Spread the love

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

Introduction:

In this exercise/article we will learn about expressions, four terms Monomial, Binomial, trinomial, Polynomial and terms, factors and coefficients of addition and subtraction of algebraic expressions find out Monomial, Binomial, trinomial, Polynomial the numerical factor of a term is called its numerical coefficient or simply coefficient.

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

Class 8 Maths Exercise 9.1 (Page-140)

Q1. Identify the terms, their coefficients for each of the following expressions.

(i) 5xyz² - 3zy

Solution :

We have given terms are 5xyz² - 3zy

So, coefficients of 5xyz² = 5

and coefficient of - 3zy = -3

(ii) 1+x+x²

Solution :

We have given terms are 1+x+x²

So, coefficient of 1 = 1

coefficient of x = 1

and coefficient of x² = 1

(iii) 4x²y²- 4x²y²z²+z²

Solution :

We have given terms are 4x²y²- 4x²y²z²+z²

So, coefficient of 4x²y² = 4

coefficient of - 4x²y²z² = -4

and coefficient of z² = 1

(iv) 3-pq+qr-rp

We have given terms are 3-pq+qr-rp

So, coefficient of 3 = 3

coefficient of -pq = -1

coefficient of qr = 1

and coefficient -rp = -1

(v) \(\displaystyle \frac{x}{2}+\frac{y}{2}-xy\)

Solution :

We have given terms are \(\displaystyle \frac{x}{2}+\frac{y}{2}-xy\)

So, coefficient of \(\displaystyle \frac{x}{2}\) = \(\displaystyle \frac{1}{2}\)

coefficient of = \(\displaystyle \frac{1}{2}\)

and coefficient of = -1

(vi) 0.3a - 0.6ab + 0.5b

Solution :

We have given terms are 0.3a - 0.6ab + 0.5b

So, coefficient of 0.3a = 0.3

coefficient of - 0.6ab = - 0.6

and coefficient of 0.5b = 0.5

Q2. 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 :

Monomials = 1000 , pqr

Binomials = x + y , 2y - 3y² , 4z - 15z² , p²q + pq² , 2p + 2q

Trinomials = 7 + y + 5x , 2y - 3y² + 4y³ , 5x - 4y + 3xy

Polynomials = x + x² + x³ + x⁴ , ab + bc + cd + da

Class 8 Maths Chapter 9 Exercise 9.1

Q3.Add the following.

(i) ab - bc , bc - ca , ca - ab

Solution :

\(\displaystyle \begin{array}{l}\,\,\,ab-bc-ca\\-ab+bc+ca\\\overline{{\underline{{\,\,\,\,\,0\,\,\,+\,\,0\,\,+\,0\,\,\,\,\,}}}}\end{array}\)

If you have any doubt then follow the step by steps Given below :

( ab - bc ) + ( bc - ca ) + ( ca - ab )

= ab - bc + bc - ca + ca - ab

= ( ab - ab ) + ( - bc + bc ) + ( - ca + ca )

= (0) + (0) + (0)

= 0 + 0 + 0

= 0

( ii ) a - b + ab , b - c + bc , c - a + ac

Solution :

\(\displaystyle \begin{array}{l}\,\,\,a-b-c+ab+bc+ac\\-a+b+\,c\,+\,0\,+\,0\,+0\\\overline{{\underline{{\,\,\,0+\,0\,+0+\,ab+bc+ac\,}}}}\end{array}\)

If you have any doubt then follow the step by steps Given below :

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

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

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

= ( 0 ) + ( 0 ) + ( 0 ) + ( ab + bc + ac )

= ab + bc + ac

(iii) 2p²q² - 3pq + 4 , 5 + 7pq - 3p²q²

Solution :

\(\displaystyle \begin{array}{l}\,\,\,\,\,\,2{{p}^{2}}{{q}^{2}}-3pq+4\text{ }\\-\text{ }3{{p}^{2}}{{q}^{2}}+7pq+5\\\overline{{\underline{{\,\,-{{p}^{2}}{{q}^{2}}+\,4pq\,+9}}}}\,\end{array}\)

If you have any doubt then follow the step by steps Given below :

2p²q² - 3pq + 4 + 5 + 7pq - 3p²q²

= ( 2p²q² - 3p²q² ) + ( - 3pq + 7pq ) + ( 4 + 5 )

= ( - p²q² ) + ( 4pq ) + ( 9 )

= - p²q² + 4pq + 9

(iv) l² + m² , m² + n² , n² + l² , 2lm + 2mn + 2nl

Solution :

\(\displaystyle \begin{array}{l}{{l}^{{2~}}}+{{m}^{2}}+{{n}^{2}}+2lm+2mn+2nl\\{{l}^{2}}+{{m}^{2}}+{{n}^{2}}\,\,+\,\,0\,\,\,+\,\,\,0\,\,\,+\,\,\,0\,\,\\\overline{{\underline{{2{{l}^{2}}+2{{m}^{2}}+2{{n}^{2}}+2lm+2mn+2nl}}}}\end{array}\)

If you have any doubt then follow the step by steps Given below :

l² + m² + m² + n² + n² + l² + 2lm + 2mn + 2nl

= ( l² + l² ) + ( m² + m² ) + ( n² + n² ) + ( 2lm + 2mn + 2nl )

= ( 2l² ) + ( 2m² ) + ( 2n² ) + ( 2lm + 2mn + 2nl )

= 2l² + 2m² + 2n² + 2lm + 2mn + 2nl

= 2 ( l² + m² + n² + lm + mn + nl )

Q4. (a) Subtract 4a - 7ab + 3b + 12 from 12a - 9ab + 5b - 3

Solution :

( 12a - 9ab + 5b - 3 ) - ( 4a - 7ab + 3b + 12 )

= 12a - 9ab + 5b - 3 - 4a + 7ab - 3b - 12

= 12a - 4a - 9ab + 7ab + 5b - 3b - 3 - 12

= 8a - 2ab + 2b - 15

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

Solution :

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

= 5xy - 2yz - 2zx + 10xyz - 3xy - 5yz + 7zx

= 5xy - 3xy - 2yz - 5yz + 7zx + 10xyz

= 2xy - 7yz + 7zx + 10xyz

(c) Subtract 4p²q - 3pq + 5pq² - 8p + 7q - 10 from 18 - 3p - 11q + 5pq - 2pq² + 5p²q

Solution :

( 18 - 3p - 11q + 5pq - 2pq² + 5p²q ) - ( 4p²q - 3pq + 5pq² - 8p + 7q - 10 )

= 18 - 3p - 11q + 5pq - 2pq² + 5p²q - 4p²q + 3pq - 5pq² + 8p - 7q + 10

= 18 + 10 - 3p + 8p - 11q - 7q + 5pq + 3pq - 2pq² - 5pq² + 5p²q - 4p²q

= 28 + 5p - 18q + 8pq - 7pq² + p²q .

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

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

Introduction:

Class 8 Maths Exercise 9.1 (Page-140)

Class 8 Maths Chapter 9 Exercise 9.1

Leave a Comment Cancel Reply