{"id":553559,"date":"2021-11-02T04:49:50","date_gmt":"2021-11-02T04:49:50","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=553559"},"modified":"2021-11-03T05:32:25","modified_gmt":"2021-11-03T05:32:25","slug":"rs-aggarwal-solutions-for-class-6-maths-chapter-8-algebraic-expressions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rs-aggarwal-solutions-for-class-6-maths-chapter-8-algebraic-expressions\/","title":{"rendered":"RS Aggarwal Solutions for Class 6 Maths Chapter 8\u2013Algebraic Expressions"},"content":{"rendered":"\n

Class 6: Maths Chapter 8 solutions. Complete Class 6 Maths Chapter 8 Notes.<\/p>\n\n\n\n

RS Aggarwal Solutions for Class 6 Maths Chapter 8\u2013Algebraic Expressions<\/h2>\n\n\n\n

RS Aggarwal 6th Maths Chapter 8, Class 6 Maths Chapter 8 solutions<\/p>\n\n\n\n

Ex 8A Solutions<\/h4>\n\n\n\n

Question 1.<\/strong>
Solution:<\/strong>
(i) x + 12
(ii) y \u2013 7
(iii) a \u2013 b
(iv) (x + y) + xy
(v) 13x (a + b)
(vi) 7y + 5x
(vii) x+y5
(viii) 4 \u2013 x
(ix) xy\u22122
(x) x2<\/sup>
(xi) 2x + y
(xii) y2<\/sup> + 3x <\/sup>
(xiii) x \u2013 2y
(xiv) y3<\/sup> \u2013 x3<\/sup>
(xv) x8\u00d7y<\/p>\n\n\n\n

Question 2.<\/strong>
Solution:<\/strong>
Marks scored in English = 80
Marks scored in Hindi = x
\u2234 Total score in the two subjects = 80 + x<\/p>\n\n\n\n

Question 3.<\/strong>
Solution:<\/strong>
We can write :
(i) b \u00d7 b \u00d7 b \u00d7\u2026.15 times = 615<\/sup>
(ii) y \u00d7 y \u00d7 y \u00d7\u2026..20 times = y20<\/sup>
(iii) 14 \u00d7 a \u00d7 a \u00d7 a \u00d7 a \u00d7 b \u00d7 b \u00d7 b= 14a4<\/sup> b3<\/sup>
(iv) 6 \u00d7 x \u00d7 x \u00d7 y \u00d7 y = 6x2<\/sup>y2<\/sup>
(v) 3 \u00d7 z \u00d7 z \u00d7 z \u00d7 y \u00d7 y \u00d7 x= 3z3<\/sup>y2<\/sup>x<\/p>\n\n\n\n

Question 4.<\/strong>
Solution:<\/strong>
We can write :
(i) x2<\/sup>y4<\/sup> = x \u00d7 x \u00d7 y \u00d7 y \u00d7 y \u00d7 y
(ii) 6y5<\/sup> = 6 \u00d7 y \u00d7 y \u00d7 y \u00d7 y \u00d7 y
(iii) 9xy2<\/sup>z = 9 \u00d7 x \u00d7 y \u00d7 y \u00d7 z
(iv) 10a3<\/sup>b3<\/sup>c3<\/sup> = 10 \u00d7 a \u00d7 a \u00d7 a \u00d7 b \u00d7 b \u00d7 b \u00d7 c \u00d7 c \u00d7 c<\/p>\n\n\n\n

Ex 8B Solutions<\/h4>\n\n\n\n

Question 1.<\/strong>
Solution:<\/strong>
(i) Substituting a = 2 and b = 3 in the , given expression, we get :
a + b = 2 + 3 = 5
(ii) Substituting a = 2 and b = 3 in the given expression, we get :
a2<\/sup> + ab = (2)2<\/sup> + 2 x 3
= 4 + 6 = 10
(iii) Substituting a = 2 and b = 3 in the given expression, we get :
ab \u2013 a2<\/sup> = 2 x 3 \u2013 (2)2<\/sup>
= 6 \u2013 4 = 2
(iv) Substituting a = 2 and b = 3 in the given expression, we get :
2a \u2013 3b = 2 x 2 \u2013 3 x 3
= 4 \u2013 9 = \u2013 5
(v) Substituting a = 2 and b = 3 in the given expression, we get :
5a2<\/sup> \u2013 2ab = 5 x (2)2<\/sup> \u2013 2 x 2 x 3
= 5 x 4 \u2013 4 x 3
= 20 \u2013 12 = 8
(vi) Substituting a = 2 and b = 3 in the given expression, we get :
a3<\/sup> \u2013 b3<\/sup> = (2)3<\/sup> \u2013 (3)3<\/sup> = 2 x 2 x 2 \u2013 3 x 3 x 3
= 8 \u2013 27 = \u2013 19<\/p>\n\n\n\n

