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

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

Edited By Ramraj Saini | Updated on Oct 09, 2023 05:20 PM IST

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

Algebraic Expressions and Identities Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by experts team at Careers360 team considering the latest syllabus and pattern of CBSE 2023-24. This chapter will introduce you to the application of algebraic terms and variables to solve various problems. 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. Also Practice NCERT solutions for class 8 maths to command the concepts.

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NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities
Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download
Algebraic Expressions and Identities Class 8 Solutions - Important Formulae
Algebraic Expressions and Identities Class 8 NCERT Solutions (Intext Questions and Exercise)
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Topics
NCERT Solutions for Class 8 Maths - Chapter Wise
NCERT Solutions for Class 8 - Subject Wise
Also Check NCERT Books and NCERT Syllabus here

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

Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download

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Algebraic Expressions and Identities Class 8 Solutions - Important Formulae

(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
(a + b)(a - b) = a² - b²
(x + a)(x + b) = x² + (a + b)x + ab
(x + a)(x - b) = x² + (a - b)x - ab
(x - a)(x + b) = x² + (b - a)x - ab
(x - a)(x - b) = x² - (a + b)x + ab
(a + b)³ = a³ + b³ + 3ab(a + b)
(a - b)³ = a³ - b³ - 3ab(a - b)

Free download NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities for CBSE Exam.

Algebraic Expressions and Identities Class 8 NCERT Solutions (Intext Questions and Exercise)

NCERT Solutions to Exercises of Chapter 9: Algebraic Expressions and Identities

what are expressions?

Question: 1 Give five examples of expressions containing one variable and five examples of expressions containing two variables.

Answer:

Five examples of expressions containing one variable are:

$x^{^{4}}, y, 3z, p^{^{2}}, -2q^{3}$

Five examples of expressions containing two variables are:

$x + y, 3p-4q,ab,uv^{2},-z^{2}+x^{3}$

Question: 2(i) Show on the number line

$x$

Answer:

x on the number line:

1643105164197

Question: 2(ii) Show on the number line :

$x-4$

Answer:

x-4 on the number line:

1643105231337

Question: 2(iii) Show on the number line :

2x+1

Answer:

2x+1 on the number line:

c360_4-1

Question: 2(iv) Show on the number line:

$3x-2$

Answer:

3x - 2 on the number line

1643105272368

Algebraic expressions and identities class 8 solutions - Topic 9.2 Terms, Factors and Coefficients

Question:1 Identify the coefficient of each term in the expression.

$x^2y^2-10x^2y+5xy^2-20$

Answer:

coefficient of each term are given below

$\\The\ coefficient\ of\ x^{2}y^{2}\ is \1\\ \\The\ coefficient\ of\ x^{2}y\ is \ -10\\ \\The\ coefficient\ of\ xy^{2}\ is \5\\$

Algebraic expressions and identities class 8 ncert solutions - Topic 9.3 Monomials, Binomials and Polynomials

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

$-z+5$

Answer:

Binomial since there are two terms with non zero coefficients.

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

$x+y+z$

Answer:

Trinomial since there are three terms with non zero coefficients.

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

$y+z+100$

Answer:

Trinomial since there are three terms with non zero coefficients.

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

$ab-ac$

Answer:

Binomial since there are two terms with non zero coefficients.

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

$17$

Answer:

Monomial since there is only one term.

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

Answer:

Three binomials with the only x as a variable are:

$\\ \\x+2,\ x +x^{2},\ 3x^{3}-5x^{4}$

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

Answer:

Three binomials with x and y as variables are:

$\\ \\x+y,\ x-7y, xy^{2} + 2xy$

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

Answer:

Three monomials with x and y as variables are

$\\ xy,\ 3xy^{4},\ -2x^{3}y^{2}$

Question: 2(d) Construct 2 polynomials with 4 or more terms .

Answer:

Two polynomials with 4 or more terms are:

$a+b+c+d, x-3xy+2y+4xy^{2}$

NCERT Solutions 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

$7xy$

Answer:

$\\Two\ terms\ like\ 7xy\ are:\\ -3xy\ and\ 5xy$

Question:(ii) Write two terms which are like

$4mn^2$

Answer:

$\\Two\ terms\ which\ are\ like\ 4mn^{2}\ are:\\ mn^{2}\ and -3mn^{2.}$

we can write more like terms

Question:(iii) Write two terms which are like

$2l$

Answer:

$\\Two\ terms\ which\ are\ like\ 2l\ are:\\ l\ and\ -3l$

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

Question:1(i) Identify the terms, their coefficients for each of the following expressions.

$5xyz^2-3zy$

Answer

following are the terms and coefficient

The terms are $5xyz^{2}\ and\ -3zy$ and the coefficients are 5 and -3.

Question: 1(ii) Identify the terms, their coefficients for each of the following expressions.

$1+x+x^2$

Answer:

the following is the solution

$\\The\ terms\ are\ 1,\ x,\ and\ x^{2}\ and\ the\ coefficients\ are\ 1,\ 1,\ and\ 1\ respectively.$

Question:1(iii) Identify the terms, their coefficients for each of the following expressions.

$4x^2y^2-4x^2y^2z^2+z^2$

Answer:

$\\The\ terms\ are\ 4x^{2}y^{2},\ -4x^{2}y^{2}z^{2}and\ z^{2}\ and\ the\ coefficients\ are\ 4,\ -4\ and\ 1\ respectively.$

Question: 1(iv) Identify the terms, their coefficients for each of the following expressions.

$3-pq+qr-rp$

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.

$\frac{x}{2}+\frac{y}{2}-xy$

Answer:

$\\The\ terms\ are\ \frac{x}{2},\ \frac{y}{2}\ and\ -xy\ and\ the\ coefficients\ are\ \frac{1}{2},\ \frac{1}{2}\ and\ -1\ respectively.$

Above are the terms and coefficients

Question: 1(vi) Identify the terms, their coefficients for each of the following expressions.

$0.3a-0.6ab+0.5b$

Answer:

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

Question: 2(a) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$x+y$

Answer:

Binomial.

Question: 2(b) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$1000$

Answer:

Monomial.

Question: 2(c) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$x+x^2+x^3+x^4$

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?

$7+y-5x$

Answer:

Trinomial.

Question: 2(e) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$2y-3y^2$

Answer:

Binomial.

Question: 2(f) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$2y-3y^2+4y^3$

Answer:

Trinomial.

Question: 2(g) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$5x-4y+3xy$

Answer:

Trinomial.

Question: 2(h) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$4z-15z^2$

Answer:

Binomial.

Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$ab+bc+cd+da$

Answer:

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

Question:2(j) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$pqr$

Answer:

Monomial.

Question: 2(k) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$p^2q+pq^2$

Answer:

Binomial.

Question: 2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

$2p+2q$

Answer:

Binomial.

Question: 3(i) Add the following.

$ab-bc , bc -ca, ca-ab$

Answer:

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

Question:3 (ii) Add the following.

