- For Problem 7, the solution is the intersection of the two half-planes defined by \(5x + 3y \geq -15\) (solid boundary) and \(2x + 6y < -9\) (dashed boundary).
- For Problem 9, the solution is the intersection of three half-planes defined by \(y < 3x - 4\) (dashed), \(2x - 3y > 6\) (dashed), and \(y - 2 \leq -0.6(x - 4)\) (solid).
- Both problems result in a specific region on the Cartesian plane determined by the intersection of the given linear inequalities.
- Final answer: \(\boxed{\text{Regional solutions defined by systems of linear inequalities}}\)
Explanation
- Analyzing Problem 7
To analyze Problem 7, we examine the system of inequalities:
- \(5x + 3y \geq -15\)
- \(2x + 6y < -9\)
For the first inequality \(5x + 3y \geq -15\), the boundary line is \(5x + 3y = -15\). Since the inequality is "greater than or equal to", the boundary line is solid. We can find two points on this line: if \(x=0\), \(y=-5\); if \(y=0\), \(x=-3\). Testing the origin \((0,0)\): \(5(0) + 3(0) = 0 \geq -15\), which is true. Thus, the solution region for this inequality is the half-plane containing the origin.
For the second inequality \(2x + 6y < -9\), the boundary line is \(2x + 6y = -9\). Since the inequality is "less than", the boundary line is dashed. Two points on this line: if \(x=0\), \(y=-1.5\); if \(y=0\), \(x=-4.5\). Testing the origin \((0,0)\): \(2(0) + 6(0) = 0 < -9\), which is false. Thus, the solution region for this inequality is the half-plane not containing the origin.
The final solution for Problem 7 is the intersection of these two half-planes.
- Analyzing Problem 9
To analyze Problem 9, we examine the system of three inequalities:
- \(y < 3x - 4\)
- \(2x - 3y > 6\)
- \(y - 2 \leq -\frac{3}{5}(x - 4)\)
For the first inequality \(y < 3x - 4\), the boundary line is \(y = 3x - 4\). This is a dashed line. Testing \((0,0)\): \(0 < 3(0) - 4 = -4\), which is false. The region is below the line.
For the second inequality \(2x - 3y > 6\), the boundary line is \(2x - 3y = 6\), or \(y = \frac{2}{3}x - 2\). This is a dashed line. Testing \((0,0)\): \(2(0) - 3(0) = 0 > 6\), which is false. The region is below the line (the side not containing the origin).
For the third inequality \(y - 2 \leq -\frac{3}{5}(x - 4)\), the boundary line is \(y = -\frac{3}{5}x + \frac{12}{5} + 2 = -0.6x + 4.4\). This is a solid line. Testing \((0,0)\): \(0 - 2 = -2 \leq -\frac{3}{5}(0 - 4) = 2.4\), which is true. The region is below the line (including the line).
The intersection of these three half-planes defines the solution region for Problem 9.
- Conclusion
In summary, Problem 7 defines a region between two lines where one boundary is solid and the other is dashed. Problem 9 defines a region bounded by three lines, forming a triangular region in the Cartesian plane where two sides are dashed and one is solid.
The solution to Problem 7 is the intersection of the half-planes \(5x + 3y \geq -15\) and \(2x + 6y < -9\).
The solution to Problem 9 is the intersection of the half-planes \(y < 3x - 4\), \(2x - 3y > 6\), and \(y - 2 \leq -0.6(x - 4)\).
These systems represent planar regions defined by linear constraints. \(\boxed{\text{Systems of linear inequalities}}\)
Examples
Understanding systems of inequalities is essential in optimization problems, often referred to as linear programming. For example, a bakery might have limited resources like flour and sugar to bake two types of bread. If each type requires different amounts of these ingredients, the set of all possible combinations of breads they can bake is constrained by linear inequalities. By graphing these, the bakery owner can visually identify the feasible production region and determine the best strategy to maximize profit.