 # RS Aggarwal Solutions for Class 6 Maths Chapter 8 Algebraic Expressions

## Class 6 RS Aggarwal Chapter 8 – Algebraic Expressions

RS Aggarwal Solutions for Class 6 Maths, Chapter 8 are given here. Students who wish to prepare thoroughly for the exam are advised to go through the RS Aggarwal Solutions for Class 6 Maths Chapter 8 to score good marks.These Solutions need a lot of practice to gain high marks in examinations. By referring RS Aggarwal Solutions, students will understand the chapter in detail. Practising textbook questions will increase the confidence of the students and help to understand the topics which are talked about, in this chapter.

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## Download PDF of RS Aggarwal Solutions for Class 6 Chapter 8 Algebraic Expressions                 ## Exercise 8A PAGE NO: 130

1. Write the following using literals, numbers and signs of basic operation:

(i) x increased by 12

(ii) y decreased by 7

(iii) The difference of a and b, when a > b

(iv) The product of x and y added to their sum

(v) One third of x multiplied by the sum of a and b

(vi) 5 times x added to seven times y

(vii) Sum of x and the quotient of y by 5

(viii) x taken away from 4

(ix) 2 less than the quotient of x by y

(x) x multiplied by itself

(xi) Twice x increased by y

(xii) Thrice x added to y squared

(xiii) x minus twice y

(xiv) x cubed less than y cubed

(xv) The quotient of x by 8 is multiplied by y

Solution

(i) x increased by 12 is written as x + 12

(ii) y decreased by 7 is written as y – 7

(iii) The difference of a and b, when a > b is (a – b)

(iv) The product of x and y is xy, added to their sum is (x + y)

Hence, the product of x and y added to their sum is (x + y) + xy

(v) One third of x = x / 3

Sum of a and b = (a + b)

∴ One third of x multiplied by the sum of a and b = x / 3 × (a + b)

= x (a + b) / 3

(vi) 5 times x = 5x, seven times y = 7y

∴ 5 times x added to seven times y is written as 7y + 5x

(vii) Sum of x = x

Quotient of y = y / 5

∴ Sum of x and the quotient of y by 5 = x + y / 5

(viii) x taken away from 4 is written as (4 – x)

(ix) Quotient of x by y = x / y

∴ 2 less than the quotient of x by y = (x / y) – 2

(x) x multiplied by itself is x × x = x2

∴ x multiplied by itself is written as x2

(xi) Twice x increased by y is written as (2x + y)

(xii) Thrice x = 3 × x = 3x and y squared = (y × y) = y2

∴ Thrice x added to y squared is written as 3x + y2

(xiii) Twice y = 2 × y = 2y

∴ x minus twice y is written as (x – 2y)

(xiv) x cubed = (x × x × x) = x3 y cubed = (y × y × y ) = y3

∴ x cubed less than y cubed is written as (y3 – x3)

(xv) The quotient of x by 8 is x / 8

∴ The quotient of x by 8 is multiplied by y is written as (x / 8) × y

2. Ranjit scores 80 marks in English and x marks in Hindi. What is his total score in the two subjects?

Solution

Marks scored by Ranjit in English = 80

Marks scored by Ranjit in Hindi = x

Total score in the two subjects = (Marks in English) + (Marks in Hindi)

= (80 + x)

∴ Total score in the two subjects scored by Ranjit is (80 + x) marks

## Exercise 8B PAGE NO: 132

1. If a = 2 and b = 3, find the value of

(i) a + b

(ii) a2 + ab

(iii) ab – a2

(iv) 2a – 3b

(v) 5a2 – 2ab

(vi) a3 – b3

Solutions

Given a = 2 and b = 3

(i) a + b

Substituting a = 2 and b = 3 in the given expression, we get

a + b = 2 + 3

= 5

∴ a + b = 5

(ii) a2 + ab

Substituting a = 2 and b = 3 in the given expression, we get

a2 + ab = (2)2 + (2) × (3)