$a-b+ab, b-c+bc, c-a+ac$

Answer:

$\\a-b+ab+b-c+bc+c-a+ac\\ =(a-a)+(b-b)+(c-c)+ab+bc+ac\\ =ab+bc+ca$

Question:3 (iii) Add the following

$2p^2q^2-3pq+4, 5+7pq-3p^2q^2$

Answer:

$\\2p^{2}q^{2}-3pq+4+5+7pq-3p^{2}q^{2}\\ =(2-3)p^{2}q^{2} +(-3+7)pq +4+5\\ =-p^{2}q^{2}+4pq+9$

Question: 3(iv) Add the following.

$l^2+m^2+n^2 , n^2+l^2, 2lm+2mn+2nl$

Answer:

$\\l^{2}+m^{2}+n^{2}+n^{2}+l^{2}+2lm+2mn+2nl\\ =2l^{2}+m^{2}+2n^{2}+2lm+2mn+2nl$

Question: 4(a) Subtract $4a-7ab+3b+12$ from $12a-9ab+5b-3$

Answer:

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

Question: 4(b) Subtract $3xy+5yz-7zx$ from $5xy-2yz-2zx+10xyz$

Answer:

$\\5xy-2yz-2zx+10xyz-(3xy+5yz-7zx)\\ =(5-3)xy+(-2-5)yz+(-2+7)zx+10xyz\\ =2xy-7yz+5zx+10xyz$

Question: 4(c) Subtract $4p^2q - 3pq + 5pq^2 - 8p + 7q - 10$ from $18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q$

Answer:

$\\18-3p-11q+5pq-2pq^{2}+5p^{2}q-(4p^{2}q-3pq+5pq^{2}-8p+7q-10)\\ =18-(-10)-3p-(-8p)-11q-7q+5pq-(-3pq)-2pq^{2}-5pq^{2}+5p^{2}q-4p^{2}q\\ =28+5p-18q+8pq-7pq^{2}+p^{2}q$

NCERT class 8 maths chapter 9 question answer - Topic 9.7.2 Multiplying Three or More Monomials

Question:1 Find $4x\times 5y\times 7z$ . First find $4x\times 5y$ and multiply it by $7z$ ; or first find $5y \times 7z$ and multiply it by $4x$ .

Answer:

$\\4x\times 5y\times 7z\\ =(4x\times 5y)\times 7z\\ =20xy\times 7z\\ =140xyz\\ \\4x\times 5y\times 7z\\ =(5y\times 7z)\times 4x\\ =35yz\times 4x\\ =140xyz$

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.

Class 8 maths chapter 9 question answer - exercise: 9.2

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

$4,7p$

Answer:

$4\times 7p=28p$

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

$-4p,7p$

Answer:

$\\-4p\times 7p\\=(-4\times 7)p\times p\\=-28p^{2}$

Question: 1(iii) Find the product of the following pairs of monomials

$-4p,7pq$

Answer:

$-4p\times 7pq\\=(-4\times 7)p\times pq\\=-28p^{2}q$

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

$4p^3,-3p$

Answer:

$\\4p^{3}\times (-3p)\\ =4\times (-3)p^{3}\times p\\=-12p^{4}$

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

$4p,0$

Answer:

$\\4p\times 0=0$

Question:2(A) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$(p,q)$

Answer:

The question can be solved as follows

$\\Area=length\times breadth\\ =(p\times q)\\ =pq$

Question:2(B) Find the areas of rectangles with the following pairs of monomials as their lengths and breadth respectively.

$(10m,5n)$

Answer:

the area is calculated as follows

$\\Area=length\times breadth\\ =10m\times 5n\\ =50mn$

Question:2(C) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$(20x^2,5y^2)$

Answer:

the following is the solution

$\\Area=length\times breadth\\ =20x^{2}\times 5y^{2}\\ =100x^{2}y^{2}$

Question:2(D) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$(4x,3x^2)$

Answer:

area of rectangles is

$\\Area=length\times breadth\\ =4x\times 3x^{2}\\ =12x^{3}$

Question:2(E) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

$(3mn,4np)$

Answer:

The area is calculated as follows

$\\Area=length\times breadth\\ =3mn\times 4np\\ =12mn^{2}p$

Question:3 Complete the table of products.

First monomial $\rightarrow$ Second monomial $\downarrow$	$2x$	$-5y$	$3x^2$	$-4xy$	$7x^2y$	$-9x^2y^2$
$2x$	$4x^2$	...	...	...	...	...
$-5y$	...	...	$-15x^2y$	...	...	...
$3x^2$	...	...	...	...	...	...
$-4xy$	...	...	...	...	...	...
$7x^2y$	...	...	...	...	...	...
$-9x^2y^2$	...	...	...	...	...	...

Answer:

First monomial $\rightarrow$ Second monomial $\downarrow$	$2x$	$-5y$	$3x^{2}$	$-4xy$	$7x^{2}y$	$-9x^{2}y^{2}$
$2x$	$4x^{2}$	$-10xy$	$6x^{3}$	$-8x^{2}y$	$14x^{3}y$	$-18x^{3}y^{2}$
$-5y$	$-10xy$	$25y^{2}$	$-15x^{2}y$	$20xy^{2}$	$-35x^{2}y^{2}$	$45x^{2}y^{3}$
$3x^{2}$	$6x^{3}$	$-15x^{2}y^{}$	$9x^{4}$	$-12x^{3}y$	$21x^{4}y$	$-27x^{4}y^{2}$
$-4xy$	$-8x^{2}y$	$20xy^{2}$	$-12x^{3}y$	$16x^{2}y^{2}$	$-28x^{3}y$	$36x^{3}y^{3}$
$7x^{2}y$	$14x^{3}y$	$-35x^{2}y^{2}$	$21x^{4}y$	$-28x^{3}y^{2}$	$49x^{4}y^{2}$	$-63x^{4}y^{3}$
$-9x^{2}y^{2}$	$-18x^{3}y^{2}$	$45x^{2}y^{3}$	$-27x^{4}y^{2}$	$36x^{3}y^{3}$	$-63x^{4}y^{3}$	$81x^{4}y^{4}$

Question:4(i) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

$5a, 3a^2, 7a^4$

Answer:

$\\Volume=length\times breadth\times height\\ =5a\times 3a^{2}\times 7a^{4}\\ =15a^{3}\times 7a^{4}\\ =105a^{7}$

Question:4(ii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

$2p,4q,8r$

Answer:

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

$\\Volume=length\times breadth\times height\\ =2p\times 4q\times 8r\\ =8pq\times 8r\\ =64pqr$

Question:4(iii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

$xy, 2x^2y, 2xy^2$

Answer:

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

$\\Volume=length\times breadth\times height\\ =xy\times 2x^{2}y\times 2xy^{2}\\ =2x^{3}y^{2}\times 2xy^{2}\\ =4x^{4}y^{4}$

Question:4(iv) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

$a, 2b, 3c$

Answer:

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

$\\Volume=length\times breadth\times height\\ =a\times 2b\times 3c\\ =2ab\times 3c\\ =6abc$

