# Mathematics – Class 8 – Chapter 9 – Algebraic Expression and Identities – Exercise 9.1 -NCERT Exercise Solution

Q1. 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 – p

(v) (x/2) + (y/2) – xy

(vi) 0.3a – 0.6ab + 0.5b

Solution:

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, p 2 q + pq 2 , 2p + 2q

Solution:

(i) ab – bc, bc – ca, ca – ab

Solution:

According to question:

= (ab – bc) + (bc – ca) + (ca-ab)

= ab – bc + bc – ca + ca – ab

= ab – ab – bc + bc – ca + ca

= 0

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

Solution:

According to question:

= (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 + ca

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

= ab + bc + ca

(iii) 2p 2q2 – 3pq + 4, 5 + 7pq – 3p 2 q2

Solution:

According to question:

= 2p 2q2 – 3pq + 4, 5 + 7pq – 3p 2 q2

= (2p 2q2  – 3pq + 4) + (5 + 7pq – 3p 2 q2)

= 2p 2q2   – 3p 2 q2 – 3pq + 7pq + 4 + 5

= – p 2q2  + 4pq + 9

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

Solution:

According to question:

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

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

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

Thus, the sum of the given expressions is 2(l2 + m2+ n2 + lm + mn + nl)

4.

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

Solution:

According to question:

= (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:

According to question:

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

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

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

= 2xy – 7yz + 5zx + 10xyz

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

Solution:

According to question:

= (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 + 5 pq+ 3pq- 2pq2– 5pq2+ 5 p2q – 4p2q

= 28 + 5p – 18q + 8pq – 7pq2 + p2 q

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