Question 2.<\/strong>
Solution:<\/strong>
(i) Substituting x = 1, y = 2 and z = 5 in the given expression, we get :
3x \u2013 2y + 4z = 3 x 1 \u2013 2 x 2 + 4 x 5
= 3 \u2013 4 + 20 = 23 \u2013 4 = 19
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 3.<\/strong>
Solution:<\/strong>
(i) Substituting p = \u2013 2, q = \u2013 1 and r = 3
in the given expression, we get :
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 4.<\/strong>
Solution:<\/strong>
(i) The coefficient of x in 13x is 13
(ii) The coefficient of y in \u2013 5y is \u2013 5
(iii) The coefficient of a in 6ab is 6b
(iv) The coefficient of z in \u2013 7xz is \u2013 7x
(v) The coefficient of p in \u2013 2pqr is \u2013 2qr
(vi) The coefficient of y2<\/sup> in 8xy2<\/sup>z is 8xz
(vii) The coefficient of x3<\/sup> in x3<\/sup> is 1
(viii) The coefficient of x2<\/sup> in \u2013 x2<\/sup> is -1<\/p>\n\n\n\n

Question 5.<\/strong>
Solution:<\/strong>
(i) The numerical coefficient of ab is 1
(ii) The numerical coefficient of \u2013 6bc is \u2013 6
(iii) The numerical coefficient of 7xyz is 7
(iv) The numerical coefficient of \u2013 2x3<\/sup>y2<\/sup>z is \u2013 2.<\/p>\n\n\n\n

Question 6.<\/strong>
Solution:<\/strong>
(i) The constant term is 8
(ii) The constant term is \u2013 9
(iii) The constant term is 35
(iv) The constant term is \u201383<\/p>\n\n\n\n

Question 7.<\/strong>
Solution:<\/strong>
(i) The given expression contains only one term, so it is monimial.
(ii) The given expression contains only two terms, so it is binomial.
(iii) The given expression contains only one term, so it is monomial.
(iv) The given expression contains three terms, so it is trinomial.
(v) The given expression contains three terms, so it is trinomial.
(vi) The given expression contains only one term, so it is monomial.
(vii) The given expression contains four terms, so it is none of monomial, binomial and trinomial.
(viii) The given expression contains only one term so it is monomial.
(ix) The given expression contains two terms, so it is binomial.<\/p>\n\n\n\n

Question 8.<\/strong>
Solution:<\/strong>
(i) The terms of the given expression 4x5<\/sup> \u2013 6y4<\/sup> + 7x2<\/sup>y \u2013 9 are :
4x5<\/sup>, \u2013 6y4<\/sup>, 7x2<\/sup>y, \u2013 9
(ii) The terms of the given expression 9x3<\/sup> \u2013 5z4<\/sup> + 7x3<\/sup>y \u2013 xyz are :
9x3<\/sup>, \u2013 5z4<\/sup>, 7x3<\/sup>y, \u2013 xyz.<\/p>\n\n\n\n

Question 9.<\/strong>
Solution:<\/strong>
(i) We have : a2<\/sup>, b2<\/sup>, \u2013 2a2<\/sup>, c2<\/sup>, 4a
Here like terms are a2<\/sup>, \u2013 2a2<\/sup>
(ii) We have : 3x, 4xy, \u2013 yz, 12 zy
Here like terms are \u2013 yz, 12 zy
(iii) We have : \u2013 2xy2<\/sup>, x2<\/sup>y, 5y2<\/sup>x, x2<\/sup>z
Here like terms are \u2013 2xy2<\/sup>, 5y2<\/sup>x
(iv) We have :
abc, ab2<\/sup>c, acb2<\/sup>, c2<\/sup>ab, b2<\/sup>ac, a2<\/sup>bc, cab2<\/sup>
Here like terms are ab2<\/sup>c, acb2<\/sup>, b2<\/sup>ac, cab2<\/sup>.<\/p>\n\n\n\n