Question:5(i) Obtain the product of

$xy,yz,zx$

Answer:

the product

$\\xy\times yz\times zx\\ =xy^{2}z\times zx\\ =x^{2}y^{2}z^{2}$

Question:5(ii) Obtain the product of

$a,-a^2,a^3$

Answer:

the product

$\\a\times (-a^{2})\times a^{3}\\ =-a^{^{3}}\times a^{3} =-a^{6}$

Question:5(iii) Obtain the product of

$2,\ 4y,\ 8y^{2},\ 16y^{3}$

Answer:

the product

$\\2\times 4y\times 8y^{2}\times 16y^{3}\\ =8y\times 8y^{2}\times 16y^{3}\\ =64y^{3}\times 16y^{3}\\ =1024y^{6}$

Question:5(iv) Obtain the product of

$a, 2b, 3c, 6abc$

Answer:

the product

$\\a\times 2b\times 3c\times 6abc\\ =2ab\times 3c\times 6abc\\ =6abc\times 6abc\\ =36a^{2}b^{2}c^{2}$

Question:5(v) Obtain the product of

$m, -mn, mnp$

Answer:

the product

$\\m\times (-mn)\times mnp\\ =-m^{2}n\times mnp\\ =-m^{3}n^{2}p$

Class 8 maths chapter 9 NCERT solutions - Topic 9.8.1 Multiplying a Monomial by a Binomial

Question:(i) Find the product

$2x(3x+5xy)$

Answer:

Using distributive law,

$2x(3x + 5xy) = 6x^2 + 10x^2y$

Question:(ii) Find the product

$a^2(2ab-5c)$

Answer:

Using distributive law,

We have : $a^2(2ab-5c) = 2a^3b - 5a^2c$

NCERT 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:

$(4p^2+5p+7)\times 3p$

Answer:

By using distributive law,

$(4p^2+5p+7)\times 3p = 12p^3 + 15p^2 + 21p$

Class 8 maths chapter 9 NCERT solutions - exercise: 9.3

Question:1(i) Carry out the multiplication of the expressions in each of the following pairs.

$4p, q+r$

Answer:

Multiplication of the given expression gives :

By distributive law,

$(4p)(q+r) = 4pq + 4pr$

Question:1(ii) Carry out the multiplication of the expressions in each of the following pairs.

$ab, a-b$

Answer:

We have ab, (a-b).

Using distributive law we get,

$ab(a-b) = a^2b - ab^2$

Question:1(iii) Carry out the multiplication of the expressions in each of the following pairs.

$a+b, 7a^2b^2$

Answer:

Using distributive law we can obtain multiplication of given expression:

$(a + b)(7a^2b^2) = 7a^3b^2 + 7a^2b^3$

Question:1(iv) Carry out the multiplication of the expressions in each of the following pairs.

$a^2-9,4a$

Answer:

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

$(a^2 - 9 )(4a) = 4a^3 - 36a$

Question:1(v) Carry out the multiplication of the expressions in each of the following pairs.

$pq+qr+rp, 0$

Answer:

Using distributive law :

$(pq + qr + rp)(0) = pq(0) + qr(0) + rp(0) = 0$

Question:2 Complete the table

	First expression	Second expression	Product
(i)	$a$	$b+c+d$	...
(ii)	$x+y-5$	$5xy$	...
(iii)	$p$	$6p^2-7p+5$	...
(iv)	$4p^2q^2$	$p^2-q^2$	...
(v)	$a+b+c$	$abc$	...

Answer:

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

	First expression	Second expression	Product
(i)	$a$	$b+c+d$	$ab + ac+ ad$
(ii)	$x+y-5$	$5xy$	$5x^2y + 5xy^2 - 25xy$
(iii)	$p$	$6p^2-7p+5$	$6p^3 - 7p^2 + 5p$
(iv)	$4p^2q^2$	$p^2-q^2$	$4p^4q^2 - 4p^2q^4$
(v)	$a+b+c$	$abc$	$a^2bc + ab^2c + abc^2$

Question:3(i) Find the product.

$(a^2)\times (2a^{22})\times (4a^{26})$

Answer:

Opening brackets :

$(a^2)\times (2a^{22})\times (4a^{26}) = (a^2\times2a^{22})\times(4a^{26}) = 2a^{24}\times4a^{26}$

or $=8a^{50}$

Question:3(ii) Find the product.

$(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2)$

Answer:

We have,

$(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2) = \frac{-3}{5}x^3y^3$

Question:3(iii) Find the product.

$(\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q)$

Answer:

We have

$(\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q) = -4p^4q^4$

Question:3(iv) Find the product.

$x \times x^2\times x^3\times x^4$

Answer:

We have $x \times x^2\times x^3\times x^4$

$x \times x^2\times x^3\times x^4 = (x \times x^2)\times x^3\times x^4$

or $(x^3)\times x^3\times x^4$

$= x^{10}$

Question:4(a) Simplify $3x(4x-5)+3$ and find its values for

(i) $\small x=3$

Answer:

(a) We have

$3x(4x-5)+3 = 12x^2 - 15x + 3$

Put x = 3,

We get : $12(3)^2 - 15(3) + 3 = 12(9) - 45 + 3 = 108 - 42 = 66$

Question:4(a) Simplify $\small 3x(4x-5)+3$ and find its values for

(ii) $\small x=\frac{1}{2}$

Answer:

We have

$\small 3x(4x-5)+3 = 12x^2 -15x + 3$

Put

$x = \frac{1}{2}$

. So We get,

$12x^2 -15x + 3 = 12(\frac{1}{2})^2 - 15(\frac{1}{2}) + 3 = 6 - \frac{15}{2} = \frac{-3}{2}$

Question:4(b) Simplify $\small a(a^2+a+1) + 5$ and find its value for

(i) $\small a =0$

Answer:

We have : $\small a(a^2+a+1) +5 = a^3 + a^2 + a +5$

Put a = 0 : $= 0^3 + 0^2 + 0 + 5 = 5$

Question:4(b) Simplify $\small a(a^2+a+1)+5$ and find its value for

(ii) $\small a=1$

Answer:

We have $\small a(a^2+a+1)+5 = a^3 + a^2 + a + 5$

Put a = 1 ,

we get : $1^3 + 1^2 + 1 + 5 = 1 + 1 + 1+ 5 = 8$

Question:4(b) Simplify $\small a(a^2+a+1)+5$ and find its value for

(iii) $\small a=-1$

Answer:

We have $\small a(a^2+a+1)+5$ .

or $\small a(a^2+a+1)+5 = a^3+a^2+a+5$

Put a = (-1)

$= (-1)^3+(-1)^2+(-1)+5 = -1 + 1 -1 +5 = 4$

Question:5(a) Add: $p(p-q),q(q-r)$ and $r(r-p)$

Answer:

(a)First we will solve each brackets individually.