= (2 × 2) + (2 × 3)

= 4 + 6

= 10

∴ a2 + ab = 10

(iii) ab – a2

Substituting a = 2 and b = 3 in the given expression, we get

ab – a2 = (2 × 3) – (2)2

= (2 × 3) – (2 × 2)

= 6 – 4

= 2

∴ ab – a2 = 2

(iv) 2a – 3b

Substituting a = 2 and b = 3 in the given expression, we get

2a – 3b = (2 × 2) – (3 × 3)

= 4 – 9

= -5

∴ 2a – 3b = -5

(v) 5a2 – 2ab

Substituting a = 2 and b = 3 in the given expression, we get

5a2 – 2ab = 5 × (2)2 – 2 (2) (3)

= (5 4) – 2 (2 × 3)

= 20 – 2 (6)

= 20 – 12

= 8

∴ 5a2 – 2ab = 8

(vi) a3 – b3

Substituting a = 2 and b = 3 in the given expression, we get

a3 – b3 = (2)3 – (3)3

= (2 × 2 × 2) – (3 × 3 × 3)

= 8 – 27

=-19

∴ a3 – b3 = -19

2. If x = 1, y = 2 and z = 5, find the value of

(i) 3x – 2y + 4z

(ii) x2 + y2 + z2

(iii) 2x2 – 3 y2 + z2

(iv) xy + yz -zx

(v) 2x2y – 5yz + xy2

(vi) x3 – y3 – z3

Solutions

Given x = 1, y = 2 and z = 5

(i) 3x – 2y + 4z

Substituting x = 1, y = 2 and z = 5 in the given expression, we get

3x – 2y + 4z = 3 (1) – 2 (2) + 4 (5)

= (3 × 1) – (2 × 2) + (4 × 5)

= 3 – 4 + 20

= 23 – 4

= 19

∴ 3x – 2y + 4z = 19

(ii) x2 + y2 + z2

Substituting x = 1, y = 2 and z = 5 in the given expression, we get

x2 + y2 + z2 = (1)2 + (2)2 + (5)2

= (1× 1) + (2 × 2) + (5 × 5)

= 1 + 4 + 25

= 30

∴ x2 + y2 + z2 = 30

(iii) 2x2 – 3y2 + z2

Substituting x = 1, y = 2 and z = 5 in the given expression, we get

2x2 – 3y2 + z2 = 2 (1)2 – 3(2)2 + (5)2

= (2 × 1) – 3 (2 × 2) + (5 × 5)

= (2) – 3 (4) + (25)

= 2 – 12 + 25

= 27 – 12

= 15

∴ 2x2 – 3y2 + z2 = 15

(iv) xy + yz –zx

Substituting x = 1, y = 2 and z = 5 in the given expression, we get

xy + yz –zx = (1) (2) + (2) (5) – (5) (1)

= (1 × 2) + (2 × 5) – (5 × 1)

= 2 + 10 -5

= 12 – 5

= 7

∴ xy + yz –zx = 7

(v) 2x2y – 5yz + xy2

Substituting x = 1, y = 2 and z = 5 in the given expression, we get

2x2y – 5yz + xy2 = 2 (1)2 (2) – 5(2) (5) + (1) (2)2

= 2 (1 × 1) (2) – 5 (2 × 5) + (1 × 2 × 2)

= 4 – 5 (10) + (4)

= 4 – 50 + 4

= 8 – 50

= – 42

∴ 2x2y – 5yz + xy2 = -42

(vi) x3 – y3 – z3

Substituting x = 1, y = 2 and z = 5 in the given expression, we get

x3 – y3 – z3 = (1)3 – (2)3 – (5)3

= (1 × 1 × 1) – (2 × 2 × 2) – (5 × 5 × 5)

= (1) – (8) – (125)