Ex 8C Solutions<\/h4>\n\n\n\n

Question 1.<\/strong>
Solution:<\/strong>
(i) The required sum
= 3x + 7x
= (3 + 7) x
= 10x
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 2.<\/strong>
Solution:<\/strong>
(i) Adding columnwise,
we get
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 3.<\/strong>
Solution:<\/strong>
(i) Arranging the like terms column wise and adding, we get :
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 4.<\/strong>
Solution:<\/strong>
(i) We have :
2x \u2013 5x = (2 \u2013 5)x = \u2013 3x
(ii) We have :
6x \u2013 y \u2013 (- xy) = 6xy + xy = 7xy
(iii) We have : 5b \u2013 3a
(iv) We have : 9y \u2013 ( \u2013 7x) = 9y + 7x
(v) We have : \u2013 7x2<\/sup> \u2013 10x2<\/sup> = ( \u2013 7 \u2013 10)x2<\/sup>
= \u2013 17x2<\/sup>
(vi) We have : b2<\/sup> \u2013 a2<\/sup> \u2013 (a2<\/sup> \u2013 b2<\/sup>)
= b2<\/sup> \u2013 a2<\/sup> \u2013 a2<\/sup> + b2<\/sup>
= b2<\/sup> + b2<\/sup> \u2013 a2<\/sup> \u2013 a2<\/sup>
= (1 + 1) b2<\/sup> + ( \u2013 1 \u2013 1) a2<\/sup>
= 2b2<\/sup> \u2013 2a2<\/sup><\/p>\n\n\n\n

Question 5.<\/strong>
Solution:<\/strong>
(i) Arranging the like terms column wise, we get :
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 6.<\/strong>
Solution:<\/strong>
(i) Rearranging and collecting the like terms, we get :
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n
\"RS<\/figure>\n\n\n\n

Question 7.<\/strong>
Solution:<\/strong>
We have:
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 8.<\/strong>
Solution:<\/strong>
We have :
A = 7x2<\/sup> + 5xy \u2013 9y2<\/sup>
B = \u2013 4x2<\/sup> + xy + 5y2<\/sup>
C = 4y2<\/sup> \u2013 3x2<\/sup> \u2013 6xy
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n


= 0+0+0 = 0
Hence the result<\/p>\n\n\n\n

Question 9.<\/strong>
Solution:<\/strong>
Required expression
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 10.<\/strong>
Solution:<\/strong>
Substituting the values of P, Q, R and S, we have :
P + Q + R + S = (a2<\/sup> \u2013 b2<\/sup> + 2ab)
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 11.<\/strong>
Solution:<\/strong>
Required expression
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 12.<\/strong>
Solution:<\/strong>
Required expression
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 13.<\/strong>
Solution:<\/strong>
Required expression
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 14.<\/strong>
Solution:<\/strong>
Required expression
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 15.<\/strong>
Solution:<\/strong>
Sum of 5x \u2013 4y + 6z and \u2013 8x + y \u2013 2z
= 5x \u2013 4y + 6z \u2013 8x + y \u2013 2z
= 5x \u2013 8x \u2013 4y + y + 6z \u2013 2z
= \u2013 3x \u2013 3y + 4z
Sum of 12x \u2013 y + 3z and \u2013 3x + 5y \u2013 8z
= 12x \u2013 y + 3z \u2013 3x + 5y \u2013 8z
= 12x \u2013 3x \u2013 y + 5y + 3z \u2013 8z
= 9x + 4y \u2013 5z
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 16.<\/strong>
Solution:<\/strong>
Required expression
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Question 17.<\/strong>
Solution:<\/strong>
Required expression
<\/p>\n\n\n\n

\"RS<\/figure>\n\n\n\n

Ex 8D Solutions<\/h4>\n\n\n\n

Simplify :<\/strong><\/p>\n\n\n\n

Question 1.<\/strong>
Solution:<\/strong>
We have : a \u2013 (b \u2013 2a)
= a \u2013 b + 2a
= a + 2a \u2013 b
= (1 + 2) a \u2013 b
= 3a \u2013 b.<\/p>\n\n\n\n