$p(p-q) = p^2 - pq$ ; $q(q-r) = q^2 - qr$ ; $r(r-p) = r^2 - rp$

Addind all we get : $p^2 - pq + q^2 - qr + r^2 - rp$

$= p^2 + q^2 + r^2 -pq-qr-rp$

Question:5(b) Add: $\small 2x(z-x-y)$ and $\small 2y(z-y-x)$

Answer:

Firstly, open the brackets:

$\small 2x(z-x-y) = 2xz -2x^2-2xy$

and $\small 2y(z-y-x) = 2yz-2y^2-2xy$

Adding both, we get :

$\small 2xz -2x^2-2xy +2yz-2y^2-2xy$

or $\small = -2x^2-2y^2-4xy + 2xz+2yz$

Question:5(c) Subtract: $\small 3l(l-4m+5n)$ from $\small 4l(10n-3m+2l)$

Answer:

At first we will solve each bracket individually,

$\small 3l(l-4m+5n) = 3l^2 - 12lm + 15ln$

and $\small 4l(10n-3m+2l) = 40ln - 12ml + 8l^2$

Subtracting:

$\small 40ln - 12ml + 8l^2 - (3l^2 - 12lm+15ln)$

or $\small = 40ln - 12ml + 8l^2 - 3l^2 + 12lm-15ln$

or $\small = 25ln + 5l^2$

Question:5(d) Subtract: $\small 3a(a+b+c)-2b(a-b+c)$ from $\small 4c(-a+b+c)$

Answer:

Solving brackets :

$3a(a+b+c)-2b(a-b+c) = 3a^2+3ab+3ac - 2ab+2b^2-2bc$

$= 3a^2+ab+3ac+ 2b^2-2bc$

and $\small 4c(-a+b+c) = -4ac +4bc + 4c^2$

Subtracting : $\small -4ac +4bc + 4c^2 -(3a^2 + ab + 3ac+2b^2-2bc)$

$\small = -4ac + 4bc+4c^2-3a^2-ab-3ac-2b^2+2bc$

$\small =-3a^2 -2b^2+4c^2-ab+ 6bc-7ac$

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities-Exercise: 9.4

Question:1(i) Multiply the binomials.

$\small (2x+5)$ and $\small (4x-3)$

Answer:

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

Question:1(ii) Multiply the binomials.

$\small (y-8)$ and $\small (3y-4)$

Answer:

We need to multiply (y - 8) and (3y - 4)
(y - 8) X (3y - 4) = (y)(3y) + (y)(-4) + (-8)(3y) + (-8)(-4)
= 3 $y^{2}$ - 4y - 24y + 32
= 3 $y^{2}$ - 28y + 32

Question:1(iii) Multiply the binomials

$\small (2.5l-0.5m)$ and $\small (2.5l+0.5m)$

Answer:

We need to multiply (2.5l - 0.5m) and (2.5l + 0..5m)
(2.5l - 0.5m) X (2.5l + 0..5m) = $(2.5l)^{2} - (0.5m)^{2}$ using $(a-b)(a+b) = (a)^{2} - (b)^{2}$
= 6.25 $l^{2}$ - 0.25 $m^{2}$

Question:1(iv) Multiply the binomials.

$\small (a+3b)$ and $\small (x+5)$

Answer:

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

Question:1(v) Multiply the binomials.

$\small (2pq+3q^2)$ and $\small (3pq-2q^2)$

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:1(vi) Multiply the binomials.

$\small (\frac{3}{4}a^2+3b^2)$ and $\small 4(a^2-\frac{2}{3}b^2)$

Answer:

Multiplication can be done as follows

$\small (\frac{3}{4}a^2+3b^2)$ X $\small (4a^2-\frac{8}{3}b^2)$ = $\frac{3a^{2}}{4} \times 4a^{2} + \frac{3a^{2}}{4} \times (-\frac{8b^{2}}{3}) + 3b^{2} \times 4a^{2} + 3b^{2} \times (-\frac{8b^{2}}{3})$

= $3a^{4} - 2a^{2}b^{2} + 12a^{2}b^{2} - 8b^{4}$

= $3a^{4} + 10a^{2}b^{2} - 8b^{4}$

Question:2(i) Find the product.

$\small (5-2x)$ $\small (3+x)$

Answer:

(5 - 2x) X (3 + x) = (5)(3) + (5)(x) +(-2x)(3) + (-2x)(x)
= 15 + 5x - 6x - 2 $x^{2}$
= 15 - x - 2 $x^{2}$

Question:2(ii) Find the product.

$\small (x+7y)(7x-y)$

Answer:

(x + 7y) X (7x - y) = (x)(7x) + (x)(-y) + (7y)(7x) + (7y)(-y)
= 7 $x^{2}$ - xy + 49xy - 7 $y^{2}$
= 7 $x^{2}$ + 48xy - 7 $y^{2}$

Question:2(iii) Find the product.

$\small (a^2+b)(a+b^2)$

Answer:

( $a^{2}$ + b) X (a + $b^{2}$ ) = ( $a^{2}$ )(a) + ( $a^{2}$ )( $b^{2}$ ) + (b)(a) + (b)( $b^{2}$ )
= $a^{3 } + a^{2}b^{2} + ab + b^{3}$

Question:2(iv) Find the product.

$\small (p^2-q^2)(2p+q)$

Answer:

following is the solution

( $p^{2}- q^{2}$ ) X (2p + q) = $(p^{2})(2p) + (p^{2})(q) + (-q^{2})(2p) + (-q^{2})(q)$
$2p^{3} + p^{2}q - 2q^{2}p - q^{3}$

Question:3(i) Simplify.

$\small (x^2-5)(x+5)+25$

Answer:

this can be simplified as follows

( $x^{2}$ -5) X (x + 5) + 25 = ( $x^{2}$ )(x) + ( $x^{2}$ )(5) + (-5)(x) + (-5)(5) + 25
= $x^{3} + 5x^{2} - 5x -25 + 25$
= $x^{3} + 5x^{2} - 5x$

Question:3(ii) Simplify .

$(a^2+5)(b^3+3)+5$

Answer:

This can be simplified as

( $a^{2}$ + 5) X ( $b^{3}$ + 3) + 5 = ( $a^{2}$ )( $b^{3}$ ) + ( $a^{2}$ )(3) + (5)( $b^{3}$ ) + (5)(3) + 5
= $a^{2}b^{3} + 3a^{2} + 5b^{3} + 15+5$
= $a^{2}b^{3} + 3a^{2} + 5b^{3} + 20$

Question:3(iii) Simplify.

$(t+s^2)(t^2-s)$

Answer:

simplifications can be

(t + $s^{2}$ )( $t^{2}$ - s) = (t)( $t^{2}$ ) + (t)(-s) + ( $s^{2}$ )( $t^{2}$ ) + ( $s^{2}$ )(-s)
= $t^{3} - ts + s^{2}t^{2} - s^{3}$

Question:3(iv) Simplify.

$(a+b)(c-d)+(a-b)(c+d)+2 (ac+bd)$

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.

$(x+y)(2x+y)+(x+2y)(x-y)$

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 $x^{2}$ + xy + 2xy + $y^{2}$ + $x^{2}$ - xy + 2xy - 2 $y^{2}$
=3 $x^{2}$ + 4xy - $y^{2}$

Question:3(vi) Simplify.

$(x+y)(x^2-xy+y^2)$

Answer:

simplification is done as follows

(x + y) X ( $x^{2} -xy + y^{2}$ ) = x X ( $x^{2} -xy + y^{2}$ ) + y ( $x^{2} -xy + y^{2}$ )
= $x^{3} -x^{2}y + xy^{2} + yx^{2} - xy^{2} + y^{3}$
= $x^{3}+ y^{3}$

Question:3(vii) Simplify.

