 # Samacheer Kalvi 8th Maths Book Solutions Term 1 Chapter 3 | Algebra Answers For Tamil Nadu Board

Samacheer Kalvi 8th Maths Book Solutions Term 1 Chapter 3 – Algebra is available here. The Samacheer Kalvi 8th Maths book answers of Term 1 Chapter 3, available at BYJU’S, contain step by step explanations designed by our mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on samacheer kalvi 8th maths solutions.

Term 1 Chapter 3 of the Samacheer Kalvi 8th Maths guide will help the students to solve problems related to the multiplication of algebraic expressions, division of algebraic expressions, algebraic identities, factorisation.

### Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.1 Questions and Solutions

Question 1: Find the product of the terms:

[i] -2mn, (2m)2, -3mn

[ii] 3x2y, -3xy3, x2y2

Solution:

[i] -2mn, (2m)2, -3mn

-2mn * (2m)2 * -3mn

= (-2mn) * (4m2) * (-3mn)

= (-) (+) (-) (2 * 4 * 3) (mn * m2 * mn)

= 24 m4n2

[ii] 3x2y, -3xy3, x2y2

3x2y * -3xy3 * x2y2

= (-) (3 * 3 * 1) (x2y * xy3 * x2y2)

= -9x5y6

Question 2: A car moves at a uniform speed of x + 30 km/hr. Find the distance covered by the car in y + 2 hours. (Hint: distance = speed × time).

Solution:

Speed of the car = (x + 30) km / hr

Time = (y + 2) hours

Distance = Speed * Time

= (x + 30) * (y + 2)

= xy + 2x + 30y + 60

### Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.2 Questions and Solutions

Question 1: Divide

(i) (32y2 – 8yz) by 2y

(ii) 10 (4x – 8y) by 5 (x – 2y)

Solution:

(i) (32y2 – 8yz) by 2y

= [32y2 / 2y] – [8yz / 2y]

= 16y – 4z

(ii) 10 (4x – 8y) by 5 (x – 2y)

= 2 (4x – 8y) / (x – 2y)

= 8 (x – 2y) / (x – 2y)

= 8

Question 2: Find Adirai’s percentage of marks who scored 25m3n2p out of 100m2np.

Solution:

Total score = 100m2np

= {[25m3n2p] / [100m2np]} * 100%

= 25 mn%

### Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.3 Questions and Solutions

Question 1: Expand (3m + 5)2.

Solution:

(3m + 5)2

On comparing (3m + 5)2 to (a + b)2, a = 3m, b = 5

(a + b)2 = a2 + b2 + 2ab

= (3m)2 + (5)2 + 2 * 3m * 5

= 9m2 + 25 + 30m

Question 2: Find the volume of the cube whose side is x + 1 cm.

Solution:

Side of the cube = (x + 1) cm

Volume of the cube = (side)3 cubic units

= (x + 1)3 cm3

(a + b)3 = a3 + 3a2b + 3ab2 + b3 cm3

(x + 1)3 = x3 + 3 * x2 * 1 + 3 * x * 12 + 13

= x3 + 3x2 + 3x + 1 cm3

### Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.4 Questions and Solutions

Question 1: Factorise the following by taking out the common factor

(i) 18xy − 12yz

(ii) (ax + ay) (bx + by)

Solution:

(i) 18xy − 12yz

= 2 * 3 * y (3x – 2z)

= 6y (3x – 2z)

(ii) (ax + ay) (bx + by)

Taking a and b as common from each of them,

= a (x + y) b (x + y)

= (a + b) (x + y)

Question 2: Factorise the expression: x2 + 14x + 49.

Solution:

x2 + 14x + 49

= x2 + 7x + 7x + 49

= x (x + 7) + 7 (x + 7)

= (x + 7) (x + 7)

= (x + 7)2

### Samacheer Kalvi 8th Maths Term 1 Chapter 3: Algebra Book Exercise 3.5 Questions and Solutions

Question 1: Find the missing term: y2 + ( _ ) x + 56 = (y + 7) (y + _ ).

Solution:

(x + a) (x + b) = x2 + (a + b) x + ab

56 = 7 * 8

= y2 + (8 + 7) x + 8 * 7

= y2 + 15x + 56

= (y + 8) (y + 7)

Question 2: Factorise x6 – 64y3

Solution:

(x6 – 64y3)

= (x2)3 – (4y)3

a3 – b3 = (a – b) (a2 + ab + b2)

= (x2 – 4y) [(x2)2 + 4x2y + 16y2]

= (x2 – 4y) (x4 + 4x2y + 16y2)