physicscatalyst.com logo








Ncert Solutions for Algebraic Expressions and Identities Class 8 CBSE Part 1



In this page we have NCERT book Solutions for Class 8th Maths:Algebraic Expressions and Identities for EXERCISE 1 . Hope you like them and do not forget to like , social share and comment at the end of the page.

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) (x/2) –(y/2) -xy

 (vi) 0.3a – 0.6ab + 0.5b

 

Answer:

No

Expression

Coefficient

1

Term: xyz2 

Term: zy 

 

5

-3

2

Term: 1

Term: x

Term x2 

 

1

1

1

3

Term: x2y2

Term: x2y2z2

Term z2

 

 

4

-4

1

4

3

pq

qr

rp

3

-1

1

-1

5

x

Y

xy

½

-1/2

-1

6

a

ab

b

.3

-.6

.5

 

 

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

 

Answer:

x + y: Binomial

1000: Monomial

x + x2 + x3 + x4: Polynomial

7 + y + 5x: Binomial

2y – 3y2: Binomial

2y – 3y2 + 4y3: Trinomial

5x – 4y + 3xy: Trinomial

4z – 15z2: Binomial

ab + bc + cd + da: Polynomial

pqr: Monomial

p2q + pq2: Binomial

2p + 2q: Binomial

 

Question 3

 Add the following.

(i) ab – bc, bc – ca, ca – ab

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

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

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

 

Answer: 

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

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

= 0

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

= a + b + c + ab + bc + ca - b - c - a

= ab + bc + ca

iii) 2p2q2 – 3pq + 4, 5 + 7pq – 3p2q2

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

= 2p2q2 - 3p2q2 - 3pq + 7pq + 4 + 5

= - p2q2 + 4pq + 9

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

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

= 2l2 + 2m2 + 2n2 + 2lm + 2mn + 2nl

 

 

Question 4.

(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

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

 (c) Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10

from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q

Answer: 

While subtracting, we need to remember signs are reversed after –sign once bracket is opened

ie. + becomes - and - becomes +

Let solve the below question keeping that in mind

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

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

= 8a - 2ab + 2b - 15

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

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

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

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

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

= 28 + 5p - 18q + 8pq - 7pq2 - p2q

 

 


Download this assignment as pdf
Go Back to Class 8 Maths Home page Go Back to Class 8 Science Home page


link to us