$(1.5x-4y)(1.5x+4y+3)-4.5x+12y$

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 $x^{2}$ + 6xy + 4.5x - 6xy - 16 $y^{2}$ - 12y -4.5x + 12 y
= 2.25 $x^{2}$ - 16 $y^{2}$

Question:3(viii) Simplify.

$(a+b+c)(a+b-c)$

Answer:

(a + b + c) X (a + b - c) = a X (a + b - c) + b X (a + b - c) + c X (a + b - c)
= $a^{2}$ + ab - ac + ab + $b^{2}$ -bc + ac + bc - $c^{2}$
= $a^{2}$ + $b^{2}$ - $c^{2}$ + 2ab

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities - Topic 9.11 Standard Identities

Question:1(i) Put -b in place of b in identity 1. Do you get identity 2?

Answer:

Identity 1 $\Rightarrow (a+b)^{2} = a^{2} + 2ab + b^{2}$
If we replace b with -b in identity 1
We get,
$a^{2} + 2a(-b) + (-b)^{2} = a^{2} - 2ab + b^{2}$
which is equal to
$(a-b)^{2}$ 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 $a=2,b=3,x=5$ .

Answer:

Identity IV
(a + x)(b + x) = $x^{2} + (a+b)x + ab$
So, it is given that a = 2, b = 3 and x = 5
Lets put these value in identity IV
(2 + 5)(3 + 5) = $5^{2}$ + (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 $\Rightarrow (a +x)(b+x) = x^{2} + (a+b)x + ab$
If a =b than

(a + x)(a + x) = $x^{2} + (a+a)x + a\times a$
$(a+x)^{2} = x^{2} + 2ax + a^{2}$
Which is identity I

Question:3 Consider, the special case of Identity (IV) with $a=-c$ and $b=-c$ What do you get? Is it related to Identity ?

Answer:

Identity IV is $\Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab$
If a = b = -c than,
(x - c)(x - c) = $x^{2} + (-c + (-c))x + (-c) \times (-c)$
$(x-c)^{2} = x^{2} + -2cx + c^{2}$
Which is identity II

Question:4 Consider the special case of Identity (IV) with $b=-a$ . What do you get? Is it related to Identity?

Answer:

Identity IV is $\Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab$
If b = -a than,

(x + a)(x - a) = $x^{2} + (a +(-a))x + (-a) \times a$
= $x^{2} - a^{2}$
Which is identity III

Class 8 algebraic expressions and identities NCERT solutions - exercise: 9.5

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

$(x+3)(x+3)$

Answer:

(x + 3) X (x +3) = $(x +3)^{2}$
So, we use identity I for this which is
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a=x and b = x
$(x+3)^{2} = x^{2} + 2(x)(3)+ 3^{2}$
= $x^{2} + 6x+ 9$

Question:1(ii) Use a suitable identity to get each of the following products in bracket.

$(2y+5)(2y+5)$

Answer:

(2y + 5) X ( 2y + 5) = $(2y +5)^{2}$
We use identity I for this which is
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
IN this a = 2y and b = 5
$(2y+5)^{2} = (2y)^{2} + 2(2y)(5) + 5^{2}$
= $(2y+5)^{2} = 4y^{2} + 20y + 25$

Question:1(iii) Use a suitable identity to get each of the following products in bracket.

$(2a-7)(2a-7)$

Answer:

(2a -7) X (2a - 7) = $(2a - 7)^{2}$
We use identity II for this which is
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
in this a = 2a and b = 7
$(2a-7)^{2} = (2a)^{2} - 2(2a)(7) + 7^{2}$
= $4a^{2} - 28a + 49$

Question:1(iv) Use a suitable identity to get each of the following products in bracket.

$(3a - \frac{1}{2}) (3a -\frac{1}{2} )$

Answer:

$(3a - \frac{1}{2}) \times (3a -\frac{1}{2} ) = ((3a - \frac{1}{2}))^{2}$
We use identity II for this which is
$(a-b)^{2} = a^{2} -2ab + b^{2}$
in this a = 3a and b = -1/2
$(3a-\frac{1}{2})^{2} = (3a)^{2} -2(3a)(\frac{1}{2}) + (\frac{1}{2})^{2}$
= $9a^{2} -3a + \frac{1}{4}$

Question:1(v) Use a suitable identity to get each of the following products in bracket.

$(1.1m - 4)(1.1m+4)$

Answer:

$(1.1m - 4)(1.1m+4)$
We use identity III for this which is
(a - b)(a + b) = $a^{2} - b^{2}$
In this a = 1.1m and b = 4
$(1.1m - 4)(1.1m+4)$ = $(1.1m)^{2} - (4)^{2}$
= 1.21 $m^{2}$ - 16

Question:1(vi) Use a suitable identity to get each of the following products in bracket.

$(a^2+b^2)(-a^2+b^2)$

Answer:

take the (-)ve sign common so our question becomes
- $-(a^{2}+b^{2})(a^{2}-b^{2})$
We use identity III for this which is
(a - b)(a + b) = $a^{2} - b^{2}$
In this a = $a^{2}$ and b = $b^{2}$

$-(a^{2}+b^{2})(a^{2}-b^{2})$ = $-((a^{2})^{2} -(b^{2})^{2}) = -a^{4} + b^{4}$

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

$(6x-7) (6x+7)$

Answer:

(6x -7) X (6x - 7) = $(6x-7)^{2}$
We use identity III for this which is
(a - b)(a + b) = $a^{2} - b^{2}$
In this a = 6x and b = 7
(6x -7) X (6x - 7) = $(6x)^{2} - (7)^{2} = 36x^{2} - 49$

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

$(-a+c)(-a+c)$

Answer:

take (-)ve sign common from both the brackets So, our question become
(a -c) X (a -c) = $(a -c)^{2}$
We use identity II for this which is
$(a-b)^{2} =a^{2} -2ab + b^{2}$
In this a = a and b = c
$(a-c)^{2} =a^{2} -2ac + c^{2}$

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

$(\frac{x}{2}+ \frac{3y}{4})(\frac{x}{2}+ \frac{3y}{4})$

Answer:

$(\frac{x}{2}+ \frac{3y}{4}) \times (\frac{x}{2}+ \frac{3y}{4}) = (\frac{x}{2}+ \frac{3y}{4})^{2}$

We use identity I for this which is
$(a+b)^{2} =a^{2}+2ab + b^{2}$
In this a = $\frac{x}{2}$ and b = $\frac{3y}{4}$

$(\frac{x}{2}+ \frac{3y}{4})^{2} = (\frac{x}{2})^{2} + 2 (\frac{x}{2})(\frac{3y}{4}) + (\frac{3y}{4})^{2}$
= $\frac{x^{2}}{4} + \frac{3xy}{4} + \frac{9y^{2}}{16}$

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

$(7a-9b)(7a-9b)$

Answer:

$(7a-9b) \times (7a-9b) = (7a-9b)^{2}$

We use identity II for this which is
$(a-b)^{2} =a^{2}-2ab + b^{2}$
In this a = 7a and b = 9b
$(7a-9b)^{2} =(7a)^{2}-2(7a)(9b) + (9b)^{2}$
= $49a^{2}-126ab + 81b^{2}$

Question:2(i) Use the identity $(x+a) (x+b) = x^2+(a+b)x+ab$ to find the following products.

