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

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

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 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 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. 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