# Class 8 ICSE Maths Solutions Chapter 11

## Class 8 ICSE Maths Solutions Chapter 11 Algebraic Expressions (Selina Concise)

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 11, Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions, Exercise 11A, 11B, 11C, 11D and 11E Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 11 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Algebraic Expressions Exercise 11(A) Solution :

Question no – (1)

Solution :

From question,

Constants are = – 7, √5, 8 – 5

Variable are = 7x + yz, √xy, 3yz/8, 4.5y – 3x, 8 – 5, 8 – 5y × p and 3y²z ÷ 4x.

Question no – (2)

Solution :

(i) 5x² + 3 × ax

The no of terms in this polynomial 2

(ii) ax ÷ 4 – 7

The no of terms in this polynomial – 2

(iii) ax – by + y × z

= ax – by + yz

The no of terms polynomial 3

(iv) 23 + a × b ÷ 2

= 23 + ab/2

The no of terms in this polynomial 2

Question no – (3)

Solution :

Binomials are = 8 – 3x

Trinomials are = 8 – 3x, 3y2 – 5y + 8, 9x – 3×2 + 15×3 – 7, 2y ÷ 7 = 3x – 7, 4 – ax2 + bx + y

Monomials are = xy2, 3x × 5y, 3x ÷ 5y.

Question no – (4)

Solution :

(i) Given, xy + 7z

= Degree is 2

(ii) x2 – 6x3 + 8

=  Degree is 3

(iii) y – 6y2 + 5y8

=  Degree is 8

(iv) xyz – 3

= Degree is 3

(v) xy + yz2 – zx3

=  Degree is 4

(vi) x5y7 – bx3y8 + 10x4y4z4

=  Degree is 12

Question no – (5)

Solution :

(i) Given, ab in 7abx

The coefficient of ab in 7abx = 7x.

(ii) 7a in 7abx,

The coefficient 7a in 7abx = bx.

(iii) 5xin 5x2 – 5x.

The coefficient 5x2 in 5x2 – 5x is = 1

(iv) 8 in a2 – 8ax + a,

The coefficient of 8 in a2 – 8ax + a = – ax.

(v) 4xy in x2 – 4xy + y2

The coefficient of 4xy in x2 – 4xy + y2 = – 1.

Question no – (6)

Solution :

(i) 5/7 xy²z³

The coefficient of 5 is = 1/7 xy²x³

(ii) 5/7 xy2z3

The coefficient of 5/7 is = xy2z3

(iii) 5/7 xy2z3

The coefficient of 5x is = 1/7 y2z3

(iv) Given,  5/7 xy2z3

The coefficient of xyis = 5/7z2

(v) 5/7 xy2z3

The coefficient of z3 is = 5/7xy2

(vi) 5/7 xy2z3

The coefficient of xz3 is =  5/7y2

(vii) 5/7 xy²z³

The coefficient of 5xy² is = 1/7z3

(viii) 5/7 xy²z³

The coefficient of 1/7yz is =  5xyz2

(ix) 5/7 xy²z³

The coefficient of z is = 5/7 xy2z2

(x) 5/7 xy²z³

The coefficient of yz2 = 5/7xy – z

(xi) 5/7 xy²z³

The coefficient of 5xyz is = 1/7yz2

Question no – (7)

Solution :

Given in the question,

3xy, – 4yx2, 2xy2, 2.5x2y, – 3.2y2x and x2y

Like terms are –  3xy, – 8yx2, 2.5x2y, x2y, 2xy2

Algebraic Expressions Exercise 11(B) Solution :

Question no – (1)

Solution :

(i) -7x2 + 18x2 + 3x2 – 5x2

= 21x2 – 12x2

= 9x2

(ii) b²y – 9b²y + 2b²y – 5b²y

= 3b²y – 14b²y

= – 11 b²y

(iii) abx – 15abx – 10abx + 32abx

= 33abx – 25abx

= 8abx

(iv) 7x – 9y + 3 – 3x – 5y + 8

= 4x – 14y + 11

(v) 3x² + 5xy – 4y² + x² – 8xy – 5y²

= 3x² – 3xy – 9y²

Question no – (2)