$(x+3)(x+7)$

Answer:

We use identity $(x+a) (x+b) = x^2+(a+b)x+ab$
in this a = 3 and b = 7
$(x+3)(x+7)$ = $x^2+(3+7)x+3 \times 7$
= $x^2+10x+ 21$

Question:2(ii) Use the identity $(x+a)(x+b)=x^2+(a+b)x+ab$ to find the following products.

$(4x+5)(4x+1)$

Answer:

We use identity $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a= 5 , b = 1 and x = 4x
$(4x+5)(4x+1)$ = $(4x)^2+(5+1)4x+(5)(1)$
= $16x^2+24x+5$

Question:2(iii) Use the identity $(x+a)(x+b)= x^2+(a+b)x+ab$ to find the following products.

$(4x-5)(4x-1)$

Answer:

We use identity $(x+a)(x+b)= x^2+(a+b)x+ab$
in this x = 4x , a = -5 and b = -1
$(4x-5)(4x-1)$ = $(4x)^2+(-5-1)4x+(-5)(-1)$
= $16x^2 - 24x+ 5$

Question:1(iv) Use the identity $(x+a)(x+b)=x^2+(a+b)x+ab$ to find the following products.

$(4x+5)(4x-1)$

Answer:

We use identity $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = 5 , b = -1 and x = 4x
$(4x+5)(4x-1)$ = $(4x)^2+(5+(-1))4x+(5)(-1)$
= $16x^2+16x- 5$

Question:2(v) Use the identity $(x+a)(x+b)=x^2+(a+b)x+ab$ to find the following products.

$(2x+5y)(2x+3y)$

Answer:

We use identity $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = 5y , b = 3y and x = 2x
$(2x+5y)(2x+3y)$ = $(2x)^2+(5y+3y)(2x)+(5y)(3y)$
= $4x^2+16xy + 15y^{2}$

Question:2(vi) Use the identity $(x+a)(x+b)=x^2+(a+b)x+ab$ to find the following products.

$(2a^2+9)(2a^2+5)$

Answer:

We use identity $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = 9 , b = 5 and x = $2a^{2}$
$(2a^{2}+9)(2a^{2}+5)$ = $(2a^{2})^2+(9+5)2a^{2}+(9)(5)$
= $4a^{4} + 28a^{2} + 45$

Question:2(vii) Use the identity $(x+a) (x+b)=x^2+(a+b)x+ab$ to find the following products.

$(xyz-4) (xyz-2)$

Answer:

We use identity $(x+a)(x+b)=x^2+(a+b)x+ab$
In this a = -4 , b = -2 and x = xyz
$(xyz-4)(xyz-2)$ = $(xyz)^2+((-4)+(-2))xyz+(-4)(-2)$
= $x^{2}y^{2}z^{2} -6xyz + 8$

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

$(b-7)^2$

Answer:

We use identity
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a =b and b = 7
$(b-7)^{2} = b^{2} - 2(b)(7) + 7^{2}$
= $b^{2} - 14b + 49$

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

$(xy+3z)^2$

Answer:

We use
$(a+b)^{2} = a^{2} + 2ab + b^{2}$

In this a = xy and b = 3z
$(xy+3z)^{2} = (xy)^{2} + 2(xy)(3z) + (3z)^{2}$
= $x^{2}y^{2} + 6xyz+ 9z^{2}$

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

$(6x^2-5y)^2$

Answer:

We use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $6x^{2}$ and b = $5y^{2}$
$(6x-5y)^{2} = (6x)^{2} - 2(6x)(5y) + (5y)^{2}$
= $36x^{2} - 60xy + 25y^{2}$

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

$(\frac{2}{3}m+\frac{3}{2}n)^2$

Answer:

we use the identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = $\frac{2m}{3}$ and b = $\frac{3n}{2}$
$(\frac{2m}{3} + \frac{3n}{2})^{2} = (\frac{2m}{3})^{2} + 2(\frac{2m}{3})( \frac{3n}{2}) + ( \frac{3n}{2})^{2}$

= $\frac{4m^{2}}{9} + 2mn + \frac{9n^{2}}{4}$

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

$(0.4p-0.5q)^2$

Answer:

we use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = 0.4p and b =0.5q
$(0.4p-0.5q)^{2} = (0.4p)^{2} - 2(0.4p)(0.5q) + (0.5q)^{2}$
= $0.16p^{2} - 0.4pq + 0.25q^{2}$

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

$(2xy+5y)^2$

Answer:

we use the identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = 2xy and b =5y
$(2xy+5y)^{2} = (2xy)^{2} + 2(2xy)(5y) + (5y)^{2}$
= $4x^{2}y^{2} + 20xy^{2} + 25y^{2}$

Question:4(i) Simplify:

$(a^2-b^2)^2$

Answer:

we use
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $a^{2}$ and b = $b^{2}$
$(a^{2}-b^{2})^{2} = (a^{2})^{2} - 2(a^{2})(b^{2}) + (b^{2})^{2}$
= $a^{4} - 2a^{2}b^{2} + b^{4}$

Question:4(ii) Simplify.

$(2x+5)^2-(2x-5)^2$

Answer:

we use
$a^{2} - b^{2} = (a-b)(a+b)$
In this a = (2x + 5) and b = (2x - 5)
$(2x + 5)^{2} - (2x - 5)^{2} = ( (2x + 5)- (2x - 5))( (2x + 5)+ (2x - 5))$
= $( 2x + 5- 2x + 5)( 2x + 5+ 2x - 5)$
= (4x)(10)
=40x

remember that

$(a+b)^2-(a-b)^2=4ab$

here a= 2x, b= 5

$4ab=4\times 2x \times 5=40x$

Question:4(iii) Simplify.

$(7m-8n)^2+(7m+8n)^2$

Answer:

we use
$(a-b)^{2} = a^{2} -2ab + b^{2}$ and $(a+b)^{2} = a^{2} +2ab + b^{2}$
In this a = 7m and b = 8n
$(7m-8n)^{2} = (7m)^{2} -2(7m)(8n) + (8n)^{2}$
= $49m^{2} -112mn + 64n^{2}$
and
$(7m+8n)^{2} = (7m)^{2} +2(7m)(8n) + (8n)^{2}$
= $49m^{2} +112mn + 64n^{2}$

So, $(7m - 8n)^{2} + (7m + 8n)^{2}$ = $49m^{2} -112mn + 64n^{2}$ + $49m^{2} +112mn + 64n^{2}$
= $2(49m^{2} + 64n^{2})$

remember that

$(a-b)^2+(a+b)^2=2(a^2+b^2)$

Question: 4(iv) Simplify.

$(4m+5n)^2+(5m+4n)^2$

Answer:

we use
$(a+b)^{2} = a^{2} +2ab + b^{2}$
1 ) In this a = 4m and b = 5n

$(4m+5n)^{2} = (4m)^{2} +2(4m)(5n) + (5n)^{2}$
= $16m^{2} +40mn + 25n^{2}$
2 ) in this a = 5m and b = 4n
$(5m+4n)^{2} = (5m)^{2} +2(5m)(4n) + (4n)^{2}$
= $25m^{2} +40mn + 16n^{2}$

So, $(4m + 5n)^{2} + (5m + 4n)^{2}$ = $16m^{2} +40mn + 25n^{2}$ + $25m^{2} +40mn + 16n^{2}$
= $41m^{2} +80mn + 41n^{2}$

Question: 4(v) Simplify.

