Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.10

Determine the region in the plane determined by the inequalities

Question 1. \(x \le 3y\), \(x \ge y\)
Answer:
To find the region for these inequalities, we first consider the boundary lines and then test points to determine the shaded areas.

First, consider the line \(x = 3y\). To draw this line, we find a few points:

To find the region for \(x \le 3y\), we test a point like (1, 1). If we substitute these values into the inequality, we get \(1 \le 3(1)\), which simplifies to \(1 \le 3\). This statement is true. The point (1, 1) lies above the line \(x = 3y\). So, all points satisfying the inequality \(x \le 3y\) will lie on or above the line \(x = 3y\). This includes the line itself because of the "equal to" part of the inequality. The line \(x=3y\) divides the plane into two sections; one side contains the origin, and the other does not. Here, (0,0) is on the line itself, so testing a point not on the line helps determine the region. The upper half-plane (above the line) satisfies the inequality.

Next, consider the line \(x = y\). To draw this line, we find a few points:

To find the region for \(x \ge y\), we test a point like (2, 1). If we substitute these values into the inequality, we get \(2 \ge 1\). This statement is true. The point (2, 1) lies below the line \(x = y\). So, all points satisfying the inequality \(x \ge y\) will lie on or below the line \(x = y\). This includes the line itself. The lower half-plane (below the line) satisfies the inequality.

The first graph below shows the region for \(x \le 3y\).

The second graph below shows the region for \(x \ge y\).

The final required region is where both inequalities are true. This is the common area where the two shaded regions overlap. The intersection forms the solution set.

🎯 Exam Tip: Always use a test point (like the origin if it's not on the line) to correctly determine which side of the boundary line represents the inequality. For "less than or equal to" or "greater than or equal to" inequalities, the boundary line itself is part of the solution.

Question 2. \(y \ge 2x\), \(-2x + 3y \le 6\)
Answer:
To determine the region, we analyze each inequality by first drawing its boundary line.

First, consider the line \(y = 2x\). We find points to plot this line:

To find the region for \(y \ge 2x\), we test a point like (1, 3). If we substitute these values into the inequality, we get \(3 \ge 2(1)\), which simplifies to \(3 \ge 2\). This statement is true. The point (1, 3) lies above the line \(y = 2x\). So, all points satisfying the inequality \(y \ge 2x\) will lie on or above the line \(y = 2x\). The line \(y = 2x\) passes through the origin. Therefore, the region includes the upper half-plane defined by this line.

The first graph below shows the region for \(y \ge 2x\).

Next, consider the line \(-2x + 3y = 6\). We find points to plot this line:

To find the region for \(-2x + 3y \le 6\), we substitute the origin (0, 0) into the inequality: \(-2(0) + 3(0) \le 6\), which simplifies to \(0 \le 6\). This statement is true. Since the origin (0, 0) satisfies the inequality, the region for \(-2x + 3y \le 6\) is the half-plane that contains the origin. This includes the line itself because of the "equal to" part of the inequality.

The graph below shows the region for \(-2x + 3y \le 6\).

The final required region is the area common to both inequalities. This means it is above the line \(y = 2x\) and on the side of the line \(-2x + 3y = 6\) that includes the origin.

🎯 Exam Tip: When dealing with multiple inequalities, it's helpful to shade each region separately (perhaps with different colored pencils on scratch paper) and then identify the area where all shadings overlap. This overlapping region is your final answer.

Question 3. \(3x + 5y \ge 45\), \(x \ge 0\), \(y \ge 0\)
Answer:
To find the region, we will consider each inequality one by one.

First, consider the line \(3x + 5y = 45\). We find points to plot this line:

To find the region for \(3x + 5y \ge 45\), we substitute the origin (0, 0) into the inequality: \(3(0) + 5(0) \ge 45\), which simplifies to \(0 \ge 45\). This statement is false. Since the origin (0, 0) does not satisfy the inequality, the region for \(3x + 5y \ge 45\) is the half-plane that does not contain the origin. This includes the line itself because of the "equal to" part of the inequality.

The inequality \(x \ge 0\) means all points to the right of the y-axis (including the y-axis itself). The inequality \(y \ge 0\) means all points above the x-axis (including the x-axis itself). Together, \(x \ge 0\) and \(y \ge 0\) represent the first quadrant of the Cartesian plane, where both x and y coordinates are non-negative.

The graph below shows the region for \(x \ge 0, y \ge 0\).

The graph below shows the region for \(3x + 5y \ge 45\).

The required region is the common area where \(x \ge 0\), \(y \ge 0\), and \(3x + 5y \ge 45\) are all true. This region is unbounded and lies in the first quadrant, above and to the right of the line \(3x + 5y = 45\).