Solution :

(i) 5a + 3b, a – 2b, 3a + 5b

= (5a + 3b) + (a – 2b) + (3a + 5b)

= 5a + 3b + a 2b + 3a + 5b

= 9a + 6b

(ii) 8x – 3y + 7z, – 4x + 5y – 4z, – x – y – 2z

= (8x – 3y) + (- 4x + 5y – 4z) + (- x – y – 2z)

= 3x + y + 2

(iii) 3b – 7c + 10, -2b + 5c – 15, 15 + 12c + b

= (3b – 7c + 10) + (- 2b + 5c – 15) + (b + 12c + 15)

= 2b + 10c + 10

(iv) a – 3b + 3, 2a + 5 – 3c, 6c – 15 + 6b

= (a – 3b + 3) + (2a + 5 – 3c) + (6c – 15 + 6b)

= a – 3b + 3 + 2a + 5 – 3c + 6c – 15 + 6b

= 3a + 3b + 3c – 7

(v) 13ab – 9cd – xy, 5xy, 15cd – 7ab, 6xy – 3cd

= (13ab – 9cd – xy) + (5xy) + (15cd – 7ab) + (6xy – 3cd)

= 6ab + 3cd + 10xy

(vi) x³ – x²y + 5xy² + y³, – x³ – 9xy² + y³, 3x²y + 9xy²

= (x³ – x²y + 5xy² + y³) + (-x³ – 9xy² + y³) + (3x²y + 9xy²)

= x³ – x²y + 5xy² + y³ -x³ – 9xy² + y³ + 3x²y + 9xy

= 2x²y + 5xy² + 2y³

Question no – (3)

Solution :

As per the given question,

Savings of a boy = (4x – 6y), ₹ (6x + 2y), and ₹(y – 2x)

Total saving = (4x – 6y) + (6x + 2y) + (4y – x) + (y – 2x)

= 4x² – 6y + 6x² + 2y + 4y – x + 2x

= (7x + y)

Question no – (4)

Solution :

(i) 4xy² from 3xy²

= 3xy² – 4xy²

= – xy²

(ii) -2x²y + 3xy² from 8x²y

= 8x²y + 2x²y – 3xy²

= 10xy² – 3xy²

(iii) 3a – 5b + c + 2d from 7a – 3b + c – 2d

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

= 7a – 3b + c – 2d – 3a + 5b – c – 2d

= 4a + 2b – 4d

(iv) x³ – 4x – 1 from 3x³ + x² + 6

= (3x³ + 4a + 6) – (x³ – 4x – 1)

= 3x³ – 4a + 6 – x³ + 4x + 1

= 2x³ – x² + 4x + 7

(v) 6a + 3 from a³ – 3a² + 4a + 1

= (a³ – 3a² + 4a + 1) – (6a + 3)

= a³ – 3a² + 4a + 1 – 6a – 3

= a³ – 3a² – 2a – 2

(vi) cab – 4cad – cbd from 3abc + 5bcd – cda

= (3abc + 5bcd – cda) – (cab – 4cad – cbd)

= 3abc + 5bcd – cda – cab + 4cad + cbd

= 2abc + 6bcd + 3cad

(vii) a² + ab + b² from 4a² – 3ab + 2b²

= (4a² – 3ab + 2b²) – (a² + ab + b²)

= 4a² – 3ab + 2b – a² – ab – b²

= 3a² – 4ab – b²

Question no – (5)

Solution :

(i) As per the question, -3x³ + 4x² – 5x + 6 from 3x³ – 4x² + 5x – 6

3x³ – 4x² + 5x – 6 + 3x³ – 4x² + 5x – 6

= 6x³ – 8x² + 10x – 12

(ii) According to the question,

= (-m + 3m + 6) – (m² + m + 4)

= – m² + 3m + 6 – m² – m – 4

= – 2m² – 2m – 1

Now, (m² + m + 1) – (- 2m² – 2m – 1)

= m² + m + 1 + 2m² – 2m – 1

= 3m² – m – 1

Question no – (6)

Solution :

According to the question,

= (5y² + y – 3 + y² – 3y + 7)

= 6y² – 2y + 4

Now, (6y² + y² – 1) – (6y² – 2y + 4)

= 6y² + y – 2 – 6y² – 2y + 4

= 3y – 6

Question no – (7)

Solution :

Given, (x⁴ + x² – 1) – (x4 + x³ + x² + x + 3)

= x⁴ + x² – 1 – x4 + x³ + x² + x + 3

= x³ – x – 4

Thus, x³ – x – 4 must be added.