Question 2.<\/strong>
Solution:<\/strong>
We have : 4x \u2013 (3y \u2013 x + 2z)
= 4x \u2013 3y + x \u2013 2z
= 4x + x \u2013 2y \u2013 2z
= 5x \u2013 3y \u2013 2z<\/p>\n\n\n\n

Question 3.<\/strong>
Solution:<\/strong>
We have :
(a2<\/sup> + b2<\/sup> + 2ab) \u2013 (a2<\/sup> + b2<\/sup> \u2013 2ab)
= a2<\/sup> + b2<\/sup> + 2ab \u2013 a2<\/sup> \u2013 b2<\/sup> + 2ab
= a2<\/sup> \u2013 a2<\/sup> + b2<\/sup> \u2013 b2<\/sup> + 2ab + 2ab
= 0 + 0 + (2 + 2) ab
= 4 ab<\/p>\n\n\n\n

Question 4.<\/strong>
Solution:<\/strong>
We have :
\u2013 3 (a + b) + 4 (2a \u2013 3b) \u2013 (2a \u2013 b)
= \u2013 3a \u2013 3b + 8a \u2013 12b \u2013 2a + b
= \u2013 3a + 8a \u2013 2a \u2013 3b \u2013 12b + b
= ( \u2013 3 + 8 \u2013 2) a + ( \u2013 3 \u2013 12 + 1) b
= 3a \u2013 14 b.<\/p>\n\n\n\n

Question 5.<\/strong>
Solution:<\/strong>
We have :
\u2013 4x2<\/sup> + {(2x2<\/sup> \u2013 3) \u2013 (4 \u2013 3x2<\/sup>)}
= \u2013 4x2<\/sup> + {2x2<\/sup> \u2013 3 \u2013 4 + 3x2<\/sup>}
[removing grouping symbol]
= \u2013 4x2<\/sup> + {5x2<\/sup> \u2013 7)
= \u2013 4x2<\/sup> + 5x2<\/sup> \u2013 7
(removing grouping symbol {})
= x2<\/sup> \u2013 7<\/p>\n\n\n\n

Question 6.<\/strong>
Solution:<\/strong>
We have :
\u2013 2 (x2<\/sup> \u2013 y2 <\/sup>+ xy) \u2013 3 (x2<\/sup> + y2<\/sup> \u2013 xy)
= \u2013 2x2<\/sup> + 2y2<\/sup> \u2013 2xy \u2013 3x2<\/sup> \u2013 3y2<\/sup> + 3xy
= \u2013 2x2<\/sup> \u2013 3x2<\/sup> + 2y2<\/sup> \u2013 3y2<\/sup> \u2013 2xy + 3xy
= ( \u2013 2 \u2013 3)x2<\/sup> + (2 \u2013 3) y2<\/sup> + ( \u2013 2 + 3)xy
= \u2013 5x2<\/sup> \u2013 y2<\/sup> + xy<\/p>\n\n\n\n

Question 7.<\/strong>
Solution:<\/strong>
a \u2013 [2b \u2013 {3a \u2013 (2b \u2013 3c)}]
= a \u2013 [2b \u2013 {3a \u2013 2b + 3c}]
[removing grouping symbol( )]
= a \u2013 [2b \u2013 3a + 2b \u2013 3c]
(removing grouping symbol {})
= a \u2013 [4b \u2013 3a \u2013 3c]
= a \u2013 4b + 3a + 3c
(removing grouping symbol [ ])
= 4a \u2013 4b + 3c<\/p>\n\n\n\n

Question 8.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
\u2013 x + [5y \u2013 {x \u2013 (5y \u2013 2x)}]
= \u2013 x + [5y \u2013 {x \u2013 5y + 2x}]
= \u2013 x + [5y \u2013 {3x \u2013 5y}]
= \u2013 x + [5y \u2013 3x + 5y]
= \u2013 x + [ 10y \u2013 3x]
= \u2013 x + 10y \u2013 3x
= \u2013 x \u2013 3x + 10y
= \u2013 4x + 10y<\/p>\n\n\n\n

Question 9.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
86 \u2013 [15x \u2013 7 (6x \u2013 9) \u2013 2 {10x \u2013 5(2 \u2013 3x)}]
= 86 \u2013 [15x \u2013 42x + 63 \u2013 2 {10x \u2013 10 + 15x}
= 86 \u2013 [ 15x \u2013 42x + 63 \u2013 2 {25x \u2013 10}]
= 86 \u2013 [15x \u2013 42x + 63 \u2013 50x + 20]
= 86 \u2013 15x + 42x \u2013 63 + 50x \u2013 20
= (86 \u2013 63 \u2013 20) \u2013 15x + 42x + 50x
= (86 \u2013 83) + (- 15 + 42 + 50) x
= 3 + 77x<\/p>\n\n\n\n