🎯 Exam Tip: Remember that \(x \ge 0\) and \(y \ge 0\) always define the first quadrant. When combining these with other inequalities, the solution region must always be entirely within this first quadrant.

Question 4. \(2x + 3y \le 35\), \(y \ge 2\), \(x \ge 5\).
Answer:
To find the region, we analyze each inequality separately.

First, consider the line \(2x + 3y = 35\). We find points to plot this line:

To find the region for \(2x + 3y \le 35\), we substitute the origin (0, 0) into the inequality: \(2(0) + 3(0) \le 35\), which simplifies to \(0 \le 35\). This statement is true. Since the origin (0, 0) satisfies the inequality, the region for \(2x + 3y \le 35\) is the half-plane that contains the origin. This includes the line itself.

The graph below shows the region for \(2x + 3y \le 35\).

Next, consider the inequality \(y \ge 2\). This represents all points on or above the horizontal line \(y = 2\). To verify, substitute the origin (0, 0): \(0 \ge 2\) is false. So, the region is the half-plane that does not contain the origin, which is above the line \(y = 2\).

The graph below shows the region for \(y \ge 2\).

Finally, consider the inequality \(x \ge 5\). This represents all points on or to the right of the vertical line \(x = 5\). To verify, substitute the origin (0, 0): \(0 \ge 5\) is false. So, the region is the half-plane that does not contain the origin, which is to the right of the line \(x = 5\).

The graph below shows the region for \(x \ge 5\).

The required region is the common area where all three inequalities are true: below the line \(2x + 3y = 35\), above the line \(y = 2\), and to the right of the line \(x = 5\).

🎯 Exam Tip: When dealing with more than two inequalities, precisely locating the vertices of the feasible region (the corners where lines intersect) is crucial for accurate shading. Always check if these intersection points satisfy all inequalities.

Question 5. Determine the region in the plane determined by the inequalities \(2x + 3y \le 6\), \(x + 4y \le 4\), \(x \ge 0\), \(y \ge 0\).
Answer:
To find the region, we will consider each inequality by drawing its boundary line.

First, consider the line \(2x + 3y = 6\). We find points to plot this line:

To find the region for \(2x + 3y \le 6\), we substitute the origin (0, 0) into the inequality: \(2(0) + 3(0) \le 6\), which simplifies to \(0 \le 6\). This statement is true. So, the region is the half-plane containing the origin, including the line itself.

The graph below shows the region for \(2x + 3y \le 6\).

Next, consider the line \(x + 4y = 4\). We find points to plot this line:

To find the region for \(x + 4y \le 4\), we substitute the origin (0, 0) into the inequality: \(0 + 4(0) \le 4\), which simplifies to \(0 \le 4\). This statement is true. So, the region is the half-plane containing the origin, including the line itself.

🎯 Exam Tip: Always make sure to include \(x \ge 0\) and \(y \ge 0\) in your analysis if the problem describes a "region in the plane" and these are typically implied for feasible regions in many contexts unless stated otherwise. These restrictions mean your solution will only be in the first quadrant.

Question 6. x - 2y ≥ 0, 2x - y ≤ -2, x ≥ 0, y ≥ 0
Answer:

First, let's consider the straight line \( x - 2y = 0 \). This line defines the boundary for the inequality \( x - 2y \ge 0 \).
When \( x = 0 \), we have \( -2y = 0 \implies y = 0 \). So, the line passes through \( (0,0) \).
When \( x = 2 \), we have \( 2 - 2y = 0 \implies 2y = 2 \implies y = 1 \). This gives us the point \( (2,1) \).
When \( x = -2 \), we have \( -2 - 2y = 0 \implies -2y = 2 \implies y = -1 \). This gives us the point \( (-2,-1) \).

To determine the region for \( x - 2y \ge 0 \), we use a test point. Since the line passes through the origin \( (0,0) \), we cannot use it as a test point. Let's try \( (3,1) \). Substituting \( x=3 \) and \( y=1 \) into \( x - 2y \ge 0 \) gives \( 3 - 2(1) \ge 0 \implies 1 \ge 0 \), which is true. The point \( (3,1) \) lies below the line \( x = 2y \) (or \( x - 2y = 0 \)). Therefore, all points satisfying \( x - 2y \ge 0 \) lie below the line \( x = 2y \), including the points on the line itself. The line \( x = 2y \) is the boundary.

Next, let's consider the straight line \( 2x - y = -2 \). This line sets the boundary for the inequality \( 2x - y \le -2 \).
When \( x = 0 \), we get \( -y = -2 \implies y = 2 \). This means the line passes through \( (0,2) \).
When \( y = 0 \), we get \( 2x = -2 \implies x = -1 \). This means the line passes through \( (-1,0) \).

