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

In this page we have NCERT Solutions for Class 8 Maths Chapter 9 : Algebraic Expressions and Identities for Exercise 9.1. 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
 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
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
(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