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

NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities- As you have learned in previous classes that expressions are formed from variables and constants. This chapter will introduce you to the application of algebraic terms and variables to solve various problems. NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities will give you a detailed explanation for every problem of this chapter. Algebra is the most important branch of mathematics which teaches how to form equations and solving them using different kinds of techniques. Important topics like the product of the equation, finding the coefficient of the variable in the equation, subtraction of the equation and creating quadratic equation by the product of its two roots, and division of the equation are covered in this chapter. You will find the questions related to these topics in CBSE NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities which will make your task easy while solving the problems. As it is a new concept, you may find difficulties while dealing with algebraic parts of mathematics but if you practice questions and go through solutions of NCERT for class 8 maths chapter 9 algebraic expressions and identities, you will find it very easy and one of the strongest parts in mathematics. You will get NCERT solutions from class 6 to 12 for science and maths by clicking on the above link.  Here you will solutions to five exercise of this chapter.

Exercise:9.1

Exercise:9.2

Exercise:9.3

Exercise:9.4

Exercise:9.5

## Topics of NCERT for class 8 maths chapter 9 - Algebraic expressions and identities:-

9.1 What are Expressions?

9.2 Terms, Factors and Coefficients

9.3 Monomials, Binomials and Polynomials

9.4 Like and Unlike Terms

9.5 Addition and Subtraction of Algebraic Expressions

9.6 Multiplication of Algebraic Expressions: Introduction

9.7 Multiplying a Monomial by a Monomial

9.7.1 Multiplying two monomials

9.7.2 Multiplying three or more monomials

9.8 Multiplying a Monomial by a Polynomial

9.8.1 Multiplying a monomial by a binomial

9.8.2 Multiplying a monomial by a trinomial

9.9Multiplying a Polynomial by a Polynomial

9.9.1 Multiplying a binomial by a binomial

9.9.2 Multiplying a binomial by a trinomial

9.10 What is an Identity?

9.11 Standard Identities

9.12 Applying Identities

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.1 what are expressions?

Five examples of expressions containing one variable are:

Five examples of expressions containing two variables are:

Question:

x on the number line: Question:

x-4 on the number line: Question:

2x+1

2x+1 on the number  line: Question:

3x - 2 on the number line ## Question:1 Identify the coefficient of each term in the expression.

coefficient of each term are given below

## Question:1(i) Classify the following polynomials as monomials, binomials, trinomials.

Binomial since there are two terms with non zero coefficients.

Trinomial since there are three terms with non zero coefficients.

Trinomial since there are three terms with non zero coefficients.

Binomial since there are two terms with non zero coefficients.

Monomial since there is only one term.

Question:2(a) Construct 3 binomials with only  as a variable;

Three binomials with the only x as a variable are:

Question:2(b) Construct 3 binomials with  and  as variables;

Three binomials with x and y as variables are:

Question:2(c) Construct 3 monomials with  and  as variables;

Three monomials with x and y as variables are

Question:2(d) Construct

Two polynomials with 4 or more terms are:

## Question:(i) Write two terms which are like

Question:(ii) Write two terms which are like

we can write more like terms

Question:(iii)

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.1

following are the terms and coefficient

The terms are  and the coefficients are 5 and -3.

the following is the solution

The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.

Above are the terms and coefficients

The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.

This polynomial does not fit in any of these three categories.

This polynomial does not fit in any of these three categories.

Question:

ab-bc+bc-ca+ca-ab=0.

Question:3

Question:4(a) Subtract   from

12a-9ab+5b-3-(4a-7ab+3b+12)
=(12-4)a +(-9+7)ab+(5-3)b +(-3-12)
=8a-2ab+2b-15

Question:

Question:4(c) Subtract   from

## Solutions for NCERT class 8 maths chapter 9 algebraic expressions and identities topic 9.7.2 multiplying three or more monomials

We observe that the result is same in both cases and the result does not depend on the order in which multiplication has been carried out.

