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
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Question 1
Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz
^{2} – 3zy
(ii) 1 + x + x
^{2}
(iii) 4x
^{2}y
^{2} – 4x
^{2}y
^{2}z
^{2} + z
^{2}
(iv) 3 – pq + qr – rp
(v) (x/2) –(y/2) xy
(vi) 0.3a – 0.6ab + 0.5b
Answer:
No

Expression

Coefficient

1

Term: xyz^{2}
Term: zy

5
3

2

Term: 1
Term: x
Term x^{2}

1
1
1

3

Term: x^{2}y^{2}
Term: x^{2}y^{2}z^{2}
Term z^{2}

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 + x
^{2} + x
^{3} + x
^{4}
7 + y + 5x
2y – 3y
^{2}
2y – 3y
^{2} + 4y
^{3}
5x – 4y + 3xy
4z – 15z
^{2}
ab + bc + cd + da
pqr
p
^{2}q + pq
^{2}
2p + 2q
Answer:
x + y: Binomial
1000: Monomial
x + x
^{2} + x
^{3} + x
^{4}: Polynomial
7 + y + 5x: Binomial
2y – 3y
^{2}: Binomial
2y – 3y^{2} + 4y^{3}: Trinomial
5x – 4y + 3xy: Trinomial
4z – 15z^{2}: Binomial
ab + bc + cd + da: Polynomial
pqr: Monomial
p^{2}q + pq^{2}: 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)
2p^{2}q
^{2} – 3pq + 4, 5 + 7pq – 3p^{2}q
^{2}
(iv)
l^{2} + m^{2}, m^{2} + n^{2}, n^{2} + l^{2}, 2lm + 2mn + 2nl
Answer:
i) (ab  bc) + (bc  ca) + (caab)
= 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) 2p^{2}q
^{2} – 3pq + 4, 5 + 7pq – 3p^{2}q
^{2}
= (2p
^{2}q
^{2}  3pq + 4) + (5 + 7pq  3p
^{2}q
^{2})
= 2p
^{2}q
^{2}  3p
^{2}q
^{2}  3pq + 7pq + 4 + 5
=  p
^{2}q
^{2} + 4pq + 9
iv) (l
^{2} + m
^{2}) + (m
^{2} + n
^{2}) + (n
^{2} + l
^{2}) + (2lm + 2mn + 2nl)
= l
^{2} + l
^{2} + m
^{2} + m
^{2} + n
^{2} + n
^{2} + 2lm + 2mn + 2nl
= 2l
^{2} + 2m
^{2} + 2n
^{2} + 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
4p^{2}q – 3pq + 5pq^{2} – 8p + 7q – 10
from
18 – 3p – 11q + 5pq – 2pq^{2} + 5p^{2}q
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  2pq
^{2} + 5p
^{2}q)  (4p
^{2}q  3pq + 5pq
^{2}  8p + 7q  10)
= 18  3p  11q + 5pq  2pq
^{2} + 5p
^{2}q  4p
^{2}q + 3pq  5pq
^{2} + 8p  7q + 10
= 28 + 5p  18q + 8pq  7pq
^{2}  p
^{2}q
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