Question

1. y ≤ (1/3)x + 2; y > (1/2)x + 5

Explanation:

Step1: Analyze the first inequality \( y \leq \frac{1}{3}x + 2 \)

The boundary line for this inequality is \( y=\frac{1}{3}x + 2 \). The slope \( m=\frac{1}{3}\) and the y - intercept \( b = 2\). Since the inequality is \( y\leq\frac{1}{3}x + 2\), we draw a solid line (because the inequality includes equality) and shade the region below the line.

Step2: Analyze the second inequality \( y>-\frac{1}{2}x + 5 \)

The boundary line for this inequality is \( y =-\frac{1}{2}x+5\). The slope \( m =-\frac{1}{2}\) and the y - intercept \( b = 5\). Since the inequality is \( y>-\frac{1}{2}x + 5\), we draw a dashed line (because the inequality does not include equality) and shade the region above the line.

Step3: Find the solution region

The solution to the system of inequalities is the region that is shaded for both inequalities. To graph this, first plot the line \( y=\frac{1}{3}x + 2\) (solid) and \( y=-\frac{1}{2}x + 5\) (dashed). Then shade the area that is below \( y=\frac{1}{3}x + 2\) and above \( y=-\frac{1}{2}x + 5\).

To find the intersection point of the two lines \( y=\frac{1}{3}x + 2\) and \( y=-\frac{1}{2}x + 5\), set them equal:

$$
LATEXBLOCK0
$$

Substitute \( x = \frac{18}{5}\) into \( y=\frac{1}{3}x + 2\):

$$ y=\frac{1}{3}\times\frac{18}{5}+2=\frac{6}{5}+2=\frac{6 + 10}{5}=\frac{16}{5}=3.2 $$

The intersection point is \( (\frac{18}{5},\frac{16}{5})\) or \( (3.6,3.2)\). The solution region is the set of all points \((x,y)\) that lie below the line \( y=\frac{1}{3}x + 2\) (including the line) and above the line \( y=-\frac{1}{2}x + 5\) (excluding the line).

(If the question was to graph the system, the steps above describe how to graph it. If it was to find the intersection or describe the solution, the intersection point and the region description are as above.)

Answer:

The system of inequalities \(

$$\begin{cases}y\leq\frac{1}{3}x + 2\\y>-\frac{1}{2}x + 5\end{cases}$$

\) has a solution region bounded by the line \( y = \frac{1}{3}x+2\) (solid), \( y=-\frac{1}{2}x + 5\) (dashed), with the intersection point at \( (\frac{18}{5},\frac{16}{5})\) (or \( (3.6,3.2)\)) and the solution being the set of points below \( y=\frac{1}{3}x + 2\) and above \( y=-\frac{1}{2}x + 5\). If graphing, follow the steps of plotting the lines (with correct line types) and shading the overlapping region.