Question 10.<\/strong>
Solution:<\/strong>
Removing the innermost grouping \u2018 symbol () first, then { } and then [ ], we have :
12x \u2013 [3x3<\/sup> + 5x2<\/sup> \u2013 {7x2<\/sup> \u2013 (4 \u2013 3x \u2013 x3<\/sup>) + 6x3<\/sup>} \u2013 3x]
= 12x \u2013 [3x3<\/sup> \u2013 5x2<\/sup> \u2013 {7x2<\/sup> \u2013 4 + 3x + x3<\/sup> + 6x3<\/sup>} \u2013 3x]
= 12x \u2013 [3x3<\/sup> + 5x2<\/sup> \u2013 {7x2<\/sup> \u2013 4 + 3x + 7x3<\/sup>} \u2013 3x]
= 12x \u2013 [3x3<\/sup> + 5x2<\/sup> \u2013 7x2<\/sup> + 4 \u2013 3x \u2013 7x3<\/sup> \u2013 3x]
= 12x \u2013 [3x3<\/sup> \u2013 7x3<\/sup> + 5x2<\/sup> \u2013 7x2<\/sup> + 4 \u2013 3x \u2013 3x]
= 12x \u2013 [ \u2013 4x3<\/sup> + 2x2<\/sup> + 4 \u2013 6x]
= 12x + 4x3<\/sup> + 2x2<\/sup> \u2013 4 + 6x
= 12x + 6x + 4x3<\/sup> + 2x2<\/sup> \u2013 4
= 18x + 4x3<\/sup> + 2x2<\/sup> \u2013 4
= 4x3<\/sup> + 2x2<\/sup> + 18x \u2013 4<\/p>\n\n\n\n

Question 11.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
5a \u2013 [a2<\/sup> \u2013 {2a (1 \u2013 a + 4a2<\/sup>) \u2013 3a (a2<\/sup> \u2013 5a \u2013 3)}] \u2013 8a
= 5a \u2013 [a2<\/sup> \u2013 {2a \u2013 2a2<\/sup> + 8a3<\/sup> \u2013 3a3<\/sup> + 15a2<\/sup> + 9a}] \u2013 8a
= 5a \u2013 [a2<\/sup> \u2013 {2a + 9a \u2013 2a2<\/sup> + 15a2<\/sup> + 8a3<\/sup> \u2013 3a3<\/sup>}] \u2013 8a
= 5a \u2013 [a2<\/sup> \u2013 {11a + 13a2<\/sup> + 5a3<\/sup>}] \u2013 8a
= 5a \u2013 [a2<\/sup> \u2013 11a \u2013 13a2<\/sup> \u2013 5a3<\/sup>] \u2013 8a
= 5a \u2013 a2<\/sup> + 11a + 13a2<\/sup> + 5a3<\/sup> \u2013 8a
= 5a + 11a \u2013 8a \u2013 a2<\/sup> + 13a2<\/sup> + 5a3<\/sup>
= 8a + 12a2<\/sup> + 5a3<\/sup>
= 5a3 + 12a2<\/sup> + 8a.<\/p>\n\n\n\n

Question 12.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
3 \u2013 [x \u2013 {2y \u2013 (5x + y \u2013 3) + 2x2<\/sup>} \u2013 (x2<\/sup> \u2013 3y)]
= 3 \u2013 [x \u2013 {2y \u2013 5x \u2013 y + 3 + 2x2<\/sup>} \u2013 x2<\/sup> + 3y]
= 3 \u2013 [x \u2013 {y \u2013 5x + 3 + 2x2<\/sup>} \u2013 x2<\/sup> + 3y]
= 3 \u2013 [x \u2013 y + 5x \u2013 3 \u2013 2x2<\/sup> \u2013 x2<\/sup> + 3y]
= 3 \u2013 [6x + 2y \u2013 3 \u2013 3x2<\/sup>]
= 3 \u2013 6x \u2013 2y + 3 + 3x2<\/sup>
= 6 \u2013 6x \u2013 2y + 3x2<\/sup><\/p>\n\n\n\n

