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 i n 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.
Topics of NCERT for c lass 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 to the exercises of Chapter 9: Algebraic Expressions and Identities
NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.1 what are expressions?
Answer:
Five examples of expressions containing one variable are:
Five examples of expressions containing two variables are:
Solutions of NCERT for class 8 maths chapter 9 algebraic expressions and identities topic 9.2 terms, factors and coefficients
Question:1 Identify the coefficient of each term in the expression.
Answer:
coefficient of each term are given below
NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.3 monomials, binomials and polynomials
Question: 1(i) Classify the following polynomials as monomials, binomials, trinomials.
Answer:
Binomial since there are two terms with non zero coefficients.
Question: 1(ii) Classify the following polynomials as monomials, binomials, trinomials.
Answer:
Trinomial since there are three terms with non zero coefficients.
Question:1(iii) Classify the following polynomials as monomials, binomials, trinomials.
Answer:
Trinomial since there are three terms with non zero coefficients.
Question: 1(iv) Classify the following polynomials as monomials, binomials, trinomials.
Answer:
Binomial since there are two terms with non zero coefficients.
Question: 1(v) Classify the following polynomials as monomials, binomials, trinomials.
Answer:
Monomial since there is only one term.
Question: 2(a) Construct 3 binomials with only as a variable;
Answer:
Three binomials with the only x as a variable are:
Question: 2(b) Construct 3 binomials with and as variables;
Answer:
Three binomials with x and y as variables are:
Question: 2(c) Construct 3 monomials with and as variables;
Answer:
Three monomials with x and y as variables are
Question: 2(d) Construct 2 polynomials with 4 or more terms .
Answer:
Two polynomials with 4 or more terms are:
Solutions of NCERT for class 8 maths chapter 9 algebraic expressions and identities topic 9.4 like and unlike terms
Question:(i) Write two terms which are like
Answer:
NCERT solutions for class 8 maths chapter 9 algebraic expressions and identitiesExercise: 9.1
Question:1(i) Identify the terms, their coefficients for each of the following expressions.
Answer
following are the terms and coefficient
The terms are and the coefficients are 5 and 3.
Question: 1(ii) Identify the terms, their coefficients for each of the following expressions.
Answer:
the following is the solution
Question:1(iii) Identify the terms, their coefficients for each of the following expressions.
Answer:
Question: 1(iv) Identify the terms, their coefficients for each of the following expressions.
Answer:
The terms are 3, pq, qr,and rp and the coefficients are 3, 1, 1 and 1 respectively.
Question:1(v) Identify the terms, their coefficients for each of the following expressions.
Answer:
Above are the terms and coefficients
Question: 1(vi) Identify the terms, their coefficients for each of the following expressions.
Answer:
The terms are 0.3a, 0.6ab and 0.5b and the coefficients are 0.3, 0.6 and 0.5.
Question: 2(c) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
Answer:
This polynomial does not fit in any of these three categories.
Question: 2(d) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
Answer:
Trinomial.
Question: 2(f) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
Answer:
Trinomial.
Question: 2(g) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
Answer:
Trinomial.
Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
Answer:
This polynomial does not fit in any of these three categories.
Question: 4(a) Subtract from
Answer:
12a9ab+5b3(4a7ab+3b+12)
=(124)a +(9+7)ab+(53)b +(312)
=8a2ab+2b15
Solutions for NCERT class 8 maths chapter 9 algebraic expressions and identities topic 9.7.2 multiplying three or more monomials
Question:1 Find . First find and multiply it by ; or first find and multiply it by .
Answer:
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.
NCERT solutions for class 8 maths chapter 9 algebraic expressions and identitiesExercise: 9.2
Question: 1(i) Find the product of the following pairs of monomials.
Answer:
Question:2(A) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
Answer:
The question can be solved as follows
Question:2(B) Find the areas of rectangles with the following pairs of monomials as their lengths and breadth respectively.
Answer:
the area is calculated as follows
Question:2(C) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
Answer:
the following is the solution
Question:2(D) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
Answer:
area of rectangles is
Question:2(E) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
Answer:
The area is calculated as follows
Question:4(ii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
Answer:
the volume of rectangular boxes with the following length, breadth and height is
Question:4(iii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
Answer:
the volume of rectangular boxes with the following length, breadth and height is
Question:4(iv) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
Answer:
the volume of rectangular boxes with the following length, breadth and height is
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
Answer:
Using distributive law,
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:
Answer:
By using distributive law,
NCERT solutions for class 8 maths chapter 9 algebraic expressions and identitiesExercise: 9.3
Question:1(i) Carry out the multiplication of the expressions in each of the following pairs.
Answer:
Multiplication of the given expression gives :
By distributive law,
Question:1(ii)
Carry out the multiplication of the expressions in each of the following pairs.
Answer:
We have ab, (ab).
Using distributive law we get,
Question:1(iii) Carry out the multiplication of the expressions in each of the following pairs.
Answer:
Using distributive law we can obtain multiplication of given expression:
Question:1(iv) Carry out the multiplication of the expressions in each of the following pairs.
Answer:
We will obtain multiplication of given expression by using distributive law :
Question:1(v) Carry out the multiplication of the expressions in each of the following pairs.
Answer:
Using distributive law :
Question:2 Complete the table
Answer:
We will use distributive law to find product in each case.