= 1 – 8 – 125

= 1 – 133

= -132

∴ x3 – y3 – z3 = -132

3. If p = -2, q = -1 and r = 3, find the value of

(i) p2 + q2 – r2

(ii) 2p2 – q2 + 3r2

(iii) p – q – r

(iv) p3 + q3 + r3 + 3pqr

(v) 3p2q + 5pq2 + 2pqr

(vi) p4 + q4 – r4

Solutions

Given p = -2, q = -1 and r = 3

(i) p2 + q2 – r2

Substituting p = -2, q = -1 and r = 3 in the given expression, we get

p2 + q2 – r2 = (-2)2 + (-1)2 – (3)2

= (-2 × -2) + (-1 × -1) – (3 × 3)

= 4 + 1 – 9

= 5 – 9

= -4

∴ p2 + q2 – r2 = -4

(ii) 2p2 – q2 + 3r2

Substituting p = -2, q = -1 and r = 3 in the given expression, we get

2p2 – q2 + 3r2 = 2 (-2)2 – (-1)2 + 3 (3)2

= 2 (-2 × -2) – (-1 × -1) + 3 (3 × 3)

= 2 (4) – (1) + 3 (9)

= 8 – 1 + 27

= 7 + 27

= 34

∴ 2p2 – q2 + 3r2 = 34

(iii) p – q – r

Substituting p = -2, q = -1 and r = 3 in the given expression, we get

p – q – r = (-2) – (-1) – (3)

= -2 + 1 – 3

= -1 – 3

= -4

∴ p – q – r = -4

(iv) p3 + q3 + r3 + 3pqr

Substituting p = -2, q = -1 and r = 3 in the given expression, we get

p3 + q3 + r3 + 3pqr = (-2)3 + (-1)3 + (3)3 + 3(-2) (-1) (3)

= (-2 × -2 × -2) + (-1 × -1 × -1) + (3 × 3 × 3) + 3 (-2 × -1 × 3)

= -8 -1 + 27 + 18

= -9 + 45

= 36

∴ p3 + q3 + r3 + 3pqr = 36

(v) 3p2q + 5pq2 + 2pqr

Substituting p = -2, q = -1 and r = 3 in the given expression, we get

3p2q + 5pq2 + 2pqr = 3 (-2)2 (-1) + 5 (-2) (-1)2 + 2 (-2) (-1) (3)

= 3 (-2 × -2 × -1) + 5 (-2 × -1 × -1) + 2 (-2 × -1 × 3)

= 3(-4) + 5(-2) + 2 (6)

= -12 – 10 + 12

= -10

∴ 3p2q + 5pq2 + 2pqr = -10

(vi) p4 + q4 – r4

Substituting p = -2, q = -1 and r = 3 in the given expression, we get

p4 + q4 – r4 = (-2)4 + (-1)4 – (3)4

= (-2 × -2 × -2 × -2) + (-1 × -1 × -1 × -1) – (3 × 3 × 3 × 3)

= (16) + (1) – (81)

