In this page we have Algebraic Expressions and Identities:NCERT Solutions for Class 8 Maths Chapter 9 for
Exercise 9.3 . Hope you like them and do not forget to like , social share
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Question 1
Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a
2– 9, 4a
(v) pq + qr + rp, 0
Answer:
- 4p(q + r) = 4pq + 4pr
- ab(a - b) = a2b - ab2
- (a + b) (7a2b2) = 7a3b2 + 7a2b3
- (a2 - 9)(4a) = 4a3 - 36a2
- (pq + qr + rp) x 0 = 0
Question 2
Find the product.
i) a
2 x (2a
22) x (4a
26)
ii) (2xy/3) ×(-9x
2y
2/10)
(iii) (-10pq
3/3) ×(6p
3q/5)
(iv) ( x) x (x
2) x (x
3) x (x
8)
Answer:
We will use the below property extensively in above questions
am x an x ao = am+n+o
i) As you know
So, we get
a2 x (
2a22) x (
4a26) =
8a48
ii) (2xy/3) ×(-9x2y2/10)
=(-3x3y3/5)
iii) (-10pq3/3) ×(6p3q/5)
=(-4p4q4)
iv) ( x) x (x
2) x (x
3) x (x
8)
=
x14
Question 3
(a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =1/2
(b) Simplify
a (a2+ a + 1) + 5 and find its value for (i)
a = 0, (ii)
a = 1 (iii)
a = – 1.
Answer
a)
i) Putting x=3 in the equation we get
12x
2 - 15x + 3
=108-45+3 = 66
(ii) putting x=1/2 in the equation we get
b)
a(a
2a+1)
=a
3+a
2+a
(i) putting a= 0 in the equation we get
0
3+0
2+0=0
(ii) putting a=1 in the equation we get
1
3 + 1
2 + 1 = 1 + 1 + 1 = 3
(iii) Putting a = -1 in the equation we get
-1
3+1
2 -1 = -1 + 1 + 1 = 1
Question 5
(a) Add:
p ( p – q), q ( q – r) and r ( r – p)
(b) Add:
2x (z – x – y) and
2y (z – y – x)
(c) Subtract:
3l (l – 4 m + 5 n) from
4l ( 10 n – 3 m + 2 l )
(d) Subtract:
3a (a + b + c ) – 2 b (a – b + c) from
4c ( – a + b + c )
Answer:
i) (p
2 - pq) + (q
2 - qr) + (r
2 - pr)
= p
2 + q
2 + r
2 - pq - qr - pr
ii) (2xz - 2x
2 - 2xy) + (2yz - 2y
2 - 2xy)
= 2xz - 4xy + 2yz - 2x
2 - 2y
2
iii) (40ln - 12lm + 8l
2) - (3l
2 - 12lm + 15ln)
= 40ln - 12lm + 8l
2 - 3l
2 - 12lm + 15ln
= 55ln - 24lm + 5l
2
iv) = (-4ac + 4bc + 4c
2) - (3a
2 + 3ab + 3ac)
= -4ac + 4bc + 4c
2 - 3a
2 - 3ab - 3ac
= -7ac + 4bc + 4c
2 - 3a
2 - 3ab
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