RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Exercise 4.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Linear Equation and Inequalities in Two Variables |

Exercise |
Ex 4.2 |

Number of Questions Solved |
2 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.2

Question 1

By graphical method, show the (RBSESolutions.com) solution set of the following inequalities :

(i) x ≥ 2

(ii) y ≤ -3

(iii) x – 2y < 0

(iv) 2x + 3y ≤ 6

Solution:

(i) Given inequality x ≥ 2

writing its equation form, x = 2

It is clear that straight line is parallaled to y-axis and will pass through point (2, 0) of x-axis and

obtained the following graph.

Graph of two meeting (RBSESolutions.com) points is as follows:

Now, inequality 2x + 3y ≤ 6 satisfied by origin (0, 0). Therefore region from line to origin, shaded part is required solution set.

Question 2

Solve the following (RBSESolutions.com) inequalities, graphically:

(i) [latex]\left| x \right| \le 3[/latex]

(ii) 3x – 2y ≤ r + y – 8

(iii) [latex]\left| x-y \right| \ge 1[/latex]

Solution:

(i) Removing modules from given [latex]\left| x \right| \le 3[/latex]

inequality [latex]\left| x \right| \le 3[/latex], we get

⇒ -3 ≤ x ≤ 3

⇒ x ≤ 3 ……(i)

and x ≥ -3 ………..(ii)

writing inequality (j) in equation form,

x = 3

line x = 3, will be right side of y – axis and

parallel to y-axis writing (RBSESolutions.com) inequality (ii) in equation form, r = -3.

Line x = -3, will be left side of y-axis and ॥ to y-axis.

Graph of both linear equations is as follows:

Thus, inequality [latex]\left| x \right| \le 3[/latex] satisfy by origin (0,0)

Therefore, shaded part from line x = 3 to

x = -3 will be (RBSESolutions.com) required solution set.

(ii) Given inequality 3x – 2y ≤ x +y – 8 writing this, in equation form, we get

3x – 2y = x + y – 8

⇒ 3x – x – 2y – y = -8

⇒ 2x – 3y = -8

Putting x = 0 in equation, we get

2 x o – 3y = -8

-3y = -8

y = [latex]\frac { 8 }{ 3 }[/latex]

point ( 0, [latex]\frac { 8 }{ 3 }[/latex]) will cut y – axis.

Now, putting y = 0,

2x – 3 x 0 = -8

x = -4

Point (-4, 0) will cut x – axis.

Graph obtained from (RBSESolutions.com) two points will be as follows:

Thus, inequality 3x – 2y ≤ x + y – 8 does not satisfied by origin (0, 0).

⇒ 3 x 0 – 2 x 0 ≤ 0 + 0 – 8 which is not true.

Therefore shaded area opposite to origin from line will be required solution set.

(iii) Given (RBSESolutions.com) inequality

[latex]\left| x-y \right| \ge 1[/latex]

Removing modulus, we get

x – y ≥ 1

and x – y ≤ -1

writing in equation form, we get

x – y = 1 ……(i)

or x – y = -1 …(ii)

In equation (i),

Putting x = 0, 0 – y = 1

y = -1

Point (0,-1) will lie on y – axis.

Now, putting y = 0,

x – 0 = 1

x = 1

Point (1, 0) will lie on x – axis.

from equation (ii),

Putting x = 0, 0 – y = -1

y = 1

Point (0, 1) will lie on y – axis.

Now, putting y = 0

x – 0 = -1

x = -1

Point (-1, 0), will lie on x – axis.

By joining (RBSESolutions.com) points (0,-1), (1, 0) and (0, 1), (-1, 0), following graph is obtained:

Now, inequality x – y ≥ 1 does not satisfied by origin (0, 0).

Its shaded (RBSESolutions.com) portion will be the region from line to opposite to origin.

Second inequality x – y ≤ -1 also not satisfied by origin (0,0)

i.e., 0 – 0 < -1 is not true.

Therefore its shaded part will be the region from line to opposite to origin.

We hope the given RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Exercise 4.2, drop a comment below and we will get back to you at the earliest.

## Leave a Reply