= 17 – 81

= -64

∴ p4 + q4 – r4 = -64

4. Write the coefficient of

(i) x in 13 x

(ii) y in – 5y

(iii) a in 6ab

(iv) z in -7xz

(v) p in – 2pqr

(vi) y2 in 8xy2z

(vii) x3 in x3

(viii) x2 in – x2

Solution

(i) The coefficient of x in 13x is 13

(ii) The coefficient of y in -5y is -5

(iii) The coefficient of a in 6ab is 6b

(iv) The coefficient of z in -7xz is -7x

(v) The coefficient of p in -2pqr is -2qr

(vi) The coefficient of y2 in 8xy2z is 8xz

(vii) The coefficient of x3 in x3 is 1

(viii) The coefficient of x2 in -x2 is -1

5. Write the numerical coefficient of

(i) ab

(ii) – 6bc

(iii) 7xyz

(iv) – 2x3 y2 z

Solutions

(i) The numerical coefficient of ab is 1

(ii) The numerical coefficient -6bc is -6

(iii) The numerical coefficient 7xyz is 7

(iv) The numerical coefficient -2x3y2z is -2

## Exercise 8C PAGE NO: 134

(i) 3x, 7x

(ii) 7y, -9y

(iii) 2xy, 5xy, -xy

(iv) 3x, 2y

(v) 2x2, -3x2, 7x2

(vi) 7xyz, -5xyz, 9xyz, -8xyz

(vii) 6a3, -4a3, 10a3, -8a3

(viii) x2 – a2, – 5x2 + 2a2, -4x2 + 4a2

Solutions

(i) The required sum = 3x + 7x

= (3 + 7) x

= 10x

(ii) The required sum = 7y + (-9y)

= 7y – 9y

= (7 -9) y

= -2y

(iii) The required sum = 2xy + 5xy + (-xy)

= 2xy + 5xy – xy

= (2x + 5x –x) y

= 6xy

(iv) The required sum = 3x + 2y

= 3x + 2y

(v) The required sum = 2x2 + (-3x2) + 7x2

= 2x2 – 3x2 + 7x2

= (2 – 3 + 7) x2

= 6x2

(vi) The required sum = 7xyz + (-5xyz) + 9xyz + (-8xyz)

= 7xyz – 5xyz + 9xyz – 8xyz

= (7 – 5 + 9 – 8) xyz

= (16 – 13) xyz

= 3xyz

(vii) The required sum = 6a3 + (-4a3) + 10a3 + (-8a3)

= 6a3 – 4a3 + 10a3 – 8a3

= (6 – 4 + 10 – 8) a3

= (16 – 12) a3

= 4a3

(viii) The required sum = (x2 – a2) + (-5x2 + 2a2) + (-4x2 + 4a2)

= x2– a2 – 5x2 + 2a2 – 4x2 + 4a2

= (1 – 5 – 4) x2 – (1 – 2 – 4) a2

= (1 – 9) x2 – (1 – 6) a2

= -8x2 + 5a2

(i) x – 3y – 2z (ii) m2 – 4m + 5 (iii) 2x2 – 3xy + y2

5x + 7y – z -2m2 + 6m – 6 – 7x2 – 5xy – 2y2

-7x – 2y + 4z -m2 – 2m – 7 4x2 + xy – 6y2

__________ __________ ____________

__________ __________ ____________

(iv) 4xy – 5yz – 7z

– 5xy +2yz + zx

– 2xy -3yz +3zx

____________

______________

Solutions

(i) x – 3y – 2z

5x + 7y – z

-7x – 2y + 4z

____________________

-x + 2y + z

____________________

(ii) m 2– 4m + 5

-2m2 + 6m – 6

– m2 – 2m – 7

_______________

-2m2 +0m – 8

= -2m2 – 8

________________

(iii) 2x2 – 3xy + y2

-7x2 – 5xy – 2y2

4x2 + xy – 6y2

___________________________

-x2 – 7xy – 7y2

_____________________________

(iv) 4xy – 5yz – 7zx

-5xy + 2yz + zx

-2xy – 3yz + 3zx

_________________

-3xy -6yz -3zx

________________

(i) 3a – 2b + 5c, 2a +5b – 7c, – a – b + C

(ii) 8a – 6ab + 5b, – 6a – ab – 8b, – 4a + 2ab + 3b

(iii) 2x3 – 3x2 + 7x – 8, – 5x3 + 2x2 – 4x + 1, 3 – 6x + 5x2 – x3

(iv) 2x2 – 8xy + 7y2 – 8xy2, 2xy2 + 6xy – y2 + 3x2, 4y2 – xy – x2 + xy2

(v) x3 + y3 – z3 + 3xyz, -x3 + y3 + z3 – 6xyz, x3 – y3 – z3 – 8xyz

(vi) 2 + x – x2 + 6x3, – 6 – 2x + 4x2 – 3x3, 2 + x2, 3-x3 + 4x – 2x2

Solution

(i) The sum of the given expressions

= (3a + 2a –a) + (-2b +5b –b) + (5c – 7c +c)

