Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.3

Last Updated : 11 Sep, 2024

Chapter 9 of the Class 8 NCERT Mathematics textbook focuses on the Algebraic Expressions and Identities. Exercise 9.3 is designed to help students practice and understand the application of algebraic identities in simplifying and solving problems. This exercise covers key concepts related to algebraic expressions and provides a range of problems to enhance comprehension and problem-solving skills.

What are Algebraic Expressions and Identities?

The Algebraic Expressions are combinations of the variables, constants and operators (such as +, -, *, /) that represent mathematical quantities. For example, 3x+5 and 4y−7 are algebraic expressions.

The Algebraic Identities are equations that hold true for the all values of the variables involved. They are fundamental in simplifying expressions and solving equations. Common identities include:

Square of a Binomial:

(a+b)²=a²+2ab+b²

Difference of Squares:

a²−b²=(a−b)(a+b)

Product of Sum and Difference:

(a+b)(a−b)=a²−b²

Question 1. Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r

Solution:

(4p) * (q + r) = 4pq + 4pr

(ii) ab, a - b

Solution:

(ab) * (a - b) = a²b - ab²

(iii) a + b, 7a²b²

Solution:

(a + b) * (7a²b²) = 7a³b² + 7a²b³

(iv) a²- 9, 4a

Solution:

(a² - 9) * (4a) = 4a³- 36a

(v) pq + qr + rp, 0

Solution:

(pq + qr + rp ) * 0 = 0
Explanation: Anything multiplied to 0 will give zero.

Question 2. Complete the table.

Solution:

	First expression	Second expression	Product
(i)	a	b + c + d	ab + ac + ad
(ii)	x + y - 5	5xy	5x²y + 5xy²- 25xy
(iii)	p	6p²- 7p + 5	6p³- 7p²+ 5p
(iv)	4p²q²	p²- q²	4p⁴q²- 4p²q⁴
(v)	a + b + c	abc	a²bc + ab²c + abc²

Question 3. Find the product.

(i) (a²) x (2a²²) x (4a²⁶)

Solution:

(1 x 2 x 4 ) (a²x a²² x a²⁶)
= (8) (a⁵⁰)
= 8a⁵⁰
Explanation: when two number numbers of same base are multiplied their power gets added up. [ a^x x a^y = a^x+y ]

(ii) (2/3 xy) x (-9/10 x²y²)

Solution:

(2/3 x -9/10) (xy x x²y²)
= (-3/5) (x³y³)
= -3/5 x³y³
Explanation: when two number numbers of same base are multiplied their power gets added up. [ a^x x a^y = a^x+y ]

(iii) (-10/3pq³) * (6/5p³q)

Solution:

(-10/3 x 6/5) (pq³ x p³q)
= (-4) (p⁴q⁴)
= -4p⁴q⁴
Explanation: when two number numbers of same base are multiplied their power gets added up. [ a^x x a^y = a^x+y]

(iv) x * x² * x³ * x⁴

Solution:

(x) (x²) (x³) (x⁴)
= x¹⁰
Explanation: when two number numbers of same base are multiplied their power gets added up. [ a^x x a^y = a^x+y]

Question 4.

(a) Simplify 3x (4x - 5) + 3 and find its values for

(i) x = 3

Solution:

First we will simplify the given equation and the put the value of x as required.
3x (4x - 5) + 3
⇒ 12x²- 15x + 3
⇒ 12 (3)² - 15 (3) + 3 [ putting the value of x = 3 ]
⇒ 12 (9) - 15 (3) + 3
⇒ 108 - 45 + 3
⇒ 66

(ii) x = 1/2

Solution:

3x (4x - 5) + 3
⇒ 12x²- 15x + 3
⇒ 12 (1/2)² - 15 (1/2) + 3
⇒ 12 (1/4) - 15 (1/2) +3
⇒ 3 - 15/2 + 3
⇒ -3/2

(b) Simplify a (a²+ a + 1) + 5 and find its value for

(i) a = 0

Solution:

a (a²+ a + 1) + 5
⇒ a³+ a²+ a + 5
⇒ (0)³+ (0)²+ (0) + 5 [ anything to the power 0 gives 0 only ]
⇒ 5

(ii) a = 1

Solution:

a (a²+ a + 1) + 5
⇒ a³+ a²+ a + 5
⇒ (1)³+ (1)²+ (1) + 5 [ 1 to the power any number gives 1 ]
⇒ 8

(iii) a = -1

Solution:

a (a²+ a + 1) + 5
⇒ a³+ a²+ a + 5
⇒ (-1)³ + (-1)² + (-1) + 5 [ if -1 has even power then it is 1 or else if it has odd power it is -1]
⇒ -1 + 1 -1 + 5
⇒ 4

Question 5.

(a) Add: p (p – q), q (q – r) and r (r – p)

Solution:

p (p – q) + q (q – r) + r (r – p)
⇒ p²- pq + q²- qr + r²- rp
⇒ p²+ q²+ r² - pq - rp - qr

(b) Add: 2x (z – x – y) and 2y (z – y – x)

Solution:

2x (z – x – y) + 2y (z – y – x)
⇒ 2xz - 2x² - 2xy + 2yz - 2y²- 2yx
⇒ 2xz – 4xy + 2yz – 2x² – 2y²

(c) Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l)

Solution:

4l (10n – 3m + 2l) - 3l (l – 4m + 5n)
⇒ 40ln - 12lm + 8l² - 3l² + 12lm - 15ln
⇒ 25ln + 5l²

(d) Subtract: 3a (a + b + c) – 2b (a – b + c) from 4c (–a + b + c)

Solution:

4c (– a + b + c) - [3a (a + b + c) – 2b (a – b + c)]
⇒ -4ac + 4bc + 4c² - [3a²+ 3ab + 3ac - 2ab + 2b²- 2bc]
⇒ -4ac + 4bc + 4c² - 3a²- ab - 3ac - 2b²+ 2bc
⇒ -3a²- 2b²+ 4c²- 7ac + 6bc - ab

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Summary

Exercise 9.3 of Chapter 9 in NCERT Class 8 Mathematics focuses on multiplication of algebraic expressions. This exercise introduces students to techniques for multiplying monomials, binomials, and trinomials. It covers the distributive property of multiplication over addition and subtraction, and teaches how to expand expressions using methods like FOIL (First, Outer, Inner, Last) for binomial multiplication. Students learn to simplify the resulting expressions by combining like terms after multiplication.\\

Class 8 NCERT Solutions- Chapter 9 Algebraic Expressions and Identities - Exercise 9.4

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Class 8 NCERT Solutions - Chapter 9 Algebraic Expressions and Identities - Exercise 9.3

What are Algebraic Expressions and Identities?

Question 1. Carry out the multiplication of the expressions in each of the following pairs.

Question 2. Complete the table.

Question 3. Find the product.

Question 4.

Question 5.

Summary

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Chapter 9: Algebraic Expressions and Identities