## Question:1(i) Find the product of the following pairs of monomials.

Question:

Question:

Question:

The question can be solved as follows

the area is calculated as follows

the following is the solution

area of rectangles  is

The area is calculated as follows

Question:3 Complete the table of products.

 First monomial  Second monomial ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...

 First monomial  Second monomial

the volume of rectangular boxes with the following length, breadth and height is

the volume of rectangular boxes with the following length, breadth and height is

the volume of rectangular boxes with the following length, breadth and height is

Question:5(i) Obtain the product of

the product

Question:5(ii) Obtain the product of

the product

Question:5(iii) Obtain the product of

the product

Question:5(iv) Obtain the product of

the product

Question:5(v) Obtain the product of

the product

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.8.1 multiplying a monomial by a binomial

Question:(i) Find the product

Using distributive law,

Question:(ii)  Find the product

Using distributive law,

We have :

## NCERT textbook solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.8.2 multiplying a monomial by a trinomial

Question:1 Find the product:

By using distributive law,

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.3

Multiplication of the given expression gives :

By distributive law,

We have ab,  (a-b).

Using distributive law we get,

Using distributive law we can obtain multiplication of given expression:

We will obtain multiplication of given expression by using distributive law :

Using distributive law :

Question:2 Complete the table

 First expression Second expression Product (i) ... (ii) ... (iii) ... (iv) ... (v) ...

We will use distributive law to find product in each case.

 First expression Second expression Product (i) (ii) (iii) (iv) (v)

Question:3(i)  Find the product.

Opening brackets :

or

Question:3(ii) Find the product.

We have,

Question:3(iii) Find the product.

We have

Question:3(iv)  Find the product.

We have

or

Question:4(a) Simplify    and find its values for

(a)  We have

Put x = 3,

We get :

Question:4(a) Simplify       and find its values for

(ii)

We have

Put

.    So We get,

Question:4(b)

We have :

Put a = 0 :

Question:4(b) Simplify   and find its value for

(ii)

We have

Put a = 1 ,

we get :

Question:4(b) Simplify  and find its value for

(iii)

We have      .

or

Put       a = (-1)

(a)First we will solve each brackets individually.

;                  ;

Firstly, open the brackets:

and

or

Question:5(c) Subtract:   from

At first we will solve each bracket individually,

and

Subtracting:

or

or

Question:5(d) Subtract:    from

Solving brackets :

and

Subtracting :

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.4

Question:1(i)  Multiply the binomials.

and

We have    (2x + 5) and (4x - 3)
(2x + 5)  X   (4x - 3)   =  (2x)(4x) + (2x)(-3) + (5)(4x) + (5)(-3)
= 8 - 6x + 20x - 15
= 8 + 14x -15

Question:1(ii) Multiply the binomials.

and

We need to multiply (y - 8) and (3y - 4)
(y - 8) X  (3y - 4)  =   (y)(3y) +  (y)(-4) + (-8)(3y) + (-8)(-4)
=   3 - 4y - 24y  + 32
= 3 - 28y + 32

Question:1(iii)  Multiply the binomials

and

We need to multiply (2.5l - 0.5m) and (2.5l + 0..5m)
(2.5l - 0.5m) X (2.5l + 0..5m)    =                                     using
= 6.25  - 0.25

Question:1(iv)  Multiply the binomials.

and

(a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5)
= ax + 5a + 3bx + 15b

Question:1(v) Multiply the binomials.

and

(2pq + 3q2) X (3pq - 2q2)  =  (2pq)(3pq) + (2pq)(-2q2) + ( 3q2)(3pq) + (3q2)(-2q2)
= 6p2q2 - 4pq3  + 9pq3 - 6q4
= 6p2q2 +5pq3 - 6q4

Question:1(vi) Multiply the binomials.

and

Multiplication can be done as follows

X          =

=

=

Question:2(i) Find the product.

(5 - 2x) X (3 + x) =  (5)(3) + (5)(x) +(-2x)(3) + (-2x)(x)
= 15 + 5x - 6x - 2
= 15  - x - 2

Question:2(ii) Find the product.