= 4a + 2b – c

(ii) The sum of the given expressions

= (8a -6a -4a) + (5b – 8b + 3b) + (-6ab –ab + 2ab)

= -2a – 5ab

(iii) The sum of the given expressions

= (2x3 – 5x3 – x3) + (-3x2 + 2x2 + 5x2) + (7x – 4x – 6x) + (-8 + 1 +3)

= – 4x3 + 4x2 – 3x – 4

(iv) The sum of the given expressions

= (2x2 + 3x2 – x2) + (-8xy + 6xy – xy) + (7y2 – y2 + 4y2) + (-8xy2 + 2xy2 + xy2)

= 4x2 – 3xy + 10 y2 – 5xy2

(v) The sum of the given expressions

= (x3 – x3 + x3) + (y3 + y3 – y3) + (-z3 + z3 – z3) + (3xyz – 6xyz – 8xyz)

= x3 + y3 – z3 – 11xyz

(vi) The sum of the given expressions

= (2 – 6 + 2 + 3) + (x – 2x + 4x) + (- x2 + 4x2 + x2 – 2x2) + (6x3 – 3x3 – x3)

= 1 + 3x + 2x2 + 2x3

4. Subtract:

(i) 5x from 2x

(ii) – xy from 6xy

(iii) 3a from 5b

(iv) – 7x from 9y

(v) 10x2 from – 7x2

(vi) a2 – b2 from b2 – a2

Solutions

(i) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

We get

Term which is subtracted = 5x

Changing the sign of each term of expression = -5x

2x – 5x = (2 – 5) x

= – 3x

(ii) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

We get

Term which is subtracted = -xy

Changing the sign of each term of expression = xy

6xy + xy = (6 + 1) xy

= 7xy

(iii) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

We get

Term which is subtracted = 3a

Changing the sign of each term of expression = – 3a

= (5b – 3a)

(iv) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

We get

Term which is subtracted = – 7x

Changing the sign of each term of expression = 7x

= (9y + 7x)

(v) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

We get

Term which is subtracted = 10x2

Changing the sign of each term of expression = – 10x2

= (- 7x2 – 10x2)

= – 17x2

(vi) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

We get

Term which is subtracted = a2 – b2

Changing the sign of each term of expression = – (a2 – b2)

= (b2 – a2) – (a2 – b2)

= (b2 – a2 – a2 + b2)

= 2b2 – 2a2

5. Subtract:

(i) 5a + 7b – 2c from 3a – 7b + 4c

(ii) a – 2b -3c from -2a + 5b – 4c

(iii) 5x2 – 3xy + y2 from 7x2 – 2xy – 4y2

(iv) 6x3 – 7x2 + 5x – 3 from 4 – 5x + 6x2 – 8x3

(v) x3 + 2x2y + 6xy2 – y3 from y3 – 3xy2 – 4x2y

(vi) – 11x2y2 + 7xy – 6 from 9x2y2 – 6xy + 9

(vii) -2a + b + 6d from 5a -2b -3c

Solutions

(i) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = 5a + 7b – 2c

Changing the sign of each term of expression = – 5a – 7b + 2c

= (3a – 7b + 4c) + (- 5a – 7b + 2c)

= 3a – 5a – 7b – 7b + 4c + 2c

= -2a – 14b + 6c

(ii) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = a – 2b – 3c

Changing the sign of each term of expression = -a + 2b + 3c

= (- 2a + 5b – 4c) + (-a + 2b + 3c)

= -2a + 5b – 4c – a + 2b + 3c

= -3a + 7b – c

(iii) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = 5x2 – 3xy + y2

Changing the sign of each term of expression = – 5x2 + 3xy – y2

= (7x2 – 2xy – 4y2) + (- 5x2 + 3xy – y2)

