Question

1. graph the solution set to the given system of inequalities.

x≥0,y≥0
x + 2y≤10
x≤5

Explanation:

Step1: Analyze \(x\geq0\) and \(y\geq0\)

These inequalities represent the first - quadrant.

Step2: Analyze \(x + 2y\leq10\)

Rewrite it as \(y\leq-\frac{1}{2}x + 5\). Find two points: when \(x = 0\), \(y=5\); when \(y = 0\), \(x = 10\). Draw the line \(y=-\frac{1}{2}x + 5\) (solid line since the inequality is \(\leq\)) and shade the region below it.

Step3: Analyze \(x\leq5\)

Draw the vertical line \(x = 5\) (solid line since the inequality is \(\leq\)) and shade the region to the left of it.

Step4: Find the intersection

The intersection of all the shaded regions (first - quadrant, below \(y=-\frac{1}{2}x + 5\), and to the left of \(x = 5\)) is the solution set.

Answer:

The solution set is the intersection of the regions defined by \(x\geq0\), \(y\geq0\), \(y\leq-\frac{1}{2}x + 5\), and \(x\leq5\). Graphically, it is a polygon in the first - quadrant bounded by the \(x\) - axis (\(y = 0\)), \(y\) - axis (\(x = 0\)), the line \(x+2y = 10\), and the line \(x = 5\).