$(2.5p-1.5q)^2-(1.5p-2.5q)^2$

Answer:

we use
$a^{2}- b^{2} = (a-b)(a+b)$
1 ) In this a = (2.5p- 1.5q) and b = (1.5p - 2.5q)
$(2.5p- 1.5q)^{2}- (1.5p- 2.5q)^{2} = ( (2.5p- 1.5q)- (1.5p- 2.5q))( (2.5p- 1.5q)+ (1.5p- 2.5q))$
= $( 2.5p- 1.5q- 1.5p + 2.5q)(2.5p- 1.5q+ 1.5p- 2.5q)$
= 4(p + q ) (p - q)
= 4 $(p^{2} - q^{2})$

Question:4(vi) Simplify.

$(ab+bc)^2-2ab^2c$

Answer:

We use identity
$(a+b)^{2} = a^{2} + 2ab + b^{2}$
In this a = ab and b = bc
$(ab+bc)^{2} = (ab)^{2} + 2(ab)(bc) + (bc)^{2}$
= $a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}$
Now, $a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}$ - $2ab^{2}c$
= $a^{2}b^{2} + b^{2}c^{2}$

Question:4(vii) Simplify.

$(m^2 -n^2m)^2+2m^3n^2$

Answer:

We use identity
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
In this a = $m^{2}$ and b = $n^{2}m$
$(m^{2}-n^{2}m)^{2} = (m^{2})^{2} - 2(m^{2})(n^{2}m) + (n^{2}m)^{2}$
= $m^{4} - 2m^{3}n^{2} + n^{4}m^{2}$
Now, $m^{4} - 2m^{3}n^{2} + n^{4}m^{2}$ + $2m^{3}n^{2}$
= $m^{4} + n^{4}m^{2}$

Question:5(i) Show that

$(3x+7)^2-84x=(3x-7)^2$

Answer:

L.H.S. = $(3x+7)^2 - 84x = 9x^2 + 42x + 49 - 84x$

$= 9x^2 - 42 x +49$

$= (3x - 7)^2$

= R.H.S.

Hence it is prooved

Question:5(ii) Show that

$(9p-5q)^2+180pq=(9p+5q)^2$

Answer:

L.H.S. = $(9p-5q)^2+180pq = 81p^2 - 90pq + 25q^2 + 180pq$ (Using $(a-b)^2 = a^2 - 2ab + b^2$ )

$= 81p^2 +90pq + 25q^2$

$= (9p + 5q)^2$ $\left ( (a+b)^2 = a^2 + 2ab + b^2 \right )$

= R.H.S.

Question:5(iii) Show that.

$(\frac{4}{3}m-\frac{3}{4}n)^2 +2mn=\frac{16}{9}m^2+\frac{9}{16}n^2$

Answer:

First we will solve the LHS :

$= (\frac{4}{3}m-\frac{3}{4}n)^2 +2mn = \frac{16}{9}m^2 - 2mn + \frac{9}{16}n^2 + 2mn$

or $= \frac{16}{9}m^2 + \frac{9}{16}n^2$

= RHS

Question:5(iv) Show that.

$(4pq+3q)^2-(4pq-3q)^2=48pq^2$

Answer:

Opening both brackets we get,

$(4pq+3q)^2-(4pq-3q)^2 = 16p^2q^2 + 24pq^2 + 9q^2 - (16p^2q^2 - 24pq^2 + 9q^2)$

$= 16p^2q^2 + 24pq^2 + 9q^2 - 16p^2q^2 + 24pq^2 - 9q^2)$

$= 48pq^2$

= R.H.S.

Question:5(v) Show that

$(a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)=0$

Answer:

Opening all brackets from the LHS, we get :

$(a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)\\\\ =\ a^2 +ab - ab- b^2 + b^2+bc - bc -c^2 + c^2 +ca - ac -a^2$

$= 0$ = RHS

Question:6(i) Using identities, evaluate.

$71^2$

Answer:

We will use the identity:

$(a + b)^2 = a^2 + 2ab + b^2$

So, $71^2 = (70 +1)^2 = 70^2 + 2(70)(1) + 1^2$

$= 4900 + 140 + 1$

$= 5041$

Question:6(ii) Using identities, evaluate.

$99^2$

Answer:

Here we will use the identity :

$(a - b)^2 = a^2 - 2ab + b^2$

So : $99^2 = (100 - 1) ^2 = 100^2 - 2(100)(1) + 1^2$

or $= 10000 - 200 + 1$

$= 9801$

Question:6(iii) Using identities, evaluate.

$102^2$

Answer:

Here we will use the identity :

$(a+b)^2 = a^2 +2ab + b^2$

So :

$102^2 = (100 + 2)^2 = 100^2 + 2(100)(2) + 2^2$

or $= 10000 + 400 + 4$

$= 10404$

Question:6(iv) Using identities, evaluate.

$998^2$

Answer:

Here we will the identity :

$998^2=(1000 - 2)^2 = 1000^2 - 2(1000)(2) + 2^2$

or $= 1000000 - 4000+ 4$

or $= 996004$

Question:6(v) Using identities, evaluate.

$5.2^2$

Answer:

Here we will use :

$(a + b)^2 = a^2 + 2ab + b^2$

Thus

$(5.2)^2 = (5.0 + 0.2)^2 = 5^2 + 2(5)(0.2) + (0.2)^2$

or $= 25 + 2 + 0.04$

$= 27.04$

Question:6(vi) Using identities, evaluate.

$297 \times 303$

Answer:

This can be written as :

$297\times303 = (300-3)\times(300+3)$

using $(a-b)(a+b)=a^2-b^2$

or $= 90000 - 9$

$= 89991$

Question:6(vii) Using identities, evaluate.

$78 \times 82$

Answer:

This can be written in form of :

$78\times82 = (80 - 2) \times(80+2)$

or $= 80^2 - 2^2$ $\because \left ( a-b \right )\left ( a+b \right ) = a^2 - b^2$

or $= 6400- 4 = 6396$

Question:6(viii) Using identities, evaluate.

$8.9^2$

Answer:

Here we will use the identity :

$(a - b)^2 = a^2 - 2ab + b^2$

Thus :

$8.9^2 = (9 - 0.1) ^2 = 9^2 - 2(9)(0.1) + 0.1^2$

or $= 81 - 1.8 + 0.01$

or $= 79.21$

Question:6(ix) Using identities, evaluate.

$10.5\times9.5$

Answer:

This can be written as :

$10.5\times9.5 = (10 +0.5)\times(10-0.5)$

or $= 10^2 - 0.5^2$ $\because (a+b)(a-b) = a^2 - b^2$

or $= 100 - 0.25$

or $= 99.75$

Question:7(i) Using $a^2-b^2=(a+b)(a-b)$ , find

$51^2-49^2$

Answer:

We know,

$a^2-b^2=(a+b)(a-b)$

Using this formula,

$51^2-49^2$ = (51 + 49)(51 - 49)

= (100)(2)

= 200

Question:7(ii) Using $a^2-b^2=(a+b)(a-b)$ , find

$(1.02)^2-(0.98)^2$

Answer:

We know,

$a^2-b^2=(a+b)(a-b)$

Using this formula,

$(1.02)^2-(0.98)^2$ = (1.02 + 0.98)(1.02 - 0.98)
= (2.00)(0.04)

= 0.08

Question:7(iii) Using $a^2-b^2=(a+b)(a-b)$ , find.

