Consider the inequations 5x - 4y + 12 < 0, x + y < 2, x < 0 and y > 0. Which one of the following points lies in the common region?
(-1, 2)
Finding the Common Region of Inequalities
We are given a set of four linear inequalities and asked to determine which of the provided points lies within the common region defined by these inequalities. The common region is the area where all the inequalities are simultaneously satisfied.
The given inequalities are:
- $\qquad 5x - 4y + 12 < 0$
- $\qquad x + y < 2$
- $\qquad x < 0$
- $\qquad y > 0$
To find which point lies in the common region, we need to test each given point by substituting its coordinates (x, y) into every inequality. If a point satisfies all four inequalities, then it lies in the common region.
Testing Each Point Against the Inequalities
Test Point 1: (0, 0)
Substitute x = 0 and y = 0 into each inequality:
- Inequality 1: $5(0) - 4(0) + 12 < 0 \implies 0 - 0 + 12 < 0 \implies 12 < 0$. This is False.
Since (0, 0) fails the first inequality, it does not lie in the common region.
Test Point 2: (-2, 4)
Substitute x = -2 and y = 4 into each inequality:
- Inequality 1: $5(-2) - 4(4) + 12 < 0 \implies -10 - 16 + 12 < 0 \implies -14 < 0$. This is True.
- Inequality 2: $-2 + 4 < 2 \implies 2 < 2$. This is False.
Since (-2, 4) fails the second inequality, it does not lie in the common region.
Test Point 3: (-1, 4)
Substitute x = -1 and y = 4 into each inequality:
- Inequality 1: $5(-1) - 4(4) + 12 < 0 \implies -5 - 16 + 12 < 0 \implies -9 < 0$. This is True.
- Inequality 2: $-1 + 4 < 2 \implies 3 < 2$. This is False.
Since (-1, 4) fails the second inequality, it does not lie in the common region.
Test Point 4: (-1, 2)
Substitute x = -1 and y = 2 into each inequality:
- Inequality 1: $5(-1) - 4(2) + 12 < 0 \implies -5 - 8 + 12 < 0 \implies -1 < 0$. This is True.
- Inequality 2: $-1 + 2 < 2 \implies 1 < 2$. This is True.
- Inequality 3: $-1 < 0$. This is True.
- Inequality 4: $2 > 0$. This is True.
Since (-1, 2) satisfies all four inequalities, it lies in the common region.
Summary of Point Testing
| Point (x, y) | $5x - 4y + 12 < 0$ | $x + y < 2$ | $x < 0$ | $y > 0$ | Lies in Common Region? |
|---|---|---|---|---|---|
| (0, 0) | $12 < 0$ (False) | N/A | N/A | N/A | No |
| (-2, 4) | $-14 < 0$ (True) | $2 < 2$ (False) | N/A | N/A | No |
| (-1, 4) | $-9 < 0$ (True) | $3 < 2$ (False) | N/A | N/A | No |
| (-1, 2) | $-1 < 0$ (True) | $1 < 2$ (True) | $-1 < 0$ (True) | $2 > 0$ (True) | Yes |
Based on the testing, the point (-1, 2) is the only one that satisfies all the given inequalities and therefore lies in the common region.
Revision Table: Understanding Inequalities
| Concept | Description |
|---|---|
| Linear Inequality | A mathematical statement involving a linear expression and an inequality symbol (<, >, ≤, or ≥). For example, $2x + 3y < 6$. |
| Solution Region | The set of all points (x, y) that satisfy a given inequality. For a linear inequality, this region is typically a half-plane. |
| Common Region (Feasible Region) | The intersection of the solution regions of all inequalities in a system. Points in this region satisfy every inequality simultaneously. |
| Graphing Inequalities | To graph a linear inequality like $Ax + By < C$, first graph the boundary line $Ax + By = C$. Use a dashed line for strict inequalities (< or >) and a solid line for non-strict inequalities ($\le$ or $\ge$). Then, pick a test point (like (0,0) if it's not on the line) and substitute it into the inequality. If the inequality is true, shade the side of the line containing the test point. If false, shade the other side. |
Additional Information: Systems of Linear Inequalities
A system of linear inequalities consists of two or more linear inequalities involving the same variables. The solution to a system of inequalities is the common region where all individual inequalities are true. This common region is often called the feasible region, especially in the context of linear programming.
Graphically, the feasible region is the area on the coordinate plane where the shaded areas of all inequalities overlap. The shape of the feasible region is a convex polygon (or an unbounded region) if it exists.
Finding points in the common region can be done by graphing or, as demonstrated in this solution, by algebraically testing points. For complex systems, graphing provides a visual representation of the common region and its vertices, while algebraic testing is useful for verifying if a specific point is within the region.
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