Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

Additional Questions and Answers

Exercise 3.1

Question 1.
Write any three expressions each having 4 terms:
Solution:
(i) 2x3 – 3x2 + 3xy + 8
(ii) 7x3 + 9y2 – 2xy2 – 6
(iii) 9x2 – 2x + 3xy – 1

Question 2.
Identify the co-efficients of the terms of the following expressions
(i) 2x – 2y
(ii) x + y +3
Solution:
(i) 2x – 2y
The co-efficient of x in 2x is 2
The co-efficient of y in – 2y is – 2

(ii) x + y + 3
The co-efficient of x is 1
The co-efficient ofy is 1
The constant term is 3

SamacheerKalvi.Guru

Question 3.
Group the like terms together from the following: 6x, 6, -5x, – 5, 1, x, 6y, y, 7y, 16x, 3
Solution:
We have 6x, -5x, x, 16x are like terms
6y, y, 7y, are like terms
6, – 5, 1, 3 are like terms

Question 4.
Give the algebraic expressions for the following cases:
(i) One half of the sum of a and b.
(ii) Numbers p and q both squared and added
Solution:
(i) \(\frac{1}{2}\) (a + b)
(ii) p2 + q2

Exercise 3.2

Question 1.
If A = 2a2 – 4b – 1 ; B = 5a2 + 3b – 8 and C = 2a2 – 9b + 3 then find the value of A – B + C.
Solution:
Given A = 2a2 – 4b – 1 ; B = 5a2 + 3b – 8 ; C = 2a2 – 9b + 3
A – B + C = (2a2 – 4b – 1) – (5a2 + 3b – 8) + (2a2 – 9b + 3)
= 2a2 – 4b – 1 + (-5a2 – 3b + 8) + 2a2 – 9b + 3
= 2a2 – 4b – 1 – 5a2 – 3b + 8 + 2a2 – 9b + 3
= 2a2 – 5a2 + 2a2 – 4b – 3b – 9b – 1 + 8 + 3
= (2 – 5 + 2) a2 + (-4 – 3 – 9) 6 + (-1 + 8 + 3)
= -a2 – 16b + 10

Question 2.
How much 2x3 – 2x2 + 3x + 5 is greater than 2x3 + 7x2 – 2x + 7?
Solution:
The required expression can be obtained as follows.
= 2x3 – 2x2 + 3x + 5 – (2x3 + 7x2 – 2x + 7)
= 2x3 – 2x2 + 3x + 5 + (-2x3 – 7x2 + 2x – 7)
= 2x3– 2x2 + 3x + 5 – 2x3 – 7x2 + 2x – 7
= (2 – 2) x3 + (-2 – 7) x2 + (3 + 2) x + (5 – 7)
= 0x3 + (-9x2) + 5x – 2 = -9x2 + 5x – 2
∴ 2x3 – 2x2 + 3x + 5 is greater than 2x3 + 7x2 – 2x + 7 by -9x2 + 5x – 2

SamacheerKalvi.Guru

Question 3.
What should be added to 2b2 – a2 to get b2 – 2a2
Solution:
The required expression is obtained by subtracting 2b2 – a2 from b2 – 2a2
b2 – 2a2 – (2b2 – a2) = b2 – 2a2 + (-2b2 + a2)
= b2 – 2a2 – 2b2 + a2
= (1 – 2) b2 + (-2 + 1) a2 = -b2 – a2
So -b2 – a2 must be added

Exercise 3.3

Question 1.
Length of one side of an equilateral triangle is 3x – 4 units. Find the perimeter.
Solution:
Equilateral triangle has three sides equal.
Perimeter = Sum of three sides
= (3x – 4) + (3x – 4) + (3x – 4) = 3x – 4 + 3x – 4 + 3x – 4
= (3 + 3 + 3)x + [(-4) + (-4) + (-4)] = 9x + (-12) = 9x – 12
∴ Perimeter = 9x – 12 units.

Question 2.
Find the perimeter of a square whose side is y – 2 units.
Solution:
Perimeter = (y – 2) + (y – 2) + (y – 2) + (y – 2)
= y – 2 + y – 2 + y – 2 + y – 2 = 4y – 8
Perimeter of the square = 4y – 8 units.

SamacheerKalvi.Guru

Question 3.
Simplify 3x – 5 – x + 9 if x = 3
Solution:
3x – 5 – x + 9 = 3(3) – 5 – 3 + 9
= 9 – 5 – 3 + 9 = 18 – 8 = 10

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