RBSE Solutions Class 10 Maths Chapter 4 दो चरों वाले रैखिक समीकरण एवं असमिकाएँ Exercise 4.2

Exercise 4.2

Question 1. निम्न असमिकाओं का आलेखीय विधि से हल समुच्चय दर्शाइये
(i) \( r \ge 2 \)
(ii) \( y \le -3 \)
(iii) \( x - 2y < 0 \)
(iv) \( 2x + 3y \le 6 \)
Answer:
(i) \( x \ge 2 \)
The first step is to draw the graph of the line \( x = 2 \). This line will be parallel to the y-axis and will be at a distance of \( +2 \) units from the origin. The origin \( (0, 0) \) does not satisfy the inequality \( x \ge 2 \) because \( 0 \ge 2 \) is false. Therefore, the region to the right of \( x = 2 \) will show the solution for \( x \ge 2 \). Points on the line \( x = 2 \) are also included in the solution region. A key understanding of inequalities is that they represent regions, not just single points.

In simple words: First, draw a line for \( x = 2 \). This line goes straight up and down, parallel to the y-axis, two steps away from the center point \( (0,0) \). The center point does not fit the rule \( x \ge 2 \) because 0 is not bigger than or equal to 2. So, the answer area is everything on the right side of the line \( x = 2 \), including the line itself.

🎯 Exam Tip: Always test a point (like the origin if it's not on the line) to determine which side of the line represents the solution region for an inequality.

(ii) \( y \le -3 \)
First, draw the graph for the line \( y = -3 \). This line will be parallel to the x-axis and will be at a distance of \( -3 \) units from the origin. The origin \( (0, 0) \) does not satisfy the inequality \( y \le -3 \) because \( 0 \le -3 \) is false. So, the line \( y = -3 \) divides the xy-plane into two areas: one below the line and one above it. The origin \( (0, 0) \) is not in the region of this inequality and does not satisfy it. Therefore, the required solution region is where the origin is not included, which means the area below the line \( y = -3 \). The points on the line itself are also part of the solution. Visualizing these regions helps in understanding the solution set.

In simple words: Start by drawing the line \( y = -3 \). This line runs flat, parallel to the x-axis, three steps down from the center. The point \( (0,0) \) does not fit the rule \( y \le -3 \) because 0 is not smaller than or equal to -3. So, the answer area is everything below the line \( y = -3 \), including the line itself.

🎯 Exam Tip: Remember that horizontal lines have equations of the form \( y = k \), and vertical lines have equations of the form \( x = k \).

(iii) \( x - 2y < 0 \)
To graph \( x - 2y < 0 \), we first consider its corresponding equation \( x - 2y = 0 \), which can be rewritten as \( 2y = x \) or \( y = \frac{1}{2}x \). We find points on this line: when \( x = 2 \), \( y = 1 \); and when \( x = 4 \), \( y = 2 \). When we draw the graph, it is clear that this line passes through the origin. Since the inequality is \( < 0 \) (not \( \le \)), the points on the line itself are not included in the solution set, so we draw a dashed line. The coordinates of the origin \( (0, 0) \) do not satisfy the inequality \( x - 2y < 0 \) because \( 0 - 2(0) < 0 \) means \( 0 < 0 \), which is false. Therefore, the solution region for the inequality will not be towards the origin, but on the opposite side.

In simple words: To draw \( x - 2y < 0 \), first draw the line \( x - 2y = 0 \), which is \( y = \frac{1}{2}x \). This line goes through the center point \( (0,0) \). Points like \( (2,1) \) and \( (4,2) \) are on this line. The center point does not fit the rule \( x - 2y < 0 \) because \( 0 < 0 \) is false. So, the answer area is NOT towards the center, and the line itself is not part of the answer because it's a "less than" sign, not "less than or equal to", so we draw it as a dashed line.

🎯 Exam Tip: Use a dashed line for strict inequalities \( (< \text{ or } >) \) and a solid line for inclusive inequalities \( (\le \text{ or } \ge) \) to correctly represent the solution set.

(iv) \( 2x + 3y \le 6 \)
To find the solution for the given inequality \( 2x + 3y \le 6 \), we first write its corresponding equation: \( 2x + 3y = 6 \). To plot this line, we find two points. If we set \( x = 0 \) in the equation, we get \( 3y = 6 \), which means \( y = 2 \). So, \( (0, 2) \) is a point. Similarly, if we set \( y = 0 \), we get \( 2x = 6 \), which means \( x = 3 \). So, \( (3, 0) \) is another point. We connect these two points to draw the straight line \( 2x + 3y = 6 \). The coordinates of the origin \( (0, 0) \) satisfy the inequality \( 2x + 3y \le 6 \) (because \( 2(0) + 3(0) \le 6 \) gives \( 0 \le 6 \), which is true). Therefore, the solution region for the inequality will be towards the origin, including the points on the line \( 2x + 3y = 6 \). Linear inequalities often define half-planes, and the origin test helps determine which half-plane is the solution.

In simple words: For the inequality \( 2x + 3y \le 6 \), first draw the line \( 2x + 3y = 6 \). To do this, find points: if \( x = 0 \), then \( y = 2 \) (point \( (0,2) \)); if \( y = 0 \), then \( x = 3 \) (point \( (3,0) \)). Connect these points. The center point \( (0,0) \) fits the rule \( 2x + 3y \le 6 \) because \( 0 \le 6 \) is true. So, the answer area is towards the center point, and the line itself is included because it's "less than or equal to".

🎯 Exam Tip: Always identify two points to plot a linear equation. The x-intercept (where y=0) and y-intercept (where x=0) are often the easiest to find.