$153^2-147^2$

Answer:

We know,

$a^2-b^2=(a+b)(a-b)$

Using this formula,

$153^2-147^2$ = (153 - 147)(153 +147)

=(6) (300)

= 1800

Question:7(iv) Using $a^2-b^2=(a+b )(a-b)$ , find

$12.1^2-7.9^2$

Answer:

We know,

$a^2-b^2=(a+b)(a-b)$

Using this formula,

$(1.02)^2-(0.98)^2$ = (1.02 + 0.98)(1.02 - 0.98)

= (2.00)(0.04)

= 0.08

Question:8(i) Using $(x+a)(x+b)=x^2+(a+b)x+ab$ $103 \times 104$

Answer:

We know,

$(x+a)(x+b)=x^2+(a+b)x+ab$

Using this formula,

$103 \times 104$ = (100 + 3)(100 + 4)

Here x =100, a = 3, b = 4

$\therefore$ $103 \times 104$ $= 100^2+(3+4)100+(3\times4)$

$= 10000+1200+12$

= 11212

Question:8(ii) Using $(x+a)(x+b)=x^2+(a+b)x+ab$ , find

$5.1\times 5.2$

Answer:

We know,

$(x+a)(x+b)=x^2+(a+b)x+ab$

Using this formula,

$5.1\times 5.2$ = (5 + 0.1)(5 + 0.2)

Here x =5, a = 0.1, b = 0.2

$\therefore$ $5.1\times 5.2$ $=5^2+(0.1 + 0.2)5+(0.1\times0.2)$

$= 25+1.5+0.02$

= 26.52

Question:8(iii) Using $(x+a)(x+b)=x^2+(a+b)x+ab$ , find

$103 \times 98$

Answer:

We know,

$(x+a)(x+b)=x^2+(a+b)x+ab$

Using this formula,

$103 \times 98$ = (100 + 3)(100 - 2) = (100 + 3){100 + (-2)}

Here x =100, a = 3, b = -2

$\therefore$ $103 \times 98$ $= 100^2+(3 + (-2))100+(3\times(-2))$

$= 10000+100-6$

= 10094

Question: 8(iv) Using $(x+a)(x+b)=x^2+(a+b)x+ab$ , find

$9.7 \times 9.8$

Answer:

We know,

$(x+a)(x+b)=x^2+(a+b)x+ab$

Using this formula,

$9.7 \times 9.8$ = (10 - 0.3)(10 - 0.2) = {10 + (-0.3)}{10 + (-0.2)}

Here x =10, a = -0.3, b = -0.2

$\therefore$ $9.7 \times 9.8$ $= 10^2+((-0.3) + (-0.2))10+((-0.3)\times(-0.2))$

$= 100-5+0.06$

= 95.

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

What are Expressions?
Terms, Factors and Coefficients
Monomials, Binomials,f and Polynomials
Like and Unlike Terms
Addition and Subtraction of Algebraic Expressions
Multiplication of Algebraic Expressions: Introduction
Multiplying a Monomial by a Monomial
Multiplying two monomials
Multiplying three or more monomials
Multiplying a Monomial by a Polynomial
Multiplying a monomial by a binomial
Multiplying a monomial by a trinomial
Multiplying a Polynomial by a Polynomial
Multiplying a binomial by a binomial
Multiplying a binomial by a trinomial
What is an Identity?
Standard Identities
Applying Identities

NCERT Solutions for Class 8 Maths - Chapter Wise

Chapter -1	Rational Numbers
Chapter -2	Linear Equations in One Variable
Chapter-3	Understanding Quadrilaterals
Chapter-4	Practical Geometry
Chapter-5	Data Handling
Chapter-6	Squares and Square Roots
Chapter-7	Cubes and Cube Roots
Chapter-8	Comparing Quantities
Chapter-9	Algebraic Expressions and Identities
Chapter-10	Visualizing Solid Shapes
Chapter-11	Mensuration
Chapter-12	Exponents and Powers
Chapter-13	Direct and Inverse Proportions
Chapter-14	Factorization
Chapter-15	Introduction to Graphs
Chapter-16	Playing with Numbers

NCERT Solutions for Class 8 - Subject Wise

Some Important Identities From NCERT Book for Class 8 Chapter 9 Algebraic Expressions And Identities

$(a+b)^2=(a^2+2ab+b^2)$

you can write

$(a+b)^2=(a+b)(a+b)$

Can be simplified as follows

$(a+b)(a+b)=a\times a+a\times b+a\times b+b\times b$

Now add each term

$a\times a+a\times b+a\times b+b\times b=a^2+ab+ab+b^2$

$=a^2+2ab+b^2$

$(a+b)^2=(a^2+2ab+b^2)$

$(a-b)^2=(a^2-2ab+b^2)$
$(a-b)(a+b)=a^2-b^2$
$(x+a)(x+b)=x^{2}+(a+b) x+a b$

You can form the above identities by yourself. These above identities have been used in many problems of NCERT solutions for class 8 maths chapter 9 algebraic expression and identities.

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. What are the important topics of NCERT syllabus chapter Algebraic Expressions and Identities ?

Addition and subtraction of the algebraic expression, multiplication of the algebraic expression, standard identities, and application of identities are important topics in this chapter.

2. Does CBSE class maths is tough ?

CBSE class 8 maths is not tough at all. It teaches a very basic and simple maths.

3. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

4. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

5. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

6. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

Also Read

NCERT Syllabus for Class 8 Hindi 2023-24 (All Chapters PDF) - Download Here

NCERT Books for Class 8 Social Science 2023 - Download Pdf

NCERT Books for Class 8 Maths 2023 - Download Pdf Here

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NCERT Books for Class 8 Science 2023 - Download PDF

NCERT Books for Class 8 English 2023 - Download PDF

NCERT Syllabus for class 8 Science 2023-24

NCERT Syllabus for Class 8 Maths 2023 - Download PDF

NCERT Syllabus for Class 8 English 2023 - Download PDF

NCERT Syllabus for Class 8 2023 (All Subjects) - Download PDF

NCERT Syllabus for Class 8 Social Science 2023

NCERT Solutions for Class 8 Maths (Updated 2023-24)

NCERT solutions for Class 8 Maths Chapter 16 Playing with Numbers

NCERT Solutions for Class 8 Maths Chapter 15 Introduction to Graphs

NCERT Solutions for Class 8 Maths Chapter 14 Factorization

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download

Algebraic Expressions and Identities Class 8 Solutions - Important Formulae

Algebraic Expressions and Identities Class 8 NCERT Solutions (Intext Questions and Exercise)

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

NCERT Solutions for Class 8 Maths - Chapter Wise

NCERT Solutions for Class 8 - Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

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