Question

station 10:

1. $y < 3x + 1$

$y \geq -2x - 3$

Explanation:

Step1: Analyze \( y < 3x + 1 \)

The inequality \( y < 3x + 1 \) represents a region below the line \( y = 3x + 1 \). The line has a slope of \( 3 \) and a y - intercept of \( 1 \). Since the inequality is strict (\(<\)), we draw a dashed line for \( y = 3x + 1 \) and shade the region below the line.

Step2: Analyze \( y \geq - 2x - 3 \)

The inequality \( y\geq - 2x - 3 \) represents a region above or on the line \( y=-2x - 3 \). The line has a slope of \(-2\) and a y - intercept of \(-3\). Since the inequality is non - strict (\(\geq\)), we draw a solid line for \( y=-2x - 3 \) and shade the region above the line.

Step3: Find the intersection

To graph the system of inequalities, we first graph each line:

• For \( y = 3x+1 \): When \( x = 0 \), \( y = 1 \); when \( y = 0 \), \( 0=3x + 1\Rightarrow x=-\frac{1}{3}\). Draw a dashed line through \((0,1)\) and \((-\frac{1}{3},0)\) and shade below.
• For \( y=-2x - 3 \): When \( x = 0 \), \( y=-3 \); when \( y = 0 \), \( 0=-2x-3\Rightarrow x =-\frac{3}{2}\). Draw a solid line through \((0, - 3)\) and \((-\frac{3}{2},0)\) and shade above.

The solution to the system is the region that is shaded by both inequalities, i.e., the region that is below \( y = 3x + 1 \) (dashed line) and above or on \( y=-2x - 3 \) (solid line).

Answer:

To graph the system \(

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

\):

1. Draw a dashed line for \( y = 3x + 1 \) (slope \( 3 \), y - intercept \( 1 \)) and shade below it.
2. Draw a solid line for \( y=-2x - 3 \) (slope \(-2\), y - intercept \(-3\)) and shade above it.
3. The solution region is the overlapping shaded region of the two inequalities.