First expression 
Second expression 
Product 
(i) 



(ii) 



(iii) 



(iv) 



(v) 



Question:5(a) Add: and
Answer:
(a)First we will solve each brackets individually.
; ;
Addind all we get :
Question:5(c) Subtract: from
Answer:
At first we will solve each bracket individually,
and
Subtracting:
or
or
NCERT solutions for class 8 maths chapter 9 algebraic expressions and identitiesExercise: 9.4
Question:1(i) Multiply the binomials.
and
Answer:
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
Answer:
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
Answer:
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
Answer:
(a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5)
= ax + 5a + 3bx + 15b
Question:1(v) Multiply the binomials.
and
Answer:
(2pq + 3q
^{
2
}
) X (3pq  2q
^{
2
}
) = (2pq)(3pq) + (2pq)(2q
^{
2
}
) + ( 3q
^{
2
}
)(3pq) + (3q
^{
2
}
)(2q
^{
2
}
)
= 6p
^{
2
}
q
^{
2
}
 4pq
^{
3
}
+ 9pq
^{
3
}
 6q
^{
4
}
= 6p
^{
2
}
q
^{
2
}
+5pq
^{
3
}
 6q
^{
4
}
Question:2(i) Find the product.
Answer:
(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.
Answer:
(x + 7y) X (7x  y) = (x)(7x) + (x)(y) + (7y)(7x) + (7y)(y)
= 7
 xy + 49xy  7
= 7
+ 48xy  7
Question:3(i) Simplify.
Answer:
this can be simplified as follows
(
5) X (x + 5) + 25 = (
)(x) + (
)(5) + (5)(x) + (5)(5) + 25
=
=
Question:3(ii) Simplify .
Answer:
This can be simplified as
(
+ 5) X (
+ 3) + 5 = (
)(
) + (
)(3) + (5)(
) + (5)(3) + 5
=
=
Question:3(iii) Simplify.
Answer:
simplifications can be
(t +
)(
 s) = (t)(
) + (t)(s) + (
)(
) + (
)(s)
=
Question:3(iv) Simplify.
Answer:
(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.
Answer:
(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.
Answer:
simplification is done as follows
(x + y) X (
) = x X (
) + y (
)
=
=
Question:3(vii) Simplify.
Answer:
(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.
Answer:
(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
Question:1(i) Put b in place of b in identity 1. Do you get identity 2?
Answer:
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 .
Answer:
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
Question:2 Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity
Answer:
Identity IV is
If a =b than
(a + x)(a + x) =
Which is identity I
Question:3 Consider, the special case of Identity (IV) with and What do you get? Is it related to Identity ?
Answer:
Identity IV is
If a = b = c than,
(x  c)(x  c) =
Which is identity II
Question:4 Consider the special case of Identity (IV) with . What do you get? Is it related to Identity?
Answer:
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 identitiesExercise: 9.5
Question:1(i) Use a suitable identity to get each of the following products.
Answer:
(x + 3) X (x +3) =
So, we use identity I for this which is
In this a=x and b = x
=
Question:1(ii) Use a suitable identity to get each of the following products in bracket.
Answer:
(2y + 5) X ( 2y + 5) =
We use identity I for this which is
IN this a = 2y and b = 5
=
Question:1(iii) Use a suitable identity to get each of the following products in bracket.
Answer:
(2a 7) X (2a  7) =
We use identity II for this which is
in this a = 2a and b = 7
=
Question:1(iv) Use a suitable identity to get each of the following products in bracket.
Answer:
We use identity II for this which is
in this a = 3a and b = 1/2
=
Question:1(v) Use a suitable identity to get each of the following products in bracket.
Answer:
We use identity III for this which is
(a  b)(a + b) =
In this a = 1.1m and b = 4
=
= 1.21
 16
Question:1(vi) Use a suitable identity to get each of the following products in bracket.
Answer:
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.
Answer:
