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

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities.

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 9
Chapter Name	Algebraic Expressions and Identities
Exercise	Ex 9.1, Ex 9.2, Ex 9.3, Ex 9.4, Ex 9.5
Number of Questions Solved	25
Category	NCERT Solutions

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

Chapter 9 Algebraic Expressions and Identities Exercise 9.1

Question 1.
Identify the terms, their coefficients for each of the following expressions :
(i) 5xyz² – 3zy
(ii) 1 + x + x²
(iii) 4x²y² – 4x²y²z² + z²
(iv) 3 – pq + qr -rp
(v)
(vi) 0.3a – 0.6ab + 0.5b
Solution:
(i) In the expression 5xyz² – 3zy, the terms are 5xyz² and -3zy.
Coefficient of xyz² in the term 5xyz² is 5.
Coefficient of zy in the term – 3yz is – 3.

(ii) In the expression 1 + x + x², the terms are 1, x and x².
Coefficient of term 1 is 1.
Coefficient of x in the term x is 1.
Coefficient of x² in the term x² is 1.

(iii) In the expression 4x y – 4xyz + z , the terms are 4x²y², – 4x²y²z² and z².
Coefficient of x²y² in the term 4x²y² is 4.
Coefficient of x²y²z² in the term – 4x²y²z² is – 4.
Coefficient of z² in the term z2 is 1.

(iv) In the expression 3 – pq + qr – rp, the terms are 3, – pq, qr and – rp.
Coefficient of term 3 is 3.
Coefficient of pq in the term – pq is -1.
Coefficient of qr in the term qr is 1.
Coefficient of rp in the term – rp is -1.

(v) In the expression , the terms are and – xy.
Coefficient of x in the term is
Coefficient of y in the term is
Coefficient of xy in the term – xy is -1.

(vi) In the expression 0.3a -0.6 ab + 0.5 b, the terms are 0.3a, – 006ab and 0.5b.
Coefficient of a in the term 0.3a is 0.3.
Coefficient of ab in the term – 0.6ab is – 0.6.
Coefficient of b in the term 0.5b is 0.5.

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z2, ab + bc+cd + da, pqr, p²q + pq², 2p + 2q.
Solution:
The given polynomials are classified as under :
Monomials : 1000, pqr
Binomials : x + y, 2y – 3y², 4z -15z², p²q + pq², 2p + 2q.
Trinomials : 7 + y + 5x, 2y – 3y² + 4y³, 5x – 4y + 3x.
Polynomials that do not fit in any categories :
x + x² + x³ + x⁴, ab + be + cd + da.

Question 3.
Add the following :
(i) ab – be, be – ca, ea – ab
(ii) a – b + ab, b-c + be, c – a + ac
(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
(iv) l² + m², m² + n², n² + l², 2Im + 2mn + 2nl
Solution:
(i) Writing the given expressions in separate rows with like terms one below the other, we have

(ii) Writing the given expressions in separate rows with like terms one below the other, we have

(iii) Writing the given expressions in separate rows with like terms one below the other, we have

(iv) Writing the given expressions in separate rows with like terms one below the other, we have

Question 4.
(a) Subtract
4a – lab +36 + 12 from 12a – 9a6 + 56-3
(b) Subtract
3xy + 5yz – Izx from 5xy – 2yz – 2zx + IQxyz
(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q.
Solution:
Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get

Chapter 9 Algebraic Expressions and Identities Exercise 9.2

Question 1.
Find the product of the following pairs of monomials :
(i) 4, 7p
(ii) -4p, 7p
(iii) – 4p, 7pq
(iv) 4p³ , – 3p
(v) 4p, 0
Solution:
(i) 4 x 7p = (4 x 7) x p = 28p
(ii) – 4 p x 7p = (- 4 x 7) x (px P)
= -28p¹ + ¹ = – 28p²
(iii) -4px 7pq = (- 4 x 7) x (p x p x q)
= -28 p¹⁺¹q = – 28p²q
(iv) 4p³ x – 3p = (4 x – 3) x (p³ x p)
= – 12p³⁺¹ = =-12p⁴
(v) 4p x 0 = (4 x 0) x p = 0 x p = 0

Question 2.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively :
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)
Solution:
We know that the area of a rectangle = l x b, where l = length and b = breadth.
Therefore, the areas of rectangles with pair of monomials (p, q); (10m, 5n); (20x², 5y²); (4x, 3x²) and (3mn, 4np) as their lengths and breadths are given by
pxq=pq
10 m x 5n = (10 x 5) x (m x n) = 50 mn
20x² x 5y² = (20 x 5) x (x² x y²) = 100x²y²
4x x 3x² = (4 x 3) x (x x x²)
= 12x²
and, 3 mn x 4np = (3×4 )x(mxnxnxp)
=12 mn²p

Question 3.
Complete the table of products :

Solution:
Completed table is as under :

Question 4.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively
(i) 5a, 3a², 7a⁴
(ii) 2p, 4q, 8r
(iii) xy, 2x²y, 2xy²
(iv) a, 2b, 3c
Solution:
(i) Required volume = 5a x 3a² x 7a⁴
= (5 x 3 x 7) x (a x a² x a⁴)
= 105a¹⁺²⁺⁴ =105a⁷