Question 13.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have :
xy \u2013 [yz \u2013 zx \u2013 {yx \u2013 (3y \u2013 xz} \u2013 (xy \u2013 zy)}]
= xy \u2013 [yz \u2013 zx \u2013 {yx \u2013 3y + xz \u2013 xy + zy}]
= xv \u2013 [yz \u2013 zx \u2013 {- 3y + xz + zy}]
= xy \u2013 [yz \u2013 zx + 3y \u2013 xz \u2013 zy]
= xy \u2013 [ \u2013 2xz + 3y]
= xy + 2xz \u2013 3y<\/p>\n\n\n\n

Question 14.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol ( ) first, then { } and then [ ], we have
2a \u2013 3b \u2013 [3a \u2013 2b \u2013 {a \u2013 c \u2013 (a \u2013 2b)}]
= 2a \u2013 3b \u2013 [3a \u2013 2b \u2013 {a \u2013 c \u2013 a + 2b}]
= 2a \u2013 3b \u2013 [3a \u2013 2b \u2013 { \u2013 c + 2b}]
= 2a \u2013 3b \u2013 [3a \u2013 2b + c \u2013 2b]
= 2a \u2013 3b \u2013 3a + 2b \u2013 c + 2b
= 2a \u2013 3a \u2013 3b + 2b + 2b \u2013 c
= \u2013 a + b \u2013 c<\/p>\n\n\n\n

Question 15.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol () first, then { } and ten [ ], we have:
\u2013 a \u2013 [a + {a + b \u2013 2a \u2013 (a \u2013 2b)} \u2013 b]
= \u2013 a \u2013 [a + {a + b \u2013 2a \u2013 a + 2b} \u2013 b]
= \u2013 a \u2013 [a + { \u2013 2a + 3b} \u2013 b]
= \u2013 a \u2013 [a \u2013 2a + 3b \u2013 b]
= \u2013 a \u2013 a + 2a \u2013 3b + b
= \u2013 2a + 2a \u2013 2b
= \u2013 2 b<\/p>\n\n\n\n

Question 16.<\/strong>
Solution:<\/strong>
Removing the innermost grouping symbol \u2018\u2014\u2019 first, then ( ), then { } and then [ ], we have
2a \u2013 [4b \u2013 {4a \u2013 (3b \u2013 2a+2b\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af\u00af)}]
= 2a \u2013 [4b \u2013 {4a \u2013 (3b \u2013 2a \u2013 2b)}]
= 2a \u2013 [4b \u2013 {4a \u2013 (b \u2013 2a)}]
= 2a \u2013 [4b \u2013 {4a \u2013 b + 2a}]
= 2a \u2013 [4b \u2013 {6a \u2013 b}]
= 2a \u2013 [4b \u2013 6a + b]
= 2a \u2013 [5b \u2013 6a]
= 2a \u2013 5b + 6a
= 8a \u2013 5b.<\/p>\n\n\n\n

Question 17.<\/strong>
Solution:<\/strong>
Removing the innermost grouping < symbol ( ) first, then { } and then [ ], we have :
5x \u2013 [4y \u2013 {7x \u2013 (3z \u2013 2y) + 4z \u2013 3(x + 3y \u2013 2z)}]
= 5x \u2013 [4y \u2013 {7x \u2013 3z + 2y + 4z \u2013 3x \u2013 9y + 6z}]
= 5x \u2013 [4y \u2013 {4x + 7z \u2013 7y}]
= 5x \u2013 [4y \u2013 4x \u2013 7z + 7y]
= 5x \u2013 [11y \u2013 4x \u2013 7z]
= 5x \u2013 11y + 4x + 7z
= 9x \u2013 11y + 7z<\/p>\n\n\n\n

RS Aggarwal Solutions for Class 6 Maths Chapter 8: Download PDF<\/strong><\/h2>\n\n\n\n

RS Aggarwal Solutions for Class 6 Maths Chapter 8\u2013Algebraic Expressions<\/p>\n\n\n\n

Download PDF<\/strong>: RS Aggarwal Solutions for Class 6 Maths Chapter 8\u2013Algebraic Expressions PDF<\/a><\/p>\n\n\n\n

Chapterwise RS Aggarwal Solutions for Class 6 Maths :<\/strong><\/h2>\n\n\n\n