(x + 7y) X (7x - y) =  (x)(7x) + (x)(-y) + (7y)(7x) + (7y)(-y)
= 7 - xy + 49xy - 7
=  7 + 48xy - 7

Question:2(iii) Find the product.

( + b) X (a + )  = ()(a) + ()() + (b)(a) + (b)()
=

Question:2(iv) Find the product.

following is the solution

() X (2p + q)  =

Question:3(i) Simplify.

this can be simplified as follows

( -5) X (x + 5) + 25  = ()(x) + ()(5) + (-5)(x) + (-5)(5)  + 25
=
=

Question:3(ii) Simplify.

This can be simplified as

( + 5) X ( + 3) + 5 = ()() + ()(3) + (5)() + (5)(3) + 5
=
=

Question:3(iii) Simplify.

simplifications can be

(t + )( - s) =  (t)() + (t)(-s)  + ()()  + ()(-s)
=

Question:3(iv) Simplify.

(a + b) X ( c -d) + (a - b) X (c + d) + 2(ac + bd )
=   (a)(c) + (a)(-d) + (b)(c) + (b)(-d) +   (a)(c) + (a)(d) + (-b)(c) + (-b)(d)  + 2(ac + bd )
= ac - ad + bc - bd + ac +ad -bc - bd   + 2(ac + bd )
=  2(ac - bd )  + 2(ac +bd )
=  2ac - 2bd + 2ac  + 2bd
= 4ac

Question:3(v) Simplify.

(x + y) X ( 2x + y) + (x + 2y) X (x - y)
=(x)(2x) + (x)(y) + (y)(2x) + (y)(y)  + (x)(x) + (x)(-y) + (2y)(x) + (2y)(-y)
= 2 + xy + 2xy +  +  - xy + 2xy - 2
=3 + 4xy -

Question:3(vi) Simplify.

simplification is done as follows

(x + y) X ()  = x X () + y ()
=
=

Question:3(vii) Simplify.

(1.5x - 4y) X (1.5x + 4y + 3) - 4.5x + 12y =   (1.5x)  X (1.5x + 4y + 3)  -4y X (1.5x + 4y + 3)   - 4.5x + 12y
=  2.25 + 6xy + 4.5x  - 6xy - 16 - 12y   -4.5x + 12 y
= 2.25    - 16

Question:3(viii) Simplify.

(a + b + c) X (a + b - c)  = a  X (a + b - c) + b  X (a + b - c) + c  X (a + b - c)
=  + ab - ac + ab +  -bc + ac + bc -
=  +   -  + 2ab

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.11 standerd identities

Identity 1
If we replace b with -b in identity 1
We get,

which is equal to
which is identity 2
So, we get identity 2 by replacing b with -b in identity 1

## NCERT free solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.11 standard identities

Question:1 Verify Identity (IV), for    .

Identity IV
(a + x)(b + x) =
So, it is given that a = 2, b = 3 and x = 5
Lets put these value in identity  IV
(2 + 5)(3 + 5) =  + (2 + 3)5 +2 X 3
7 X 8 = 25 + 5 X 5 + 6
56 = 25 + 25 + 6
= 56
L.H.S. = R.H.S.
So,  by this we can say that identity IV satisfy with given value of a,b and x

Identity IV is
If a =b than

(a + x)(a + x) =

Which is identity I

Identity IV is
If a = b = -c  than,
(x - c)(x - c) =

Which is identity II

Identity IV is
If b = -a than,

(x + a)(x - a) =
=
Which is identity III

## NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.5

(x + 3) X (x +3) =
So, we use identity I for this which is

In  this  a=x and b = x

=

(2y + 5) X ( 2y + 5) =
We use identity I for this which is

IN this  a = 2y and b = 5

=

(2a -7) X (2a - 7)  =
We use identity II for this which is

in this a = 2a and b = 7

=

We use identity II for this which is

in this a = 3a and b = -1/2

=

We use identity III for this which is
(a - b)(a + b) =
In this a = 1.1m  and b = 4
=
=  1.21 - 16

take the (-)ve sign common so our question becomes
-
We use identity III for this which is
(a - b)(a + b) =
In this a =  and b =

=

Question:1(vii)  Use a suitable identity to get each of the following.