= 7x2 – 2xy – 4y2 – 5x2 + 3xy – y2

= 2x2 + xy – 5y2

(iv) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = 6x3 – 7x2 + 5x – 3

Changing the sign of each term of expression = -6x3 + 7x2 – 5x + 3

= (4 – 5x + 6x2 – 8x3) + (- 6x3 + 7x2 – 5x + 3)

= 4 – 5x + 6x2 – 8x3 – 6x3 + 7x2 – 5x + 3

= 4 + 3 – 5x – 5x + 6x2 + 7x2 – 8x3 – 6x3

= 7 – 10x + 13x2 – 14x3

(v) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = x3 + 2x2y + 6xy2 – y3

Changing the sign of each term of expression = -x3 – 2x2y – 6xy2 + y3

= (y3 – 3xy2 – 4x2y) + (- x3 – 2x2y – 6xy2 + y3)

= y3 – 3xy2 – 4x2y – x3 – 2x2y – 6xy2 + y3

= y3 + y3 – 3xy2 – 6xy2 – 4x2y – 2x2y – x3

= 2y3 – 9xy2 – 6x2y – x3

(vi) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = -11x2y2 + 7xy – 6

Changing the sign of each term of expression = 11x2y2 – 7xy + 6

= (9x2y2 – 6xy + 9) + (11x2y2 – 7xy + 6)

= 9x2y2 – 6xy + 9 + 11x2y2 – 7xy + 6

= 9x2y2 + 11x2y2 – 6xy – 7xy + 9 + 6

= 20x2y2 – 13xy + 15

(vii) Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = -2a + b + 6d

Changing the sign of each term of expression = 2a – b – 6d

= (5a – 2b – 3c) + (2a – b – 6d)

= 5a – 2b – 3c + 2a – b – 6d

= 5a + 2a – 2b – b – 3c – 6d

= 7a – 3b – 3c – 6d

6. Simplify:

(i) 2p3 – 3p2 + 4p – 5 – 6p3 + 2p2 – 8p -2 + 6p +8

(ii) 2x2 – xy + 6x – 4y + 5xy – 4x + 6x2 + 3y

(iii) x4 – 6x3 + 2x – 7 + 7x3 – x + 5x2 + 2 – x4

Solution

(i) Given

2p3 – 3p2 + 4p – 5 – 6p3 + 2p2 – 8p – 2 + 6p + 8

Rearranging and collecting the like terms, we get:

= 2p3 – 6p3 – 3p2 + 2p2 + 4p – 8p + 6p – 5 + 2 + 8

= (2 – 6) p3 – (3 – 2) p2 + (4 – 8 + 6) p – (5 – 2 – 8)

= (- 4) p3 – (1) p2 + (10 – 8) p – (7 – 8)

= (-4) p3 – p2 + (2) p – (-1)

= – 4p3 – p2 + 2p + 1

(ii) Given

2x2 – xy + 6x – 4y + 5xy – 4x + 6x2 + 3y

Rearranging and collecting the like terms, we get:

= 2x2 + 6x2 – xy + 5xy + 6x – 4x – 4y + 3y

= (2 + 6) x2 – (1 – 5) xy + (6 – 4) x – (4 – 3) y

= (8) x2 – (- 4) xy + (2) x – (1) y

= 8x2 + 4xy + 2x – y

(iii) Given

x4 – 6x3 + 2x – 7 + 7x3 – x + 5x2 + 2 – x4

Rearranging and collecting the like terms, we get:

= x4 – x4 – 6x3 + 7x3 + 5x2 + 2x – x – 7 + 2

= (1 – 1) x4 – (6 – 7) x3 + 5x2 + (2 – 1) x – 7 + 2

= – (-1) x3 + 5x2 + (1) x – 7 + 2

= x3 + 5x2 + x – 5

7. From the sum of 3x2 – 5x + 2 and -5x2 – 8x + 6, subtract 4x2 – 9x + 7.

Solution

To find the sum

Add 3x2 – 5x + 2 and -5x2 – 8x + 6

(3x2 – 5x + 2) + (-5x2 – 8x + 6)

