Question

1. $y < \frac{1}{2}x + 2$

$y > -2x - 3$

1. $y \leq \frac{1}{3}x - 2$

$y \leq x + 8$

1. $x - 3y > -3$

$5x - 3y \leq 9$

1. $x - 3y \geq -3$

$4x - 3y > 6$

1. $x \leq 2$

$x + y \leq 1$

1. $x - y < 1$

$y \geq -3$

Explanation:

Let's solve problem 5: \( y < \frac{1}{2}x + 2 \) and \( y \geq -2x - 3 \)

Step 1: Analyze the first inequality \( y < \frac{1}{2}x + 2 \)

This is a linear inequality. The boundary line is \( y = \frac{1}{2}x + 2 \), which has a slope of \( \frac{1}{2} \) and a y-intercept at \( (0, 2) \). Since the inequality is \( y < \frac{1}{2}x + 2 \), the line should be dashed (because it's a strict inequality, \( < \) not \( \leq \)) and we shade below the line.

Step 2: Analyze the second inequality \( y \geq -2x - 3 \)

The boundary line is \( y = -2x - 3 \), with a slope of \( -2 \) and a y-intercept at \( (0, -3) \). Since the inequality is \( y \geq -2x - 3 \), the line should be solid (because of the \( \geq \) sign) and we shade above the line.

Step 3: Graph the lines and find the intersection

First, graph \( y = \frac{1}{2}x + 2 \) as a dashed line. Plot the y-intercept \( (0, 2) \), then use the slope \( \frac{1}{2} \) (rise 1, run 2) to find another point, e.g., \( (2, 3) \). Draw the dashed line through these points and shade below it.

Next, graph \( y = -2x - 3 \) as a solid line. Plot the y-intercept \( (0, -3) \), then use the slope \( -2 \) (rise -2, run 1) to find another point, e.g., \( (1, -5) \). Draw the solid line through these points and shade above it.

The solution to the system is the region where the two shaded areas overlap.

Answer:

To graph the system \(

$$\begin{cases} y < \frac{1}{2}x + 2 \\ y \geq -2x - 3 \end{cases}$$

\):

1. Draw the dashed line \( y = \frac{1}{2}x + 2 \) and shade below it.
2. Draw the solid line \( y = -2x - 3 \) and shade above it.
3. The overlapping shaded region (where both inequalities are satisfied) is the solution.