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

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

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

Additional Questions and Answers

Exercise 3.1

Question 1.
If 4x² + y² = 40 and xy = b find the value of 2x + y.
Solution:
We have (a + b)² = a² + 2ab + b²
(2x + y)² = (2x)² + (2 × 2x × y) + y² = (4x² + y²) + 4xy = 40 + 4 × 6 = 40 + 24
(2x + y)² = 64
(2x + y)² = 82
2x + y = 8

Exercise 3.2

Question 1.
Solve 2x + 5 < 15 where x is a natural number and represent the solution in a number line.
Solution:
2x + 5 < 15
Subtracting 5 on both sides 2x + 5 – 5 < 15 – 5
2x < 10
Dividing by 2 on both the sides
2x/2 < 10/2
Since x is a natural number and it is less than 5, the solution is 4, 3, 2 and 1. It is shown in the number line as below.

Question 2.
Solve 2c + 4 < 14, where c is a whole number.
Solution:
2c + 4 < 14
Subtracting 4 on both sides 2c + 4 – 4 < 14 – 4
2c < 10
Dividing by 2 on both the sides
2c/2 < 10/2
c < 5
Since the solutions are whole numbers which are less than r equal to 5, the solution set is 0, 1, 2, 3, 4 and 5.

Question 3.
Solve -8 < -2n + 4, n is a natural number.
Solution:
-8 < – 2n + 4
Subtracting 4 on both sides
-8 -4 < -2n + 4 – 4
– 12 < – 2 n
÷ by -2, we have
−2n/−2 < −12/−2 [ ∵ Dividing by negative number, the inequation get reversed]
n < 6
Since the solutions are natural numbers which are less then 6, we have the solution as 1, 2, 3, 4 and 5.