NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

 

In NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 with mcq questions and answers for the class 8 including notes as worksheet in pdf which free to download for CBSE exam

Algebraic Expressions and Identities: Here you get NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 with Answers Pdf free download.In NCERT Solution Class 8 Maths with Answers you get question-answer based on latest exam pattern. By providing NCERT Chapter-wise Class 6 Maths Questions with Answers our target to help students get concept of the lesson very well.


You can also visit NCERT Questions for Class 8 Maths Chapter-wise Questions with Answers to revise your syllabus thoroughly and enhance the chances of securing high marks in your board exams. It also helps in your CBSE NCERT Class 8 online test, online exam for admission well.

 

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 Board                    CBSE


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TextBook                NCERT


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CLASS                    Class 8


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SUBJECT                 Maths


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CHAPTER               Chapter 9


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SUBJECT           Algebraic Expressions and Identities                       


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Visit           NCERT SOLUTIONS 

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 Class 8 Maths Algebraic Expressions and Identities Exercise 9.1

Class 8 Maths Algebraic Expressions and Identities Exercise 9.2

Class 8 Maths Algebraic Expressions and Identities Exercise 9.3

Class 8 Maths Algebraic Expressions and Identities Exercise 9.4

Class 8 Maths Algebraic Expressions and Identities Exercise 9.5


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 NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 with Answers



Exercise 9.3


Question 1.

Carry out the multiplication of the expressions in each of the following pairs:

(i) 4p, q + r

(ii) ab, a – b

(iii) a+b,7a2b2

(iv) a2−9,4a

(v) pq + qr + rp, 0


Answer 1: 

Solution:

(i) 4p × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr


(ii)  ab × (a – b) = (ab × a) – (ab × b) = a²b – ab²


(iii) (a + b) × 7a²b² = (a × 7a²b²) + (b × 7a²b²) = 7a³b² + 7a²b³


(iv) (a² – 9) × 4a = (a² × 4a) – (9 × 4a) = 4a³ – 36a


(v) (pq + qr + rp) × 0 = 0

[∵ Any number multiplied by 0 is = 0]




Complete the table.



First Expression

Second Expression

Product

(i)

a

b + c + d

a × (b + c + d)

 = (a × b) + (a × c) + (a × d) 

= ab + ac + ad

(ii) 

x + y – 5

5xy

(x + y – 5) (5xy)

 = (x × 5xy) + (y × 5xy) – (5 × 5xy) 

= 5x²y + 5xy² – 25xy

(iii) 

p

6p² – 7p + 5

p × (6p² – 7p + 5) 

= (p × 6p²) – (p × 7p) + (p × 5) 

= 6p³ – 7p² + 5p

(iv) 

4p²q²

p² – q²

4p²q² × (p² – q²) 

= 4p²q² × p² – 4p²q²× q² = 4p⁴q² – 4p²q⁴

(v) 

a + b + c

abc

(a + b + c) × (abc) 

= (a × abc) + (b × abc) + (c × abc) 

= a²bc + ab²c + abc²




Question 3.

Find the products.


Answer:

 

Algebraic Expressions and Identities

Algebraic Expressions and Identities




 

 

 

Question 4.

(a) Simplify: 3x(4x – 5) + 3 and find its values for (i) x = 3 (ii) x = 12.

(b) Simplify: a(a² + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1


Answer:

Algebraic Expressions and Identities

 
Algebraic Expressions and Identities

 





 

 

 

Question 5.

(a) Add: p(p – q), q(q – r) and r(r – p)


(b) Add: 2x(z – x – y) and 2y(z – y – x)


(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)


(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)


Answer:

(a) p(p – q) + q(q – r) + r(r – p)

= (p × p) – (p × q) + (q × q) – (q × r) + (r × r) – (r × p)

= p² – pq + q² – qr + r² – rp

= p² + q² + r² – pq – qr – rp


(b) 2x(z – x – y) + 2y(z – y – x)

= (2x × z) – (2x × x) – (2x × y) + (2y × z) – (2y × y) – (2y × x)

= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy

= -2x² – 2y² + 2xz + 2yz – 4xy

= -2x² – 2y² – 4xy + 2yz + 2xz


(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)

= (4l × 10n) – (4l × 3m) + (4l × 2l) – (3l × l) – (3l × -4m) – (3l × 5n)

= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln

= (40ln – 15ln) + (-12lm + 12lm) + (8l² – 3l²)

= 25ln + 5l²

= 5l² + 25ln


(d) [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]

= (-4ac + 4bc + 4c²) – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)

= -4ac + 4bc + 4c² – 3a² – 3ab – 3ac + 2ab – 2b² + 2bc

= -3a² – 2b² + 4c² – ab + 6bc – 7ac


NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 with Answers

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