Question no – (8)

Solution :

(i) (5x² + 10xy – y²) – (2x² + 4xy + 2y²)…(according to the question)

= 5x² + 10xy – y² – 2x² + 4xy + 2y²

= 3x² + 6xy – 3y²

(ii) 2a² + 1 is than 3a² – 6

(3a² – 6) – (2a² + 1)

= a² – 7

Question no – (9)

Solution :

According to the question we know,

x = 6a + 8b + 9c;

y = 2b – 3a – 6c

z = c – b + 3a

(i) x + y + z,

= 6a = 8b + 9c + (- 3a + 2b – 6c) + (3a – b + c)

= 6a + 8b + 9c – 3a + 2b – 6c + 3a – b + c

= 6a + 9b + 4c

(ii) x – y + z,

= (6a + 8b + 9c) – (2b – 3a – 6c) + (c – 6 + 3a)

= 6a + 8b + 9c – 2b + 3a + 6c + c – b + 3a

= 12a + 5b + 16c

(iii) 2x – y – 3z

= 2 (6a + 8b + 9c) – (2b – 3a – 6c) + (c – b + 3a)

= 12a + 16b + 18c – 2b + 3a + 6c + c – b + 3a

= 6a + 17b + 21c

(iv) 3y – 2z – 5x

= 3 (2b – 3a – 6c) – 2 (c – b + 3a) – 5 (6a + 8b + 9c)

= 6b – 9a – 18c – 2c + 2b – 6a – 30a – 40b – 45c

= – 45a – 32b – 65c

Question no – (10)

Solution :

Given, Sides of triangle = x² – 3xy + 8, 4x² + 5xy – 3 and 6 – 3x² + 4xy

Required perimeter,

= x² – 3xy + 8 + 4x² + 5xy – 3 + 6 – 3x² + 4xy

= 2x² + 6xy + 11

Therefore, its perimeter will be 2x² + 6xy + 11

Question no – (11)

Solution :

Triangle perimeter = 8y² – 9y + 4

Sun of two sides = 3y² – 5 + 4y² + 12

= 7y² – 5y + 12

Third sides = (8y² – 9y + 4) – (7y² – 5y + 12)

= 8y² – 9y + 4 – 7y² + 5y + 12

= y² – 4y – 8

Therefore, its third side will be y² – 4y – 8.

Question no – (12)

Solution :

From the question we get,

Two adjacent sides of rectangle = 2x² – 5xy + 3z² and 4xy – x² – z²

Perimeter,

= 2 (2x² – 5xy + 3z² + 4xy – x² – z²)

= 4x² – 10xy + 6z² + 8xy – 2x² – 2z²

= 2x² – 2xy + 4z²

Therefore, its perimeter will be x² – 2xy + 4z².

Question no – (13)

Solution :

Must be subtracted,

= (19x⁴ + 2x³ + 30x – 37) – (8x⁴ + 22x³ – 7x – 60)

= 19x⁴ + 2x³ + 30x – 37 – 8x⁴ + 22x³ – 7x – 60

= 11x⁴ – 20x³ + 37x + 23

Thus, 11x⁴ – 20x³ + 37x + 23 must be subtracted.