(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.
Answer:
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
Question:1(ix) Use a suitable identity to get each of the following product.
Answer:
We use identity I for this which is
In this a =
and b =
=
Question:1(x) Use a suitable identity to get each of the following products.
Answer:
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.
Answer:
We use identity
in this a = 3 and b = 7
=
=
Question:2(ii) Use the identity to find the following products.
Answer:
We use identity
In this a= 5 , b = 1 and x = 4x
=
=
Question:2(iii) Use the identity to find the following products.
Answer:
We use identity
in this x = 4x , a = 5 and b = 1
=
=
Question:1(iv) Use the identity to find the following products.
Answer:
We use identity
In this a = 5 , b = 1 and x = 4x
=
=
Question:2(v) Use the identity to find the following products.
Answer:
We use identity
In this a = 5y , b = 3y and x = 2x
=
=
Question:2(vi) Use the identity to find the following products.
Answer:
We use identity
In this a = 9 , b = 5 and x =
=
=
Question:2(vii) Use the identity to find the following products.
Answer:
We use identity
In this a = 4 , b = 2 and x = xyz
=
=
Question:3(i) Find the following squares by using the identities.
Answer:
We use identity
In this a =b and b = 7
=
Question:3(ii) Find the following squares by using the identities.
Answer:
We use
In this a = xy and b = 3z
=
Question:3(iii) Find the following squares by using the identities.
Answer:
We use
In this a =
and b =
=
Question:3(iv) Find the following squares by using the identities.
Answer:
we use the identity
In this a =
and b =
=
Question:3(v) Find the following squares by using the identities.
Answer:
we use
In this a = 0.4p and b =0.5q
=
Question:3(vi) Find the following squares by using the identities.
Answer:
we use the identity
In this a = 2xy and b =5y
=
Question:4(ii) Simplify.
Answer:
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.
Answer:
we use
and
In this a = 7m and b = 8n
=
and
=
So,
=
+
=
or
remember that
Question: 4(iv) Simplify.
Answer:
we use
1 ) In this a = 4m and b = 5n
=
2 ) in this a = 5m and b = 4n
=
So,
=
+
=
Question: 4(v) Simplify.
Answer:
we use
1 ) In this a = (2.5p 1.5q) and b = (1.5p  2.5q)
=
= 4(p + q ) (p  q)
= 4
Question:7(i) Using , find
Answer:
We know,
Using this formula,
= (51 + 49)(51  49)
= (100)(2)
= 200
Question:7(ii) Using , find
Answer:
We know,
Using this formula,
= (1.02 + 0.98)(1.02  0.98)
= (2.00)(0.04)
= 0.08
Question:7(iii) Using , find.
Answer:
We know,
Using this formula,
= (153  147)(153 +147)
=(6) (300)
= 1800
Question:7(iv) Using , find
Answer:
We know,
Using this formula,
= (1.02 + 0.98)(1.02  0.98)
= (2.00)(0.04)
= 0.08
Question:8(i) Using
Answer:
We know,
Using this formula,
= (100 + 3)(100 + 4)
Here x =100, a = 3, b = 4
= 11212
Question:8(ii) Using , find
Answer:
We know,
Using this formula,
= (5 + 0.1)(5 + 0.2)
Here x =5, a = 0.1, b = 0.2
= 26.52
Question:8(iii) Using , find
Answer:
We know,
Using this formula,
= (100 + 3)(100  2) = (100 + 3){100 + (2)}
Here x =100, a = 3, b = 2
= 10094
Question: 8(iv) Using , find
Answer:
We know,
Using this formula,
= (10  0.3)(10  0.2) = {10 + (0.3)}{10 + (0.2)}
Here x =10, a = 0.3, b = 0.2
= 95.
NCERT Solutions For Class 8 Maths: Chapterwise
NCERT Solutions For Class 8: SubjectWise
Some important identities from NCERT solutions for class 8 chapter 9 algebraic expressions and identities
you can write
Can be simplified as follows
Now add each term
You can form the above identities by yourself. These above identities have been used in many problems of solutions of NCERT for class 8 maths chapter 9 algebraic expression and identities.
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