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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.10

Determine the region in the plane determined by the inequalities:

Question 1.

x ≤ 3y, x ≥ y

Solution:

Given in equation are x ≤ 3y,x ≥ y

Suppose x = 3y ⇒ \(\frac{x}{3}=\) = y

If x = y

Question 2.

y ≥ 2x, -2x + 3y ≤ 6

Solution:

Suppose y = 2x

-2x + 3y = 6

-2x = 6 – 3y

Question 3.

3x + 5y ≥ 45, x ≥ 0, y ≥ 0.

Solution:

If 3x + 5y = 45

x ≥ 0 is nothing but the positive portion of Y-axis and y ≥ 0 is the positive portion of X-axis.

Shaded region is the required portions.

Question 4.

2x + 3y ≤ 35, y ≥ 2, x ≥ 5

Solution:

If 2x + 3y = 35 then

y = 2 is a line parallel to X-axis at a distance 2 units

x = 5 is a line parallel to Y-axis at a distance of 5 units

The required region is below 2x + 3y = 35, above y = 2 and to the right of x = 5

Question 5.

2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0.

Solution:

If 2x + 3y = 6

x + 4y = 4

x ≥ 0, y ≥ 0 represents the area in the 1 quadrant.

The required region is below 2x + 3y = 6 and below x + 4y = 4 bounded by x-axis and y-axis.

Question 6.

x – 2y ≥ 0, 2x – y ≤ -2, x ≥ 0, y ≥ 0

Solution:

If x – 2y = 0

2x – y = -2

x ≥ 0, y ≥ 0 represents the portion in the 1 quadrant only.

Question 7.

2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6.

Solution:

2x + y = 8

x + 2y = 8

x + y = 6

### Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.10 Additional Questions

Question 1.

3x + 2y ≤ 12, x ≥ 1, y ≥ 2

Solution:

The given inequality is 3x + 2y ≤ 12.

Draw the graph of the line 3x + 2y ≤ 12

Table of values satisfying the equation

3x + 2y ≤ 12

Putting (0, 0) in the given inequation, we have 3 × 0 + 2 × 0 ≤ 12

∴ Half plane of 3x + 2y ≤ 12 is towards origin

Also the given inequality is x ≥ 1.

Draw the graph of the line x = 1.

Putting (0, 0) in the given inequation, we have 0 ≥ 1 which is false.

∴ Half the plane of x ≥ 1 is away from the origin.

The given inequality is y ≥ 2.

Putting (0, 0) in the given inequation, we have 0 ≥ 2 which is false.

∴ Half plane of y ≥ 2 is away from the origin.

Question 2.

x + y ≥ 4, 2x – y > 0

Solution:

The given inequality is x + y ≥ 4.

Draw the graph of the line x + y = 4.

Table of values satisfying the equation

x + y = 4

Putting (0, 0) in the given inequation, we have 0 + 0 ≥ 4 ⇒ 0 ≥ 4, which is false.

∴ Half plane of x + y ≥ 4 is away from the origin.

Also the given inequality is 2x – y > 0.

Draw the graph of the line 2x – y = 0.

Table of values satisfying the equation

2x – y = 0

Putting (3, 0) in the given inequation, we have 2 × 3 – 0 > 0 ⇒ 6 > 0, which is true.

∴ Half plane of 2x – y > 0 containing (3, 0)

Question 3.

x + y ≤ 9, y > x, x ≥ 0

Solution:

The given inequality is x + y ≤ 9.

Draw the graph of the line x + y = 9.

Table of values satisfying the equation

x + y = 9

Putting (0, 0) in the given inequation, we have 0 + 0 ≤ 9⇒ 0 ≤ 9, is towards is the origin.

∴ Half plane of x + y ≤ 9 is away from the origin.

Also, the given inequality is x – y < 0.

Draw the graph of line x -y = 0.

Table of values satisfying the equation

x – y = 0

Putting (0, 3) in the given inequation, we have 0 – 3 – 0 < 0 ⇒ -3 < 0, which is true.

∴ Half plane of x – y < 0 containing the points (0, 3).

Question 4.

5x + 4y ≤ 20, x ≥ 1, y ≥ 2

Solution:

The given inequality is 5x + 4y ≤ 20.

Draw the graph of the line 5x + 4y = 20.

Table of values satisfying the equation

5x + 4y = 20

Putting (0, 0) in the given inequation, we have 5 × 0 + 4 × 0 ≤ 20 ⇒ 0 ≤ 20, which is true.

∴ Half plane of 5x + 4y ≤ 20 is away from the origin.

Also, the given inequality is x ≥ 1.

Draw the graph of the line x = 1.

Putting (0, 0) in the given inequation, we have 0 ≥ 1, which is false.

∴ Half plane of x ≥ 1 is y ≥ 2.

Draw the graph of the line y = 2.

Putting (0, 0) in the given inequation, we have 0 ≥ 2, which is false.

∴ Half plane of y ≥ 2 is away from the origin.

Question 5.

3x +4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

Solution:

The given inequality is 3x + 4y ≤ 60.

Draw the graph of the line 3x + 4y = 60.

Table of values satisfying the equation

3x + 4y = 60

Putting (0, 0) in the given inequation, we have 3 × 0 + 4 × 0 ≤ 60 ⇒ 0 ≤ 60, which is true.

∴ Half the plane of 3x + 4y ≤ 60 is towards the origin.

Also, the given inequality is x + 3y ≤ 30.

Draw the graph of the line x + 3y = 30.

Table of values satisfying the equation.

x + 3y = 30

Putting (0, 0) in the given inequation, we have 0 + 3 × 0 ≤ 30 ⇒ 0 ≤ 30, which is true.

∴ Half plane of x + 3y ≤ 30 is towards the origin.