Rearranging and collecting the like terms, we get:

= 3x2 – 5x2 – 5x – 8x + 2 + 6

= (3 – 5) x2 – (5 + 8) x + 2 + 6

= – 2x2 – 13x + 8

Now Subtract 4x2 – 9x + 7 from -2x2 – 13x + 8

Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = 4x2 – 9x + 7

Changing the sign of each term of expression = – 4x2 + 9x – 7

= – 2x2 – 13x + 8 – 4x2 + 9x – 7

= -2x2 – 13x + 8 – 4x2 – 9x – 7

= -2x2 – 4x2 – 13x – 9x + 8 – 7

= (-2 – 4) x2 – (13 + 9) x + 8 – 7

= – 6x2 – 4x + 1

8. If A = 7x2 + 5xy – 9y2, B = – 4x2 + xy + 5y2 and C = 4y2 – 3x2 – 6xy then show that A + B + C = 0.

Solution

Given

A = 7x2 + 5xy – 9y2

B = – 4x2 + xy + 5y2

C = 4y2 – 3x2 – 6xy

To show A + B + C = 0

Substitute the value of A, B and C in A + B + C

A + B + C = (7x2 + 5xy – 9y2) + (- 4x2 + xy + 5y2) + (4y2 – 3x2 – 6xy)

= 7x2 + 5xy – 9y2 – 4x2 + xy + 5y2 + 4y2 – 3x2 – 6xy

Rearranging and collecting the like terms, we get:

= 7x2 – 4x2 – 3x2 + 5xy + xy – 6xy – 9y2 + 5y2 + 4y2

= (7 – 4 – 3) x2 + (5 + 1 – 6) xy – (9 – 5 – 4) y2

= 0 x2 + 0 xy – 0 y2

= 0 + 0 + 0

= 0

Hence,

A + B + C = 0

9. What must be added to 5x3 – 2x2 + 6x + 7 to make the sum x3 + 3x2 – x + 1

Solution

Let X be the expression to be added to 5x3 – 2x2 + 6x + 7

(5x3 – 2x2 + 6x + 7) + X = x3 + 3x2 – x + 1

X = (x3 + 3x2 – x + 1) – (5x3 – 2x2 + 6x + 7)

Changing the sign of each term of expression to be subtracted and add it to the expression from which subtraction is to be made

Term which is subtracted = 5x3 – 2x2 + 6x + 7

Changing the sign of each term of expression = – 5x3 + 2x2 – 6x – 7

X = (x3 + 3x2 – x + 1) + (- 5x3 + 2x2 – 6x – 7)

= x3 + 3x2 – x + 1 – 5x3 + 2x2 – 6x – 7

Rearranging and collecting the like terms, we get:

= x3 – 5x3 + 3x2 + 2x2 – x – 6x + 1 – 7

= (1 – 5) x3 + (3 + 2) x2 – (1 + 6) x + 1 – 7

= – 4 x3 + 5 x2 + 7 x – 6

∴ – 4 x3 + 5 x2 + 7 x – 6 must be added to 5x3 – 2x2 + 6x + 7 to make the sum x3 + 3x2 – x + 1

## Exercise 8D PAGE NO: 136

Simplify:

1. a – (b -2a)

Solution

Here, (-) sign precedes the second parenthesis, so we remove it and change the sign of each term within

∴ a – (b – 2a)

= a – (b – 2a)

= a – b + 2a

= 3a – b

2. 4x – (3y – x + 2z)

Solution

Here, (-) sign precedes the second parenthesis, so we remove it and change the sign of each term within

∴ 4x – (3y – x + 2z)

