 5a^{3}

= 8a + 12a^{2} + 5a^{3}

= 5a3 + 12a^{2} + 8a.

**Question 12.****Solution:**

Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :

3 – [x – {2y – (5x + y – 3) + 2x^{2}} – (x^{2} – 3y)]

= 3 – [x – {2y – 5x – y + 3 + 2x^{2}} – x^{2} + 3y]

= 3 – [x – {y – 5x + 3 + 2x^{2}} – x^{2} + 3y]

= 3 – [x – y + 5x – 3 – 2x^{2} – x^{2} + 3y]

= 3 – [6x + 2y – 3 – 3x^{2}]

= 3 – 6x – 2y + 3 + 3x^{2}

= 6 – 6x – 2y + 3x^{2}

**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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