To find the region for \( 2x - y \le -2 \), let's test the origin \( (0,0) \). Substituting \( x=0 \) and \( y=0 \) into \( 2x - y \le -2 \) gives \( 2(0) - 0 \le -2 \implies 0 \le -2 \), which is false. This means the origin \( (0,0) \) is not part of the solution region. So, we shade the half-plane that does not contain the origin. Since \( 2x - y \le -2 \) implies that the region includes points on the line \( 2x - y = -2 \), the line is a solid boundary.

The conditions \( x \ge 0 \) and \( y \ge 0 \) together define the first quadrant of the Cartesian plane. This means we are only interested in the upper-right portion of the graph where both x and y values are positive or zero. This ensures that our solution lies in a physically meaningful area.

The required region is the area where all these conditions overlap. We need to find the common region bounded by \( x - 2y \ge 0 \), \( 2x - y \le -2 \), \( x \ge 0 \), and \( y \ge 0 \). This final graph shows the area that satisfies all the given inequalities at the same time.
In simple words: We draw the boundary lines for each rule. Then, for each rule, we figure out which side of its line is correct. The final answer is the area where all the correct sides overlap, and it must also be in the top-right part of the graph because x and y must be positive.

🎯 Exam Tip: When testing points for inequalities, always choose a point not on the boundary line, such as the origin (0,0), if the line does not pass through it. This makes it easy to see which side of the line satisfies the inequality.

Question 7. 2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6
Answer:

First, let's consider the straight line \( 2x + y = 8 \). This line defines the boundary for the inequality \( 2x + y \ge 8 \).
When \( x = 0 \), we have \( 2(0) + y = 8 \implies y = 8 \). So, the line passes through \( (0,8) \).
When \( y = 0 \), we have \( 2x + 0 = 8 \implies 2x = 8 \implies x = 4 \). So, the line passes through \( (4,0) \).

To determine the region for \( 2x + y \ge 8 \), let's test the origin \( (0,0) \). Substituting \( x=0 \) and \( y=0 \) into \( 2x + y \ge 8 \) gives \( 2(0) + 0 \ge 8 \implies 0 \ge 8 \), which is false. This means the origin \( (0,0) \) is not part of the solution region. Since the line is included in the inequality, we shade the half-plane that does not contain the origin. The line \( 2x + y = 8 \) acts as the boundary.

Next, let's consider the straight line \( x + 2y = 8 \). This line defines the boundary for the inequality \( x + 2y \ge 8 \).
When \( x = 0 \), we have \( 0 + 2y = 8 \implies y = 4 \). So, the line passes through \( (0,4) \).
When \( y = 0 \), we have \( x + 2(0) = 8 \implies x = 8 \). So, the line passes through \( (8,0) \).

To find the region for \( x + 2y \ge 8 \), let's test the origin \( (0,0) \). Substituting \( x=0 \) and \( y=0 \) into \( x + 2y \ge 8 \) gives \( 0 + 2(0) \ge 8 \implies 0 \ge 8 \), which is false. This means the origin \( (0,0) \) is not part of the solution region. We shade the half-plane that does not contain the origin. This boundary line is included in the solution.

Finally, let's consider the straight line \( x + y = 6 \). This line defines the boundary for the inequality \( x + y \le 6 \).
When \( x = 0 \), we have \( 0 + y = 6 \implies y = 6 \). So, the line passes through \( (0,6) \).
When \( y = 0 \), we have \( x + 0 = 6 \implies x = 6 \). So, the line passes through \( (6,0) \).

To find the region for \( x + y \le 6 \), let's test the origin \( (0,0) \). Substituting \( x=0 \) and \( y=0 \) into \( x + y \le 6 \) gives \( 0 + 0 \le 6 \implies 0 \le 6 \), which is true. This means the origin \( (0,0) \) is part of the solution region. We shade the half-plane that contains the origin. The line \( x + y = 6 \) is part of the solution boundary.

The required region is the common area that satisfies all three inequalities: \( 2x + y \ge 8 \), \( x + 2y \ge 8 \), and \( x + y \le 6 \). This region is a triangle formed by the intersection of these three half-planes. Finding the exact corner points helps to define this common solution space.
In simple words: For each rule, we draw its line. Then, we find the side of the line that makes the rule true. The final answer is the shape where all the "true" sides overlap, forming a single area.

🎯 Exam Tip: When dealing with multiple inequalities, carefully identify the vertices of the feasible region by finding the intersection points of the boundary lines. These points define the corners of your solution area.