Question no – (14)

Solution :

= (22x – 20y – 13z + 26) – (15x – 18y + 19z)…(according to question)

= 22x – 20y – 13z + 26 – 15x – 18y + 19z

= 7x – 2y – 32z + 26

Question no – (15)

Solution :

= (5x²y² – 18xy² – 10x²y) – (- 5x² + 6x²y – 7xy)…(according to question)

= 5x²y² – 18xy² – 10x²y + 5x² + 6x²y – 7xy

= 5x²y² – 18xy² – 15x²y + 5x² + 7xy

Algebraic Expressions Exercise 11(C) Solution :

Question no – (1)

Solution :

(i) Given, 8ab² by – 4a³b⁴

(8 × -4) ab² × a³b⁴

= -32a⁴b⁶

(ii) 2/3ab by -1/4a²b

2/3ab × a²b – 1/4a²b

= (3/2 × -1/4) ab × a²b

= -1/6a³b²

(iii) -5cd² by -5cd²

-5cd² × -5cd²

= (- 5 × -5) cd² × cd²

= + 25c²d⁴

(iv) 4a and 6a + 7

4a (6a + 7)

= 24a² + 28a

(v) -8x and 4 – 2x – x²

– 8x (4 – 2x – x²)

= – 32x + 16x² + 8x³

(vi) 2a² – 5a – 4 and – 3a

– 3a (2a² – 5a – 4)

= – 6a³ + 15a² + 12a

(vii) x + 4 by x – 5

(x + 4) (x – 5)

= x² – 5x + 4x – 20

= x² – x – 20

(viii) 5a – 1 by 7a – 3

(5a – 1) (7a – 3)

= 35a² – 15 – 7a + 3

= 35a² – 22a + 3

(ix) 12a + 5b by 7a – b

84a² – 12ab + 35a – 5b²

= 84a² + 23ab – 5b²

(x) x² + x + 1 by 1 – x

(x² + x + 1) (1 – x)

= x² + x + 1 – 1 – x³ – x² – x

= 1 – x³

(xi) 2m² – 3m – 1 and 4m² – m – 1

(2m² – 3m – 1) (4m² – m – 1)

= 8m⁴ – 2m³ – 2m² – 12m³ + 3m² + 3m – 4m² + m + 1

= 8m⁴ – 14m³ – 6m² + 3m² + 4m + 1

= 8m⁴ – 14m³ – 3m² + 4m + 1

(xii) a², ab and b²

a² × ab × b²

= a³b³

(xiii) abx, – 3a²x and 7b²x³

abx × – 3a²x – 7b²x

= – 21a³b³x⁵

(xiv) -3bx, – 5xy and – 7b³y²

-3bx × – 5xy × – 7b³y²

= – 105b⁴x²y³

(xv) -3/2x⁵y³ and 4/9a²x³y

(-3/2x⁵y³) (4/9a²x³y)

= (-3/2 × 4/9) a²x⁸y⁴

= -2/3 a²x⁸y⁴

(xvi) -2/3a⁷b² and -9/4ab⁵

(- 2/3a⁷b²) (- 9/4ab⁵)

= (- 2/3 × – 9/4) a⁸b⁷

= 3/2a⁸b⁷

(xvii) 2a³ – 3a²b and – 1/2 ab²

(2a³ – 3a²b) (-1/2 ab²)

= 2a³ × -1/2ab² – 3a²b × -1/2ab²

= -a⁴b² + 3/2a³b³

(xviii) 2x + 1/2y and 2x – 1/2y

(2x + 1/2y) (2x – 1/2y)

= 4x² – 1/4y²

Question no – (2)

Solution :

(i) 5x² – 8xy + 6y² – 3 by – 3xy

= (5x² – 8xy + 6y² – 3) × (- 3xy)

= – 15x³ y + 24xy – 18x²y² + 9xy³

(ii) 3 – 2/3xy + 5/7xy – 16/21x²y by – 21x²y²

= (3 – 2/3xy + 5/7xy – 16/21x²y) × (- 21x²y²)

= – 63x²y² + 14x³y³ – 15x³y⁴ + 16x⁴y³

(iii) 6x³ – 5x + 10 by 4 – 3x²

= (6x³ – 5x + 10) (4 – 3x²)

= – 18x⁵ + 39x³ – 30x² – 20x + 40

(iv) 2y – 4y³ + 6y⁵ by y² + y – 3

= (2y – 4y³ + 6y⁵) (y² + y – 3)

= 2y³ + 2y² – 6y – 4y⁵ – 4y⁴ – 12y³ + 6y⁷ + 6y⁶ – 18y⁵

= 6y⁷ + 6y⁶ – 22y⁵ – 4y⁴ + 14y³ + 2y² – 6y

(v) 5p² + 25pq + 4q² by 2p – 2pq + 3q²

= (5p² + 25pq + 4q²) (2p – 2pq + 3q²)

