Algebraic Expressions and Identities:NCERT Solutions for Class 8 Maths Chapter 9 Exercise 9.3

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) a²– 9, 4a
 (v) pq + qr + rp, 0
Answer:

1. 4p(q + r) =  4pq + 4pr
2. ab(a - b) = a²b - ab²
3. (a + b) (7a²b²) = 7a³b² + 7a²b³
4. (a² - 9)(4a) = 4a³ - 36a²
5. (pq + qr + rp) x 0 = 0

Question 2
 Find the product.
i) a² x (2a²²) x (4a²⁶)
ii) (2xy/3) ×(-9x²y²/10)
      (iii)    (-10pq³/3) ×(6p³q/5)
      (iv)    ( x) x (x²) x (x³) x (x⁸)
Answer: We will use the below property extensively in above questions
a^m x aⁿ x a^o = a^m+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 (x²) x (x³) x (x⁸)
= 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
12x² - 15x + 3 =108-45+3 = 66
(ii) putting x=1/2 in the equation we get

b)
 a(a²a+1)
=a³+a²+a
(i) putting a= 0 in the equation we get
0³+0²+0=0
(ii) putting a=1 in the equation we get
1³ + 1² + 1 = 1 + 1 + 1 = 3
(iii) Putting a = -1 in the equation we get
-1³+1² -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) (p² - pq) + (q² - qr) + (r² - pr)
= p² + q² + r² - pq - qr - pr
ii) (2xz - 2x² - 2xy) + (2yz - 2y² - 2xy)
= 2xz - 4xy + 2yz - 2x² - 2y²
iii) (40ln - 12lm + 8l²) - (3l² - 12lm + 15ln)
= 40ln - 12lm + 8l² - 3l² - 12lm + 15ln
= 55ln - 24lm + 5l²
iv) = (-4ac + 4bc + 4c²) - (3a² + 3ab + 3ac)
= -4ac + 4bc + 4c² - 3a² - 3ab - 3ac
= -7ac + 4bc + 4c² - 3a² - 3ab

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