= 4x – 3y + x – 2z

= 4x + x – 3y – 2z

= 5x – 3y – 2z

3. (a2 + b2 + 2ab) – (a2 + b2 -2ab)

Solution

Here, (-) sign precedes the second parenthesis, so we remove it and change the sign of each term within

∴ (a2 + b2 + 2ab) – (a2 + b2 -2ab)

= a2 + b2 + 2ab – a2 – b2 + 2ab

= a2 – a2 + b2 – b2 + 2ab + 2ab

= 4ab

4. – 3(a + b) + 4 (2a – 3b) – (2a – b)

Solution

Here, (-) sign precedes the first and third parenthesis, so we remove it and change the sign of each term within

∴ – 3(a + b) + 4 (2a – 3b) – (2a – b)

= -3a – 3b + 4 (2a) – 4 (3b) – 2a + b

= -3a – 3b + 8a – 12b – 2a + b

= – 3a + 8a – 2a – 3b – 12b + b

= 8a – 5a – 15b + b

= 3a -14b

5. -4x2 + {(2x2 – 3) – (4 – 3x2)}

Solution

We first remove the innermost grouping symbol ( ) and then { },

We have:

4x2 + {(2x2 – 3) – (4 – 3x2)}

= – 4x2 + {2x2 – 3 – 4 + 3x2}

= – 4x2 + 2x2 – 3 – 4 + 3x2

= – 4x2 + 2x2 + 3x2 – 3 – 4

= – 4x2 + 5x2 – 7

= x2 – 7

6. – 2(x2 – y2 + xy) – 3(x2 + y2 – xy)

Solution

Here, (-) sign precedes both the parenthesis, so we remove it and change the sign of each term within

∴ – 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

7. a – [2b – {3a – (2b -3c)}]

Solution

We first remove the innermost grouping symbol ( ), { } and [ ]

We have,

a – [2b – {3a – (2b -3c)}]

= a – [2b – {3a – 2b + 3c}]

= a – [2b – 3a + 2b – 3c]

= a – 2b + 3a – 2b + 3c

= a – 3a – 2b – 2b + 3c

= 4a – 4b + 3c

8. – x + [5y – {x – (5y – 2x)}]

Solution

We first remove the innermost grouping symbol ( ), { } and [ ]

We have

– x + [5y – {x – (5y – 2x)}]

= – x + [5y – {x – 5y + 2x}]

= – x + [5y – x + 5y – 2x]

= – x + 5y – x + 5y – 2x

= – x – x – 2x + 5y + 5y

= (-1 -1 -2) x + (5 + 5) y

= – 4x + 10y

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

Solution

We first remove the innermost grouping symbol ( ), { } and [ ]

We have

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

= 86 – [15x – 42x + 63 – 2{10x – 10 + 15x}]

= 86 – [15x – 42x + 63 – 20x + 20 – 30x]

= 86 – 15x + 42x – 63 + 20x – 20 + 30x

= 86 – 63 – 20 – 15x + 42x + 20x + 30x

= 3 + 77x

### RS Aggarwal Solutions for Class 6 Chapter 8 Algebraic Expressions

Chapter 8 – Algebraic Expressions consists of 4 exercises. RS Aggarwal Solutions have been solved in detail for each question in every exercise. Let’s have a glance at the topics included in this chapter

• Operations on Literals and Numbers
• Algebraic Expressions
• Operations on Algebraic Expressions
• Use of grouping symbols

### Chapter Brief of RS Aggarwal Solutions for Class 6 Maths Chapter 8 – Algebraic Expressions

An expression having both variable and constant along with algebraic operations such as addition, subtraction, multiplication and division. For example, 6x – 4,

here x is the variable whose value is unknown, 6 is known as coefficient of x and 4 is the constant term having a definite value. There are five types of algebraic expressions.

(i) Monomials – expression having one term known as monomials

(ii) Binomials – expression having two terms known as binomials

(iii) Trinomials – expression having three terms known as trinomials