= (10p⁴ + 10p³q + 15p²q² + 50p³q – 50p²q² + 75pq³ + 8p²q² – 8pq³ + 12q⁴

= 10p⁴ + 40p³q – 27p²q² + 67pq³ + 12q⁴

Question no – (3)

Solution :

(i) (7x – 8) (3x + 2)

= 21x² + 14x – 24x – 16

= 21x² – 10x – 16…(Simplified)

(ii) px – q) (px + q)

= (px)² – (q)²

= p²x² – q²…(Simplified)

(iii) (5a + 5b – c) (2b – 3c)

= 10a – 15ac + 10b² – 15bc – 2bc + 3c²

= 10ab + 10b² – 17bc – 15ac + 3c²…(Simplified)

(iv) (4x – 5y) (5x – 4y)

= 20x² – 16xy – 25xy + 20y²

= 20x² – 41xy + 20y²…(Simplified)

(v) (3x + 4z) (3y – 4z) + (2y + 7z) (y + z)

= (3y)² – (4z)² + (2y² + 2yz + 7yz + 7z²)

= 9y²- 16z² + 2y + 9yz + 7z²

= 11y² – 9z² + 9yz…(Simplified)

Question no – (4)

Solution :

In the question,

Adjacent sides of a rectangle are x² – 4xy + 7y² and x³ – 5xy².

Required area,

= (x² – 4xy + 7y²) (x³ – 5xy²)

= (x⁵ – 5x³y² – 4x⁴y + 20x²y³ + 7x³y² – 35xy⁴)

= (x⁵ – 4x⁴y + 2x³y² + 20x²y³ – 35xy⁴)

Thus, the area is (x⁵ – 4x⁴y + 2x³y² + 20x²y³ – 35xy⁴).

Question no – (5)

Solution :

In the given question,

Base and the altitude of a triangle = (3x – 4y) and (6x + 5y)

Required area,

= 1/2 (3x – 4y) (6x + 5y)

= 1/2 (18x² + 15xy – 24xy – 20y²)

= 1/2 (18x² – 9xy – 20y²)

Hence, the area is 1/2 (18x² – 9xy – 20y²)

Question no – (6)

Solution :

(-4xy³) × (6x²)…As per the question,

= – 24x³y⁴

= – 24 × (2)³ × (1)³ …[Given, x = 2, y = 1]

= – 192

Question no – (7)

Solution :

= (3x³) × (-5xy²) × (2x²yz³)

= – 30x⁶y³z³

= – 30 × 1⁶ × 2³ × 3³ ..[Given, x = 1, y = 2, z = 3]

= – 6480

Therefore, the value will be -6480.

Question no – (8)

Solution :

Given, (3x⁴y²) (2x²y³)

= 6x⁶y⁵

= 6 × 1⁶ × 2⁵ …[x = 1, y = 2]

= 192

Question no – (9)

Solution :

Given, x⁵ × (3x²) × (-2x) …For x =1

= 1⁵ × (3 × 1) × (- 5 × 2 × 1)

= -6

Question no – (10)

Solution :

Given, x = 2 and y = 1

(-4x²y³) × (- 5x²y⁵)

= – 16 × – 20

= 320

Therefore, the value will be 320

Question no – (11)

Solution :

(i) (3x – 2) (x + 5) for x = 2.

= (3x – 2) (x + 5)

= (3 × 2 – 2) (2 + 5) …[2 or x = 2]

= 4 × 7

= 28

(ii) x = 2; y = 3

(2x – 5y) (2x + 3y)

= (2 × 2 – 5 × 3) (2 × 2 + 3 × 3)

= (4 – 15) (4 + 9)

= – 11 × 13

= – 143

(iii) x = 2; y = 1, z = 1

xz (x² + y²)

= 2 × 1 (2² + 1²)

= 2 (4 + 1)

= 10

Question no – (12)

Solution :

(i) x (x – 5) + 2 for x = 1.

