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

CBSE Solutions | Maths Class 8

Check the below NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Pdf free download. NCERT Solutions Class 8 Maths were prepared based on the latest exam pattern. We have Provided Algebraic Expressions and Identities Class 8 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 8 Maths

NCERT Solutions Class 8 Maths
Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	8th
Subject:	Maths
Chapter:	9
Chapters Name:	Algebraic Expressions and Identities
Medium:	English

Algebraic Expressions and Identities | Class 8 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1

Ex 9.1 Class 8 Maths 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) \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\)
(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 \(\frac { x }{ 2 } +\frac { y }{ 2 } -xy\), the terms are \(\frac { x }{ 2 } ,\frac { 7 }{ 2 } \) and – xy.
Coefficient of x in the term \(\frac { x }{ 2 } \) is \(\frac { 1 }{ 2 } \)
Coefficient of y in the term \(\frac { y }{ 2 } \) is \(\frac { 1 }{ 2 } \)
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.

Ex 9.1 Class 8 Maths 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.

Ex 9.1 Class 8 Maths 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
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 1

(ii) Writing the given expressions in separate rows with like terms one below the other, we have
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 2

(iii) Writing the given expressions in separate rows with like terms one below the other, we have
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 3

(iv) Writing the given expressions in separate rows with like terms one below the other, we have
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 4

Ex 9.1 Class 8 Maths 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

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2

Ex 9.2 Class 8 Maths 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

Ex 9.2 Class 8 Maths 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

Ex 9.2 Class 8 Maths Question 3.

Complete the table of products :
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 6

Solution:

Completed table is as under :
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 7

Ex 9.2 Class 8 Maths 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

Ex 9.2 Class 8 Maths 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:

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

Ex 9.3 Class 8 Maths 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

Ex 9.3 Class 8 Maths Question 2.

Complete the table :
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 9

Solution:

Completed table is as under :
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 10

Ex 9.3 Class 8 Maths Question 3.

Find the product:
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 11

Solution:

Ex 9.3 Class 8 Maths Question 4.

(a) Simplify : 3x (4x – 5) + 3 and find its value for
(i) x = 3,
(ii) x = \(\frac { 1 }{ 2 } \)
(b) Simplify : a (a² + a + 1) + 5 and find its value for
(i) a = 0,
(ii) a = 1,
(iii) a = – 1

Solution:

Ex 9.3 Class 8 Maths 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 (1 – 4m + 5n) from 41 (10n – 3m + 21)
(d) Subtract : 3a (a + b + c) – 2b (a – 6 + c) from 4c (-a + b + c)

Solution:

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4

Ex 9.4 Class 8 Maths 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) \(\left( \frac { 3 }{ 4 } { a }^{ 2 }+3{ b }^{ 2 } \right) and\quad 4\left( { a }^{ 2 }-\frac { 2 }{ 3 } { b }^{ 2 } \right) \)

Solution:

Ex 9.4 Class 8 Maths 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:

Ex 9.4 Class 8 Maths 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:

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 21

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Ex 9.5 Class 8 Maths 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) \(\left( 3a-\frac { 1 }{ 2 } \right) \left( 3a-\frac { 1 }{ 2 } \right) \)
(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) \(\left( \frac { x }{ 2 } +\frac { 3y }{ 4 } \right) \left( \frac { x }{ 2 } +\frac { 3y }{ 4 } \right) \)
(x) (7a – 9b)(7a – 9b)

Solution:

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities 24

Ex 9.5 Class 8 Maths 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:

Ex 9.5 Class 8 Maths Question 3.

Find the following squares by using the identities :
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6x² – 5y²)
(iv) \(\left( \frac { 2 }{ 3 } m+\frac { 3 }{ 2 } n \right) ^{ 2 }\)
(v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²

Solution:

Ex 9.5 Class 8 Maths 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:

Ex 9.5 Class 8 Maths Question 5.

Show that :
(i) (3x + 7)² – 84x = (3x – 7)²
(ii) (9p – 5q)² + 180pq = (9p + 5q)²
(iii) \(\left( \frac { 4 }{ 3 } m-\frac { 3 }{ 4 } n \right) ^{ 2 }+2mn=\frac { 16 }{ 9 } { m }^{ 2 }+\frac { 9 }{ 16 } { n }^{ 2 }\)
(iv) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(v) (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0

Solution:

Ex 9.5 Class 8 Maths 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