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
2y
2 – 4x
2y
2z
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: 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 + 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
2q + 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 – 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)
2p2q
2 – 3pq + 4, 5 + 7pq – 3p2q
2
(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) 2p2q
2 – 3pq + 4, 5 + 7pq – 3p2q
2
= (2p
2q
2 - 3pq + 4) + (5 + 7pq - 3p
2q
2)
= 2p
2q
2 - 3p
2q
2 - 3pq + 7pq + 4 + 5
= - p
2q
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
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 - 2pq
2 + 5p
2q) - (4p
2q - 3pq + 5pq
2 - 8p + 7q - 10)
= 18 - 3p - 11q + 5pq - 2pq
2 + 5p
2q - 4p
2q + 3pq - 5pq
2 + 8p - 7q + 10
= 28 + 5p - 18q + 8pq - 7pq
2 - p
2q
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