- Identify boundary lines by converting inequalities to equalities.
- Determine shading regions by testing points against the inequalities.
- Locate the intersection of the shaded regions to define the solution set.
- The intersection for System 1 is defined by \(y < 0.5x - 3\) and \(y \geq -x + 2\).
- The intersection for System 2 is defined by \(y > \frac{1}{3}x - 2\) and \(y > -2x + 1.5\).
Explanation
- Analysis of the Problem
To solve systems of linear inequalities, we identify the boundary lines and the regions defined by the inequalities. A system of linear inequalities represents the set of points \((x, y)\) that satisfy all conditions simultaneously. We approach each system by first graphing the corresponding linear equations as boundary lines. For strict inequalities (like \(<\) or \(>\)), the lines are dashed; for non-strict inequalities (\(\leq\) or \(\geq\)), the lines are solid. We then determine the appropriate side of each line to shade based on test points, and the final solution is the intersection (overlap) of these shaded regions.
- Solving System 1
For the first system:
\[\left\{\begin{array}{l}y < \frac{1}{2} x - 3 \ y \geq -x + 2\end{array}\right.\]
- Boundary 1: \(y = \frac{1}{2} x - 3\). Since it is a strict inequality (\(<\)), we draw a dashed line. Testing the origin \((0, 0)\): \(0 < \frac{1}{2}(0) - 3 \implies 0 < -3\), which is false. Therefore, we shade the side not containing the origin (below the line).
- Boundary 2: \(y = -x + 2\). Since it is a non-strict inequality (\(\geq\)), we draw a solid line. Testing the origin \((0, 0)\): \(0 \geq -0 + 2 \implies 0 \geq 2\), which is false. Therefore, we shade the side not containing the origin (above the line).
The solution is the intersection of these regions. Since the lines are not parallel, they intersect at \(\frac{1}{2}x - 3 = -x + 2 \implies 1.5x = 5 \implies x = \frac{10}{3} \approx 3.33\). Substituting back, \(y = -(\frac{10}{3}) + 2 = -\frac{4}{3} \approx -1.33\). The solution is the region between these lines where the two conditions overlap.
- Solving System 2
For the second system:
\[\left\{\begin{array}{l}x - 3y < 6 \ 2x + y > 1.5\end{array}\right.\]
- Boundary 1: Rewrite \(x - 3y < 6\) as \(-3y < -x + 6\), which gives \(y > \frac{1}{3}x - 2\). This is a dashed line. Since the inequality is \(>\), we shade above the line.
- Boundary 2: \(2x + y > 1.5\) becomes \(y > -2x + 1.5\). This is also a dashed line. Since the inequality is \(>\), we shade above the line.
The lines intersect where \(\frac{1}{3}x - 2 = -2x + 1.5\):
\[\frac{7}{3}x = 3.5 \implies x = 1.5\]
Substituting \(x = 1.5\), \(y = -2(1.5) + 1.5 = -1.5\). The solution is the region where both inequalities are satisfied, which corresponds to the upper quadrant defined by the intersection of these two lines.
Examples
In logistics, consider two constraints on a transport vehicle: its volume limit and its weight capacity. If \(y < 0.5x - 3\) represents a trade-off between volume and speed, and \(y \geq -x + 2\) represents a requirement for stability relative to weight, graphing these inequalities helps a logistics planner identify the safe, operational range of the vehicle. By finding the intersection of these regions, they can determine the optimal load configurations that satisfy both safety and efficiency requirements simultaneously.