(6x -7) X (6x - 7) =
We use identity III for this which is
(a - b)(a + b) =
In this a = 6x and b = 7
(6x -7) X (6x - 7) =

Question:1(viii) Use a suitable identity to get each of the following product.

take (-)ve sign common from both the brackets So, our question become
(a -c) X (a -c) =
We use identity II for this which is

In this a = a and b = c

We use identity I for this which is

In this a =  and b =

=

We use identity II for this which is

In this a = 7a and b = 9b

=

Question:2(i) Use the identity     to find the following products.

We use identity
in this a = 3 and b = 7
=
=

Question:2(ii) Use the identity     to find the  following products.

We use identity
In this a= 5 , b = 1 and x = 4x
=
=

Question:2(iii) Use the identity      to find the following products.

We use  identity
in this x = 4x , a = -5 and b = -1
=
=

Question:1(iv)  Use the identity    to find the following products.

We use identity
In this a = 5 , b = -1 and x = 4x
=
=

Question:2(v) Use the identity     to find the following products.

We use identity
In this a = 5y , b = 3y and x = 2x
=
=

Question:2(vi) Use the identity      to find the following products.

We use identity
In this a = 9 , b = 5 and x =
=
=

Question:2(vii)  Use the identity      to find the following products.

We use identity
In this a = -4 , b = -2 and x = xyz
=
=

Question:3(i) Find the following squares by using the identities.

We use identity

In this a =b and b = 7

=

Question:3(ii) Find the following squares by using the identities.

We use

In this a = xy and b = 3z

=

Question:3(iii)Find the following squares by using the identities.

We use

In this a =  and b =

=

Question:3(iv)  Find the following squares by using the identities.

we use the identity

In this a =   and b =

=

Question:3(v) Find the following squares by using the identities.

we use

In this a = 0.4p  and b =0.5q

=

Question:3(vi) Find the following squares by using the identities.

we use the identity

In this a = 2xy  and b =5y

=

Question:4(i) Simplify:

we use

In this a =  and b =

=

Question:4(ii) Simplify.

we use

In this a = (2x + 5) and b = (2x - 5)

=
= (4x)(10)
=40x

or

remember that

here a= 2x, b= 5

Question:4(iii)  Simplify.

we use
and
In this a = 7m and b = 8n

=
and

=

So,     =     +
=

or

remember that

Question: 4(iv) Simplify.

we use

1 )   In this a = 4m and b = 5n

=
2 )      in this  a = 5m and b = 4n

=

So,        =     +
=

Question: 4(v) Simplify.

we use

1 )   In this a = (2.5p- 1.5q) and b = (1.5p - 2.5q)

=
= 4(p + q ) (p - q)
= 4

Question:4(vi) Simplify.

We use identity

In this a =  ab and b = bc

=
Now,     -
=

Question:4(vii)  Simplify.

We use identity

In this a =   and b =

=
Now,      +
=

Question:5(i) Show that

L.H.S. =

=  R.H.S.

Hence it is prooved

Question:5(ii) Show that

L.H.S. =                  (Using )

= R.H.S.

Question:5(iii) Show that.

First we will solve the LHS :

or

=  RHS

Question:5(iv)  Show that.

Opening both brackets we get,

= R.H.S.

Question:5(v) Show that

Opening all brackets from the LHS, we get :

=  RHS

Question:6(i) Using identities, evaluate.

We will use the identity:

So,

Question:6(ii)  Using identities, evaluate.

Here we will use the identity :

So :

or

Question:6(iii) Using identities, evaluate.

Here we will use the identity :

So :

or

Question:6(iv) Using identities, evaluate.

Here we will the identity :

or

or

Question:6(v) Using identities, evaluate.

Here we will use :

Thus

or

Question:6(vi) Using identities, evaluate.