Question

1. graph:

$4x - 3y < 9$
$x + 3y > 6$

Explanation:

Step1: Rewrite first inequality to slope-intercept

Rearrange $4x - 3y < 9$ to solve for $y$:

$$\begin{align*} -3y &< -4x + 9 \\ y &> \frac{4}{3}x - 3 \end{align*}$$

(Reverse inequality when dividing by negative)

Step2: Rewrite second inequality to slope-intercept

Rearrange $x + 3y > 6$ to solve for $y$:

$$\begin{align*} 3y &> -x + 6 \\ y &> -\frac{1}{3}x + 2 \end{align*}$$

Step3: Identify boundary lines

1. For $y > \frac{4}{3}x - 3$: Boundary is $y=\frac{4}{3}x - 3$ (dashed line, since inequality is strict). Plot points: when $x=0$, $y=-3$; when $y=0$, $x=\frac{9}{4}=2.25$.
2. For $y > -\frac{1}{3}x + 2$: Boundary is $y=-\frac{1}{3}x + 2$ (dashed line, strict inequality). Plot points: when $x=0$, $y=2$; when $y=0$, $x=6$.

Step4: Shade solution regions

1. For $y > \frac{4}{3}x - 3$: Shade above the dashed line.
2. For $y > -\frac{1}{3}x + 2$: Shade above the dashed line.
3. The final solution is the overlapping shaded area of both regions.

Answer:

The graph consists of two dashed lines:

1. Dashed line $y=\frac{4}{3}x - 3$, with shading above the line
2. Dashed line $y=-\frac{1}{3}x + 2$, with shading above the line

The overlapping shaded area (above both lines) is the solution set to the system of inequalities.