Question 2. निम्न असमिकाओं का आलेखीय विधि से हल ज्ञात कीजिए-
(i) \( \left| x \right| \le 3 \)
(ii) \( 3x - 2y \le x + y -8 \)
(iii) \( \left| x-y \right| \ge 1 \)
Answer:
(i) \( \left| x \right| \le 3 \)
The inequality \( \left| x \right| \le 3 \) means that \( x \) is between \( -3 \) and \( 3 \), inclusive. This can be broken into two inequalities: \( x \le 3 \) and \( x \ge -3 \). For \( x \le 3 \), the corresponding equation is \( x = 3 \). This is a line parallel to the y-axis, 3 units to the right of the origin. The origin \( (0,0) \) satisfies \( x \le 3 \), so the solution region is to the left of \( x = 3 \), including the line. For \( x \ge -3 \), the corresponding equation is \( x = -3 \). This is a line parallel to the y-axis, 3 units to the left of the origin. The origin \( (0,0) \) satisfies \( x \ge -3 \), so the solution region is to the right of \( x = -3 \), including the line. The solution for \( \left| x \right| \le 3 \) is the shaded area between the lines \( x = -3 \) and \( x = 3 \), including both lines. Understanding absolute value inequalities as a range is fundamental.

In simple words: \( \left| x \right| \le 3 \) means \( x \) is anywhere from \( -3 \) to \( 3 \), including \( -3 \) and \( 3 \). So, draw two vertical lines: one at \( x = 3 \) and one at \( x = -3 \). The answer is the area between these two lines, and the lines themselves are also part of the answer.

🎯 Exam Tip: Remember that \( |x| \le a \) translates to \( -a \le x \le a \), defining a region between two parallel vertical lines.

(ii) \( 3x - 2y \le x + y -8 \)
First, simplify the inequality \( 3x - 2y \le x + y -8 \) to \( 2x - 3y + 8 \le 0 \). The corresponding equation is \( 2x - 3y + 8 = 0 \). To plot this line, we find its intercepts. If we set \( x = 0 \), then \( -3y + 8 = 0 \implies 3y = 8 \implies y = \frac{8}{3} \), giving the point \( (0, \frac{8}{3}) \). If we set \( y = 0 \), then \( 2x + 8 = 0 \implies 2x = -8 \implies x = -4 \), giving the point \( (-4, 0) \). Connect these points to draw the line. Note that the inequality is \( \le \), which means it should be a solid line. Next, test the origin \( (0,0) \) in the inequality \( 2x - 3y + 8 \le 0 \). We get \( 2(0) - 3(0) + 8 \le 0 \), which simplifies to \( 8 \le 0 \). This is false. Therefore, the origin is not in the solution region. The solution area for the given inequality is on the side opposite to the origin, including the points on the line \( 2x - 3y + 8 = 0 \).

In simple words: First, make the inequality simpler: \( 2x - 3y + 8 \le 0 \). Then, draw the line \( 2x - 3y + 8 = 0 \). Find points for this line: \( (0, \frac{8}{3}) \) and \( (-4, 0) \). Join these points with a solid line. Now, check if the center point \( (0,0) \) fits the rule. \( 8 \le 0 \) is false. So, the answer area is on the side of the line opposite to the center point, including the line itself.

🎯 Exam Tip: Always simplify inequalities before graphing to avoid confusion and ensure correct identification of the boundary line.

(iii) \( \left| x-y \right| \ge 1 \)
The inequality \( \left| x-y \right| \ge 1 \) means that \( x-y \ge 1 \) or \( x-y \le -1 \).
For the inequality \( x-y \le -1 \):
The corresponding equation is \( x-y = -1 \).
To find points on this line:
If \( x = 0 \), then \( -y = -1 \implies y = 1 \). So, point \( (0, 1) \).
If \( y = 0 \), then \( x = -1 \). So, point \( (-1, 0) \).
Connect these points to draw the line \( x-y = -1 \).
Test the origin \( (0,0) \) in \( x-y \le -1 \): \( 0 - 0 \le -1 \implies 0 \le -1 \), which is false. So the solution region for this part is away from the origin, including the line.

For the inequality \( x-y \ge 1 \):
The corresponding equation is \( x-y = 1 \).
To find points on this line:
If \( x = 0 \), then \( -y = 1 \implies y = -1 \). So, point \( (0, -1) \).
If \( y = 0 \), then \( x = 1 \). So, point \( (1, 0) \).
Connect these points to draw the line \( x-y = 1 \).
Test the origin \( (0,0) \) in \( x-y \ge 1 \): \( 0 - 0 \ge 1 \implies 0 \ge 1 \), which is false. So the solution region for this part is also away from the origin, including the line.

The combined solution for \( \left| x-y \right| \ge 1 \) is the area outside the two parallel lines \( x-y = -1 \) and \( x-y = 1 \), including both lines.

In simple words: The rule \( \left| x-y \right| \ge 1 \) means that either \( x-y \) is greater than or equal to \( 1 \), or \( x-y \) is less than or equal to \( -1 \). Draw the line for \( x-y = -1 \). Points like \( (0,1) \) and \( (-1,0) \) are on it. The center point \( (0,0) \) does not fit the rule \( 0 \le -1 \). So, the answer area for this line is away from the center. Draw the line for \( x-y = 1 \). Points like \( (0,-1) \) and \( (1,0) \) are on it. The center point \( (0,0) \) does not fit the rule \( 0 \ge 1 \). So, the answer area for this line is also away from the center. Together, the solution is the area outside of these two parallel lines, including the lines themselves.

🎯 Exam Tip: Absolute value inequalities like \( |A| \ge k \) split into two separate inequalities: \( A \ge k \) or \( A \le -k \), leading to two distinct regions on the graph.