# RS Aggarwal Class 6 Solutions Chapter 8 - Algebraic Ex 8B(8.2)

## RS Aggarwal Class 6 Chapter 8 - Algebraic Ex 8B(8.2) Solutions Free PDF

Question 1:

a – (b – 2a)

Here, sign precedes the parenthesis.

So, we will remove it and change the sign of each term within the parenthesis.

=a – b + 2a

=3a – b

Question 2:

4x – (3y – x + 2z)

Here, 1-1 sign precedes the parenthesis.

So, we will remove it and change the sign of each term within the parenthesis.

= 4x – 3y + x – 2z

= 5x – 3y – 2z

Question 3:

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

Here, ‘-‘ sign precedes the second parenthesis. So we will remove it and change the sign of each term within the parenthesis.

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

Rearranging and collecting the like terms:

$a^{2}$$a^{2}$ + $b^{2}$$b^{2}$ + 2ab + 2ab

=(1- 1) $a^{2}$ + (1- 1) $b^{2}$ + (2 + 2)ab

=0 + 0 + 4ab

= 4ab

Question 4:

-3(a + b) + 4(2a – 3b) – (2a – b)

Here, 1-1 sign precedes the first and the third parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.

= -3a – 3b + (4x 2a )-(4x 3b) – 2a + b

= – 3a â€“ 3b + 8a – 12b – 2a + b

Rearranging and collecting the like terms:

-3a + 8a – 2a – 3b – 12b + b

= (-3 + 8 – 2)a + (-3 – 12 + 1)b

= 3a -14b

Question 5:

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

We will first remove the innermost grouping symbol ) and then { }.

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

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

= -4$x^{2}$ + {5$x^{2}$ – 7}

= -4$x^{2}$ + 5$x^{2}$ – 7

= $x^{2}$ â€“ 7

Question 6:

-2($x^{2}$$y^{2}$ + xy) -3($x^{2}$ + y2 – xy)

Here a sign precedes both the parenthesis. So, we will remove them and change the sign of each term within the two parenthesis.

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

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

= -5$x^{2}$$y^{2}$ + xy

Question 7:

a – [2b – {3a – (2b – 3c)}]

We will first remove the innermost grouping symbol ( ), followed by { and then [ ].

a – [2b – {3a – (2b – 3c)}]

= a – [2b – {3a – 2b + 3c}]

= a – [2b – 3a + 2b – 3c]

= a – [4b – 3a – 3c]

= a – 4b + 3a + 3c

= 4a – 4b + 3c

Question 8:

-x + [5y – {x – (5y – 2x)}]

We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].

-x + [5y – {x – (5y – 2x)}]

= -x + [5y – {x – 5y + 2x}]

= -x + [5y – {3x – 5y}]

= -x + [5y – 3x + 5y]

= -x + [10y – 3x]

=-x+10y-3x

= -4x+10y

Question 9:

86 – [15x – 7(6x – 9) -2{10x – 5(2 – 3x)}]

We will first remove the innermost grouping symbol ( ), followed by { and then [ ].

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 – [- 77x + 83]

= 86 + 77x – 83

= 77x + 3

Question 10:

12x – [3$x^{3}$ + 5$x^{2}$ – {7$x^{2}$ – (4 – 3x – $x^{3}$) + 6$x^{3}$} – 3x]

We will first remove the innermost grouping symbol ( ), followed by { } and then [ ]

12x – [3$x^{3}$ + 5$x^{2}$ – (7$x^{2}$ – (4 – 3x – $x^{3}$) + 6$x^{3}$} – 3x]

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

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

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

= 12x – [- 2$x^{2}$ + 4 – 4$x^{3}$ – 6x]

= 12x + 2$x^{2}$ – 4 + 4$x^{3}$ + 6x

= 4$x^{3}$ + 2$x^{2}$ +18x â€“ 4

Question 11:

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

We will first remove the innermost grouping symbol ( ), followed by { and then [ ].

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

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

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

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

= 5a – [ – 5$a^{3}$ – 12$a^{2}$ – 11a] – 8a

= 5a + 5$a^{3}$ + 12$a^{2}$ + 11a – 8a

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

Question 12:

3 â€” [x â€” {2y â€” (5x + y â€” 3) + 2×2} â€” (x2 â€” 3y)]

We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].

3 â€” [x â€” {2y â€” (5x + y â€” 3) + 2$x^{2}$} â€” ($x^{2}$ â€” 3y)]

= 3 â€” [x â€” {2y â€” 5x – y + 3 + 2$x^{2}$} â€” x2 + 3y]

= 3 â€” [x â€” {y â€” 5x + 3 + 2$x^{2}$} â€” $x^{2}$ + 3y]

=3â€”[xâ€” y+5x-3-2$x^{2}$â€” $x^{2}$+3y]

= 3 â€” [ 6x â€” 3 â€” 3$x^{2}$ + 2y]

= 3 â€“ 6x + 3+ 3$x^{2}$ â€”2y

= 3$x^{2}$ â€” 2y â€” 6x + 6

Question 13:

xy â€“ [yz – zx – {yx – (3y – xz ) – (xy – zy)}]

We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].

xy – [yz – zx – {yx – (3y – xz) – (xy – zy)}]

= xy – [yz – zx – {yx – 3y + >tz – xy + zy}]

= xy – [yz – zx – {- 3y + xz + zy}] (xy=yx)

= xy – [yz – zx + 3y – xz – zy]

= xy – [- 2zx + 3y]

= xy + 2zx – 3y

Question 14:

2a – 3b – [3a – 2b – {a – c – (a – 2b)}]

We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].

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-4b+c]

= 2a – 3b – 3a + 4b – c

=-a+b-c

Question 15:

-a – [a + {a + b – 2a – (a – 2b)} – b]

We will first remove the innermost grouping symbol ( ), followed by { } and then [ ].

-a – [a + {a + b – 2a – (a – 2b)} – b]

= -a – [a + {a + b – 2a – a + 2b} – b]

= -a – [a + {3b – 2a } – b]

= -a – [a + 3b – 2a – b]

= -a – [2b – a ]

= -a – 2b + a

= -2b

Question 16:

2a – [4b – (4a – (3b – 2a + 2b)]

We will first remove the innermost grouping symbol bar bracket. Next, we will remove ( ), followed by { } and then [ ].

2a-[4b-(4a-(3b-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:

5x – [4y- 7x – (3z – 2y) + 4z – 3(x + 3y- 2z)}]

We will first remove the innermost grouping symbol ( ), followed by { } and then [ ]

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

#### Practise This Question

If “a” chocolates are distributed amongst “b” students equally, then what is the share of chocolates obtained by each student?