(ii) Required volume = 2p x 4q x 8r
= (2 x 4 x 8 )x p x q x r = 64 pqr

(iii) Required volume =xy x 2x²y x 2xy²
= (1 x 2 x 2) x (x x x² x x x y x y x y²)
= 4x¹⁺²⁺¹ y¹⁺¹⁺² = 4x⁴y⁴

(iv) Required volume =a x 2b x 3c
= (1 x 2 x 3) x (a x b x c)
= 6abc

Question 5.
Obtain the product of
(i) xy, yz, zx
(ii) a, – a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 26, 3c, 6a6c
(v) m, – mn, mnp
Solution:

Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Question 1.
Carry out the multiplication of the expressions in each of the following pairs :
(i) 4p, q r + r
(ii) ab, a – -b
(iii) a + b, 7a²b²
(iv) a²– 9, 4a
(v) pq + qr + rp, 0
Solution:
(i) 4p x (q + r) = 4px q + 4p x r
= 4pq + 4pr
(ii) ab x (a – b) = ab x a – ab
= a²b – ab²
(iii) (a + b) x 7a²b² = a x 7a²b² + b x 7a²b²
= 7a³b² + 7a²b³
(iv) (a² -9)x 4a = a² x 4a – 9 x 4a
= 4a³ – 36a
(v) (pq + qr + rp) x 0 = 0

Question 2.
Complete the table :

Solution:
Completed table is as under :

Question 3.
Find the product:

Solution:

Question 4.
(a) Simplify : 3x (4x – 5) + 3 and find its value for
(i) x = 3,
(ii) x =
(b) Simplify : a (a² + a + 1) + 5 and find its value for
(i) a = 0,
(ii) a = 1,
(iii) a = – 1
Solution:

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 : 31 (l – 4m + 5n) from 41 (lOn – 3m + 21)
(d) Subtract : 3a (a + b + c) – 2b (a – 6 + c) from 4c (-a + b + c)
Solution:

Chapter 9 Algebraic Expressions and Identities Exercise 9.4

Question 1.
Multiply the binomials :
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q²) and (3pq – 2q²)
(vi)
Solution:

Question 2.
Find the product :
(i) (5 – 2x)(3 + x)
(ii) (x + 7y)(7x – y)
(iii) (a² + b)(a + b² )
(iv) (p² – q²)(2p + q)
Solution:

Question 3.
Simplify :
(i) (x² – 5)(x + 5) + 25
(ii) (a² +5)(b³ +3)+ 5
(iii) (t +s²)(² – s)
(iv) (a + b)(c – d) + (a – b)(c + d) + 2 (ac + bd)
(v) (x + y)(2x + y)+ (x + 2y)(x – y)
(vi) (x + y)(x² – xy + y²)
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + b – c)
Solution:

Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Question 1.
Use a suitable identity to get each of the following products,
(i) (x + 3) (x + 3)
(ii) (2y + 5)(2y + 5)
(iii) (2a – 7) (2a – 7)
(iv)
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a² + b²) (- a² + b²)
(vii) (6x – 7) (6x + 7)
(viii) (- a + c) (- a + c)
(ix)
(x) (7a – 9b)(7a – 9b)
Solution:

Question 2.
Use the identity (x + a) (x + b) = x² + (a + b)x + ab to find the following products.
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4² – 5) (4x – 1)
(iv) (4x + 5) (4² – 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a² + 9) (2a² + 5)
(vii) (xyz – 4) (xyz – 2)
Solution:

Question 3.
Find the following squares by using the identities :
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6x² – 5y²)
(iv)
(v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²
Solution:

Question 4.
Simplify :
(i) (a² – b²)²
(ii) (2x + 5)² – (2x – 5)²
(iii) (7m – 8n)² + (7m + 8n)²
(iv) (4m + 5n)² + (5m + 4n)²
(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
(vi) (ab + bc)² – 2ab²c
(vii) (m² – n²m)² + 2m³n²
Solution:

Question 5.
Show that :
(i) (3x + 7)² – 84x = (3x – 7)²
(ii) (9p – 5q)² + 180pq = (9p + 5q)²
(iii)
(iv) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(v) (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0
Solution:

Question 6.
Using identities, evaluate : .
(i) 71²
(ii) 99²
(iii) 102²
(iv) 998²
(v) 5.2²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.9²
(ix) 1.05 x 9.5
Solution:

Question 7.
Using a² – b² = (a + b) (a – b), find
(i) 51² – 49²
(ii) (1.02)² – (0.98)²
(iii) 153² – 147²
(iv) 12.1² – 7.9²
Solution:

Question 8.
Using (x + a) (x + b) = x + (a + b)x + ab, find
(i) 103 x 104
(ii) 5.1 x 5.2
(iii) 103 x 98
(iv) 9.7 x 9.8
Solution:

We hope the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities, drop a comment below and we will get back to you at the earliest.