The point which does not lie in the solution region of \( 2x - y \leq 4 \) is:
\( (5,1) \)
Understanding Linear Inequalities and Solution Regions
The question asks us to identify the point that is not part of the solution region for the linear inequality \( 2x - y \leq 4 \). The solution region of a linear inequality includes all points \( (x, y) \) in the coordinate plane that satisfy the inequality.
To determine if a specific point \( (x, y) \) lies in the solution region, we need to substitute the coordinates of the point into the inequality \( 2x - y \leq 4 \) and check if the inequality holds true. If the inequality is true for a given point, that point is in the solution region. If the inequality is false, the point is not in the solution region.
Testing Each Point in the Inequality
Let's test each given point by substituting its \(x\) and \(y\) values into the inequality \( 2x - y \leq 4 \):
Point 1: \( (1, -1) \)
Substitute \( x=1 \) and \( y=-1 \) into the inequality:
\( 2(1) - (-1) \leq 4 \)
\( 2 + 1 \leq 4 \)
\( 3 \leq 4 \)
This statement is true. Therefore, the point \( (1, -1) \) lies in the solution region of \( 2x - y \leq 4 \).
Point 2: \( (5, 1) \)
Substitute \( x=5 \) and \( y=1 \) into the inequality:
\( 2(5) - (1) \leq 4 \)
\( 10 - 1 \leq 4 \)
\( 9 \leq 4 \)
This statement is false. Therefore, the point \( (5, 1) \) does not lie in the solution region of \( 2x - y \leq 4 \).
Point 3: \( (0, 2) \)
Substitute \( x=0 \) and \( y=2 \) into the inequality:
\( 2(0) - (2) \leq 4 \)
\( 0 - 2 \leq 4 \)
\( -2 \leq 4 \)
This statement is true. Therefore, the point \( (0, 2) \) lies in the solution region of \( 2x - y \leq 4 \).
Point 4: \( (1, 1) \)
Substitute \( x=1 \) and \( y=1 \) into the inequality:
\( 2(1) - (1) \leq 4 \)
\( 2 - 1 \leq 4 \)
\( 1 \leq 4 \)
This statement is true. Therefore, the point \( (1, 1) \) lies in the solution region of \( 2x - y \leq 4 \).
Conclusion
Based on our testing, the points \( (1, -1) \), \( (0, 2) \), and \( (1, 1) \) all satisfy the inequality \( 2x - y \leq 4 \) and thus lie in the solution region. The point \( (5, 1) \) does not satisfy the inequality, meaning it does not lie in the solution region.
| Point \( (x, y) \) | Substitute into \( 2x - y \leq 4 \) | Result | In Solution Region? |
|---|---|---|---|
| \( (1, -1) \) | \( 2(1) - (-1) = 3 \) | \( 3 \leq 4 \) (True) | Yes |
| \( (5, 1) \) | \( 2(5) - (1) = 9 \) | \( 9 \leq 4 \) (False) | No |
| \( (0, 2) \) | \( 2(0) - (2) = -2 \) | \( -2 \leq 4 \) (True) | Yes |
| \( (1, 1) \) | \( 2(1) - (1) = 1 \) | \( 1 \leq 4 \) (True) | Yes |
The point which does not lie in the solution region of \( 2x - y \leq 4 \) is \( (5, 1) \).
Revision Table: Linear Inequalities
| Concept | Description | How to Test a Point |
|---|---|---|
| Linear Inequality | An inequality involving linear expressions, like \( Ax + By \leq C \), \( Ax + By \geq C \), \( Ax + By < C \), or \( Ax + By > C \). | Substitute the point's coordinates \( (x, y) \) into the inequality. |
| Solution Region | The set of all points \( (x, y) \) that satisfy the linear inequality. Graphically, it's a half-plane. | If the inequality is true for the point, it's in the solution region. |
| Boundary Line | The line defined by changing the inequality sign to an equals sign (e.g., \( 2x - y = 4 \)). | If the inequality includes \( \leq \) or \( \geq \), points on the line are part of the solution region (solid line). If it includes \( < \) or \( > \), points on the line are not (dashed line). |
Additional Information: Graphing Linear Inequalities
While not required to solve this specific problem, understanding how to graph linear inequalities is helpful. For the inequality \( 2x - y \leq 4 \):
- First, graph the boundary line \( 2x - y = 4 \). You can find two points on this line, for example:
- If \( x=0 \), then \( -y=4 \), so \( y=-4 \). Point: \( (0, -4) \).
- If \( y=0 \), then \( 2x=4 \), so \( x=2 \). Point: \( (2, 0) \).
- Next, choose a test point not on the line. The origin \( (0, 0) \) is often easy.
- Substitute the test point \( (0, 0) \) into the original inequality \( 2x - y \leq 4 \): \( 2(0) - (0) \leq 4 \Rightarrow 0 \leq 4 \).
- This is true. So, the solution region is the half-plane containing the test point \( (0, 0) \). Shade that side of the line \( 2x - y = 4 \).
- Any point within the shaded region (or on the solid boundary line) is in the solution region. Any point in the unshaded region is not. Our test confirmed \( (5, 1) \) is not in the region, which aligns with it being in the unshaded area if you were to graph it.
Similar Questions
Corner points of a feasible bounded region are (0,10), (4,2), (3,7) and (10,6). Maximum value 50 of objective function \( z = ax + by \) occurs at two points (0,10) and (10,6). The value of \( a \) and \( b \) are:
The corner points of the feasible region determined by system of linear constraints are (60,0), (120,0), (40,20), and (60,30). Let \( z = ax + by \), where \( a, b > 0 \) be the objective function. Find condition on \( a \) and \( b \) so that the maximum of \( z \) occurs at (120,0) and (60,30).
The feasible region corresponding to an LPP represented by the constraints \( x \geq 7, y \geq 4, x + 2y \geq 8 \) is:
Important Questions from Linear Programming
The corner points of the feasible region determined by:
x + y ≤ 8, 2x + y ≥ 8, x ≥ 0, y ≥ 0
are A(0, 8), B(4, 0), and C(8, 0). If the objective function Z = ax + by has its maximum value on the line segment AB, then the relation between a and b is:
An objective function Z = ax + by is maximum at points (8, 2) and (4, 6). If a ≥ 0, b ≥ 0, and ab = 25, then the maximum value of the function is:
Which of the following cannot be the direction ratios of the straight line:
\( \frac{x - 3}{2} = \frac{2 - y}{3} = \frac{z + 4}{-1} \)
Corner points of a feasible bounded region are (0,10), (4,2), (3,7) and (10,6). Maximum value 50 of objective function \( z = ax + by \) occurs at two points (0,10) and (10,6). The value of \( a \) and \( b \) are:
The corner points of the feasible region determined by system of linear constraints are (60,0), (120,0), (40,20), and (60,30). Let \( z = ax + by \), where \( a, b > 0 \) be the objective function. Find condition on \( a \) and \( b \) so that the maximum of \( z \) occurs at (120,0) and (60,30).
