Question

1. graph the system of inequality \\(\

$$\begin{cases}5x + 2y \\leq 10 \\\\ y > -x + 1\\end{cases}$$

\\) to show the solution set.

Explanation:

Step1: Rewrite first inequality

Rewrite \(5x + 2y \leq 10\) in slope - intercept form (\(y=mx + b\)). Solve for \(y\):
\(2y\leq - 5x + 10\), then \(y\leq-\frac{5}{2}x + 5\). The boundary line \(y =-\frac{5}{2}x+5\) is a solid line (because the inequality is \(\leq\)) with slope \(m =-\frac{5}{2}\) and \(y\) - intercept \(b = 5\). To find two points on this line, when \(x = 0\), \(y=5\); when \(y = 0\), \(0=-\frac{5}{2}x + 5\), \(\frac{5}{2}x=5\), \(x = 2\). So the line passes through \((0,5)\) and \((2,0)\). Shade the region below this line (since \(y\leq-\frac{5}{2}x + 5\)).

Step2: Analyze second inequality

For the inequality \(y>-x + 1\), the boundary line \(y=-x + 1\) is a dashed line (because the inequality is \(>\)) with slope \(m=-1\) and \(y\) - intercept \(b = 1\). When \(x = 0\), \(y = 1\); when \(y=0\), \(0=-x + 1\), \(x = 1\). So the line passes through \((0,1)\) and \((1,0)\). Shade the region above this line (since \(y>-x + 1\)).

Step3: Find the solution set

The solution set of the system of inequalities is the region that is shaded by both inequalities. It is the region that is below the solid line \(y =-\frac{5}{2}x + 5\) and above the dashed line \(y=-x + 1\).

To graph:

1. Draw the solid line \(y =-\frac{5}{2}x+5\) through \((0,5)\) and \((2,0)\) and shade below it.
2. Draw the dashed line \(y=-x + 1\) through \((0,1)\) and \((1,0)\) and shade above it.
3. The overlapping shaded region is the solution set.

Answer:

The graph has a solid line \(y =-\frac{5}{2}x + 5\) (passing through \((0,5)\) and \((2,0)\)) with the region below it shaded, and a dashed line \(y=-x + 1\) (passing through \((0,1)\) and \((1,0)\)) with the region above it shaded. The solution set is the overlapping region of these two shaded areas.