Question

graph the solution set of the system of linear inequalities.
2x + y ≥ 4
x ≤ 8
x ≥ 0
y ≤ 0

use the graphing tool on the right to graph the system.
click to enlarge graph

Explanation:

Step1: Analyze \(2x + y \geq 4\)

Rewrite as \(y \geq -2x + 4\). The line \(y=-2x + 4\) has slope \(-2\) and y - intercept \(4\). Since the inequality is \(\geq\), draw a solid line and shade above the line. But we also have \(y\leq0\), so the intersection with \(y\leq0\) for this line: set \(y = 0\), \(0=-2x + 4\Rightarrow x = 2\). So the part of \(y\geq -2x + 4\) with \(y\leq0\) is from \(x = 2\) (when \(y = 0\)) and to the left (but also considering other inequalities).

Step2: Analyze \(x\leq8\)

Draw a vertical solid line at \(x = 8\) and shade to the left of the line (since \(x\) values are less than or equal to \(8\)).

Step3: Analyze \(x\geq0\)

Draw a vertical solid line at \(x = 0\) (the y - axis) and shade to the right of the line (since \(x\) values are greater than or equal to \(0\)).

Step4: Analyze \(y\leq0\)

Draw a horizontal solid line at \(y = 0\) (the x - axis) and shade below the line (since \(y\) values are less than or equal to \(0\)).

Step5: Find the intersection

The solution set is the region that satisfies all four inequalities. So we look for the area that is: to the right of \(x = 0\), to the left of \(x = 8\), below \(y = 0\), and above (or on) \(y=-2x + 4\) (with the constraint \(y\leq0\)). The feasible region is a polygon (or a line segment in this case) bounded by the intersection of these half - planes. The intersection of \(y\geq - 2x+4\), \(x\leq8\), \(x\geq0\), and \(y\leq0\) occurs where \(y = 0\), \(x\) is between \(2\) (from \(y=-2x + 4,y = 0\)) and \(8\)? Wait, no. When \(y = 0\), \(x\) for \(y=-2x + 4\) is \(x = 2\). For \(x>2\) and \(y = 0\), \(2x+0=2x\geq4\) (since \(x\geq2\) when \(y = 0\)). And \(x\) is between \(2\) and \(8\) (because \(x\geq0\), \(x\leq8\), \(y\leq0\) and \(y\geq - 2x + 4\)). But also, since \(y\leq0\), the region is on or below the x - axis, on or to the right of \(x = 0\), on or to the left of \(x = 8\), and on or above \(y=-2x + 4\) (with \(y\leq0\)). So the feasible region is the line segment from \((2,0)\) to \((8,0)\)? Wait, no. Let's check for \(x = 8\) and \(y = 0\): \(2(8)+0=16\geq4\), which satisfies \(2x + y\geq4\). For \(x = 2\) and \(y = 0\): \(2(2)+0 = 4\geq4\), which satisfies \(2x + y\geq4\). For any \(x\) between \(2\) and \(8\) and \(y = 0\), \(2x+0=2x\geq4\) (since \(x\geq2\)). And since \(y = 0\) satisfies \(y\leq0\), \(x\) between \(0\) and \(2\) with \(y = 0\) does not satisfy \(2x + y\geq4\) (e.g., \(x = 1,y = 0\): \(2(1)+0=2<4\)). So the solution set is the line segment connecting the points \((2,0)\) and \((8,0)\) (since at \(y = 0\), \(x\) ranges from \(2\) to \(8\), \(x\geq0\), \(x\leq8\), and \(y\leq0\) and \(2x + y\geq4\)). To graph it, we plot the line segment from \((2,0)\) to \((8,0)\) (since all points on this segment satisfy \(2x+0\geq4\) (because \(x\geq2\)), \(x\leq8\), \(x\geq0\), and \(y = 0\leq0\)).

Answer:

The solution set is the line segment on the x - axis from the point \((2,0)\) to the point \((8,0)\) (and the region is this line segment as it satisfies all four inequalities: \(2x + y\geq4\) (since \(x\geq2\) and \(y = 0\) gives \(2x\geq4\)), \(x\leq8\), \(x\geq0\), and \(y\leq0\)). When graphing, we draw the solid line segment between \((2,0)\) and \((8,0)\) (as this is the intersection of all four half - planes).