NCERT  Solution for Class 8 Chapter – 9  Algebraic expression Exercise – 9.1

NCERT  Solution for Class 8 Chapter – 9  Algebraic expression Exercise – 9.1

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

  • 5xyz2 – 3zy
  • 1 + x + x2
  • 4x2y2 – 4x2y2z2 + z2
  • 3 – pq + qr – p
  • (x/2) + (y/2) – xy
  • 3 a – 0.6ab + 0.5b

 

Solution:-

 

SI . No Expression Term Coefficient
5xyz2 – 3zy Term: 5xyz2

Term: -3zy

5 – 3

 

1 + x + x2

 

4x2y2 – 4x2y2z2 + z2

Term: 1

Term: x

Term: x2

Term: 4x2y2

Term: -4x2 y2z2

Term: z2

111

 

3 – pq + qr – p Term: 3 – pq qr – p 4-4 1
(x/2) + (y/2) – xy Term: x/2 y/2 -xy 3 – 1 1 – 1
0.3a – 0.6ab + 0.5b Term: 0.3a -0.6ab 0.5b 1/2 1/2 – 1
0.3 – 0.6 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:-

 

Let us first define the classifications of these 3 polynomials:

Monomials , Contain only one term.

Bionomials, Contain only two terms.

Trinomials, Contain only three terms.

 

x + y Two term Binomial
1000 One term Monomial
X + x2 + x3 + x4 Four term Polynomial , and it does not fit in listed three categories
2y – 3y2 Two term Binomial
2y – 3y2 + 4y3 Three term Trinomial
5x – 4y +  3xy Three term Trinomial
4z – 15z2 Two term Binomial
Ab + bc + cd + da Four term Polynomial , and it does not fit in listed in three categories
pqr One  term Monomial
P2q +pq2 Two term Binomial
2p + 2q Two term Binomial
7 + y + 5x Three term Trinomial

 

 

 

 

 

 

 

 

 

Q3. Add the following.

  • Ab – bc, bc – ca, ca – ab

Solution:-

 

(ab – bc) + (bc – ca) + (ca – ab)

= 0

 

  • A-b + ab , b – c + bc, c – a + ac

Solution:-

 

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

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

= 0 + 0 + 0 + ab + bc + ca

= ab + bc + ca

 

  • 2p2q2 – 3pq + 4 , 5 + 7pq – 3p2q2

Solution:-

 

2p2q2 – 3p2q2 –  3pq + 7pq + 4 + 5

-p2q2 + 4pq + 9

 

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

Solution:-

 

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

 

Q4. Subtract

  • 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

Solution:-

 

12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12

= 8a – 2ab + 2b – 15

 

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

Solution:-

 

5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx

= 2xy – 7yz + 5zx + 10xyz

 

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

Solution:-

 

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

= 28 + 5p – 18q + 8pq – 7pq2 + p2q

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