= x (x – 5) + 2

= 1 (1 – 5) + 2

= 1 × – 4 + 2

= – 4 + 2

= – 2

(ii) xy² (x – 5y) + 1 for x = 1 and y = 1.

= xy² (x – 5y) + 1

= 2 × 1² (2 – 5 × 1) + 1

= 2 (- 3) + 1

= – 6 + 1

= – 5

(iii) 2x(3x – 5 ) – 5(x – 2) – 18 for x = 2.

= 2 × 2 (3 × 2 – 5) – 5 (2 – 2) – 18

= 4 (1) – 18

= -14

Question no – (13)

Solution :

Given, – 3x ²y² (x – 2y)

= – 3 × 1 × 2 (1 – 2 × 2)

= – 12 (1 – 4)

= – 12 × – 3

= 36

Question no – (14)

Solution :

(i) 2x² – 4x + 5 by x² + 3x – 7

(2x² – 4x + 5) (x² + 3x – 7)

= 2x⁴ + 6x³ – 14x² – 4x³ – 12x² + 28x + 5x² + 15x – 35

= 2x⁴ + 2x³ – 21x² + 43x – 35

(ii) (ab – 1) (3 – 2ab)

= 3ab – 2a²b² – 3 + 2ab

= – 2a²b² + 5ab – 3

Question no – (15)

Solution :

Given, (5 – x) (6 – 5x) (2 – x).

[(5 – x) (6 – 5x) (2 – x)]

= (30 – 25x – 6x + 5x) (2 – x)

= (5x – 31x + 30) (2 – x)

= 10x² – 62x + 60 – 5x³ + 31x² – 30x

= -5x² + 41x² – 92x + 60

Algebraic Expressions Exercise 11(D) Solution :

Question no – (1)

Solution :

(i) -70a³ by 14a²

= -70a³/14a²

= -5a

(ii) 24x³ y³ by -8y²

24x³y³/8y²

= -3x³y

(ii) 15a⁴b by -5a³b

15a⁴b/-5a³b

= -3a

(iv) -24x⁴d³ by -2x²d⁵

-24x⁴d³/-2x²d⁵

= 12x²/d²

(iv) 63a⁴b⁵c⁶ by -9a²b⁴c³

63a⁴b⁵c⁶/- 9a²b⁴c³

= -7a²bc b⁴c³

(vi) 8x – 10y + 6c by 2

8x – 10y + 6c/2

= 4x – 5y + 3c

(vii) 15a³b⁴ – 10a⁴b³ – 25a³b⁶ by – 5a³b²

15a³b⁴ – 10a⁴b³ – 25a³b⁶/- 5a³b²

= (15a³b⁴/-5a³b²) – (10a⁴b³/-5a³b²) – (25a³b⁶/-5a³b²)

= -3b² + 2ab + 5b⁴

(viii) -14x⁶y³ – 21x⁴y⁵ + 7x⁵y⁴ by 7x²y²

-14x⁶yc – 21x⁴y⁵ + 7x⁵y⁴/7x²y²

= -2x⁴y – 3x²y³ + x³y²

(ix) a² + 7a + 12 by a + 4 (x) x² + 3x – 54 by x – 6 (xi) 12x² + 7xy – 12y² by 3x + 4y (xii)  x⁶ – 8 by x² – 2 (xiii) 6x³ – 13x² – 13x + 30 by 2x² – x – 6 (xiv) 4a² + 12ab + 9b² – 25c² by 2a + 3b + 5c (xv) 16 + 8x + x⁶ – 8x³ – 2x⁴ + x² by x + 4 – x³ Question no – (2)

Solution :

(i) As per the question,

a³ + 5a² + 8a + 15 is divided by a + 1.

a + 1/a³ + 5a² + 8a + 15

Quotient = a² – 6a + 14

Remainder = 1.

For more better understanding : (ii) Given, 3x⁴ + 6x³ – 6x² + 2x – 7 is divided by x – 3

x – 3 / 3x⁴ + 6x³ – 6x² + 2x – 7

Quotient = 3x³ + 15x² + 39x + 119

Remainder = 350.

For more better understanding : (iii) Given, 6x² + x – 15 is divided by 3x + 5.

