Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.2.

RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.2.

Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.2 solution. Solution is provided in this post. Here we have provide the solutions of RBSE Boards Books according to chapter wise.

RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.2.

Question 1

By graphical method, show the 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 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 inequalities, graphically:
(i) 
(ii) 3x – 2y ≤ r + y – 8
(iii) 
Solution:
(i) Removing modules from given 
inequality , 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 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  satisfy by origin (0,0)
Therefore, shaded part from line x = 3 to
x = -3 will be 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 = 
point ( 0, ) 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 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 inequality

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 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 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.

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