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

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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

Question 2.
If x² + \(\frac{1}{x^{2}}\) =23 find x + \(\frac { 1 }{ x } \)
Solution:
We have (a + b)² = a² + 2ab + b²
So (x + \(\frac { 1 }{ x } \))² = x² + 2 × x × \(\frac { 1 }{ x } \) + \(\frac{1}{x^{2}}\)
= x² + 2 + \(\frac{1}{x^{2}}\) = x² + \(\frac{1}{x^{2}}\) + 2 = 23 + 2
∵ x² + \(\frac{1}{x^{2}}\) = 23
(x + \(\frac { 1 }{ x } \))² = 25
(x + \(\frac { 1 }{ x } \))² = 52
x + \(\frac { 1 }{ x } \) = 5

Question 3.
Find the product of (\(\frac { 2 }{ 3 } \) x² + 5y²) (\(\frac { 2 }{ 3 } \) x² + 5y²)
Solution:
(\(\frac { 2 }{ 3 } \) x² + 5y²) (\(\frac { 2 }{ 3 } \) x² + 5y²) = (\(\frac { 2 }{ 3 } \) x² + 5y²)²
We have (a + b)² = a² + 2ab + b²
Here a = \(\frac { 2 }{ 3 } \) x² b = 5y²
(\(\frac { 2 }{ 3 } \) x² + 5y²)² = (\(\frac { 2 }{ 3 } \) x²)² + 2 × \(\frac { 2 }{ 3 } \) x² × 5y² + (5y²)²
= (\(\frac { 2 }{ 3 } \))² (x²)² + \(\frac{20 x^{2} y^{2}}{3}\) + 52 (y²)²
(\(\frac { 2 }{ 3 } \) x² + 5y²)² = \(\frac { 4 }{ 9 } \) x⁴ + \(\frac{20 x^{2} y^{2}}{3}\) + 25y⁴

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
\(\frac { 2x }{ 2 } \) < \(\frac { 10 }{ 2 } \)
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
\(\frac { 2c }{ 2 } \) < \(\frac { 10 }{ 2 } \)
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
\(\frac { -2n }{ -2 } \) < \(\frac { -12 }{ -2 } \) [ ∵ 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.