3x + 5 / 6x² + x – 15

Quotient = 2x – 3

Remainder = None

Fore more better understanding : Question no – (3)

Solution :

Area of rectangle = x³ – 8x² + 7

One of its sides = x – 1

length of the adjacent side= ?

= x³ – 8x² + 7 / x – 1 Other side length is (x² – 7x – 7)

Question no – (4)

Solution

As per the question we know,

Product of two number = 16x⁴ – 1.

One number = 2x – 1,

Other number = ?

Other number,

= 16x⁴ – 1 / 2x – 1 The other number is 8x³ + 4x² + 2x + 1.

Question no – (5)

Solution :

= (x² + xy + y²) (x – y)…(as per the question)

= x³ – y³

Now, Algebraic Expressions Exercise 11(E) Solution :

Question no – (1)

Solution :

Given, a² – 2a + {5a² – (3a – 4a²)}

= a² – 2a + (5a² – 3a + 4a²)

= a² – 2a + 5a² – 3a + 4a²

= 10a² – 5a

Question no – (2)

Solution :

Given, x – y – {x – y – (x + y) – x – y}

= x – y (x – y – x – y – x + y)

= x – y – x + y + x + y + x – y

= 2x

Question no – (4)

Solution :

2{m – 3(n + m – 2n)}

= 2{m – 3 (n + m – 2n}

= 2 {m – 3 (m – n)}

= 2 (m – 3m + 3n)

= 2 {3n – 2m}

= (6n – 4m)

Question no – (5)

Solution :

Given, 3x – [3x – {3x – (3x – 3x – y)}]

= 3x – [3x – {3x – (3x – 3x + y)}]

= 3x – [3x – {3x – y}]

= 3x – [3x – 3x + y]

= 3x – y

Question no – (6)

Solution :

Given, p²x – 2{px – 3x(x² – 3a – x²)}

= p²x – 2{px – 3x(x² – 3a + x²)}

= p²x – 2{px – 3x (2x – 3a)}

= p²x – 2{10x – 6x³ + 9ax}

= p²x – 2px + 12x³ – 18ax

Question no – (7)

Solution :

Given, 2[6 + 4{m – 6(7 – n + p) +q}]

= 2[6 + 4{m – 6(7 – n – p) +q}]

= 2[6 + 4 (m – 42 + 6n + 6p + q)]

= 2 [6 + 4m – 168 + 24n + 24p + 4q]

= 2 [4m + 24n = 24p + 4q – 162]

= 8m + 48n + 48p + 8q – 324

Question no – (8)

Solution :

Given, a – [a b + a – {a – (a – b – a)}]

= a – [a b + a – {a – (a – b + a)}]

= a – [a – b – a – {a – (a – b + b)}]

= a – [ – b – b + a]

= a + b + b – a

= 2b

Question no – (9)

Solution :

Given, 3x – [4x – 3x – 5y – 3{2x – (3x – 2x – 3y)}]

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

= 3x – [4x – 3x + 5y – 3{2x – x – 3y}]

= 3x – [x + 5y – 6x + 3x + 9y]

= 3x – [- 2x + 14y]

= 3x + 2x – 14y

= 5x – 14y

Question no – (10)

Solution :

Given, a⁵ ÷ a³ + 3a × 2a

= a^5-3 + 6a²

= a² + 6a²

= 7a²

Question no – (11)

Solution :

Given, x⁵ ÷ (x² × y²) × y³

= x⁵/ x² × y² × y³

= x³y

Question no – (13)

Solution :

Given, (y³ – 5y²) ÷ y × (y – 1)

= y³ – 5y²/y × y – 1

= y³ – 6y² + 5y

Question no – (14)

Solution :

Given, 3a × [8b ÷ 4 – 6{a – (5a – 3b – 2a)}]

= 3a × [2b – 6 {a – (5a – 3b + 2a)}]

= 3a × [2b – 6 (a – 7a + 3b)}]

= 3a [2b – 6(- 6a + 3b)}]

= 3a [2b + 36a – 18b]

= 3a [36a – 16b]

= 108a² – 48ab

Next Chapter Solution :

Updated: June 21, 2023 — 4:58 am