Question

1-3. graph the inequality in the coordinate plane. determine whether the given inequality.
$y \geq -\frac{2}{3}x - 3$ $4x - 3y \leq 16$
then there are tables to determine if points are solutions, and 4-6. write an inequality that represents the graph.

Explanation:

Step1: Analyze the inequality \( y \geq -\frac{2}{3}x - 3 \)

First, identify the boundary line. The equation of the boundary line is \( y = -\frac{2}{3}x - 3 \). Since the inequality is \( \geq \), the line should be solid (not dashed) and we shade above the line.
To find two points on the line, when \( x = 0 \), \( y = -3 \), so the y - intercept is \( (0, -3) \). When \( y = 0 \), \( 0 = -\frac{2}{3}x - 3 \), solving for \( x \): \( \frac{2}{3}x=-3 \), \( x = -\frac{9}{2}=-4.5 \), so the x - intercept is \( (-4.5, 0) \).

Step2: Test the points for \( y \geq -\frac{2}{3}x - 3 \)

• For \( (-3, 3) \): Substitute \( x=-3 \), \( y = 3 \) into the inequality. Right - hand side: \( -\frac{2}{3}(-3)-3=2 - 3=-1 \). Since \( 3\geq - 1 \), this point is a solution (Yes).
• For \( (-1, 4) \): Substitute \( x = - 1 \), \( y = 4 \). Right - hand side: \( -\frac{2}{3}(-1)-3=\frac{2}{3}-3=\frac{2 - 9}{3}=-\frac{7}{3}\approx - 2.33 \). Since \( 4\geq-\frac{7}{3} \), this point is a solution (Yes).
• For \( (0, 0) \): Substitute \( x = 0 \), \( y = 0 \). Right - hand side: \( -\frac{2}{3}(0)-3=-3 \). Since \( 0\geq - 3 \), this point is a solution (Yes).
• For \( (0, - 3) \): Substitute \( x = 0 \), \( y=-3 \). Right - hand side: \( -\frac{2}{3}(0)-3=-3 \). Since \( - 3=-3 \) (and the inequality is \( \geq \)), this point is a solution (Yes).
• For \( (3, 3) \): Substitute \( x = 3 \), \( y = 3 \). Right - hand side: \( -\frac{2}{3}(3)-3=-2 - 3=-5 \). Since \( 3\geq - 5 \), this point is a solution (Yes).

Step3: Analyze the inequality \( 4x-3y\leq16 \)

Rewrite it in slope - intercept form: \( - 3y\leq - 4x + 16 \), \( y\geq\frac{4}{3}x-\frac{16}{3}\) (note that when we divide by a negative number, the inequality sign flips). The boundary line is \( y=\frac{4}{3}x-\frac{16}{3} \), and since the original inequality was \( \leq \) (after rearrangement \( y\geq\)), the line is solid and we shade above the line.
To find points on the line \( 4x - 3y=16 \), when \( x = 0 \), \( - 3y=16 \), \( y=-\frac{16}{3}\approx - 5.33 \). When \( y = 0 \), \( 4x=16 \), \( x = 4 \).

Step4: Test the points for \( 4x-3y\leq16 \)

• For \( (-3, 3) \): Substitute \( x=-3 \), \( y = 3 \) into \( 4x-3y \). \( 4(-3)-3(3)=-12 - 9=-21 \). Since \( - 21\leq16 \), this point is a solution (Yes).
• For \( (-1, 4) \): Substitute \( x=-1 \), \( y = 4 \). \( 4(-1)-3(4)=-4 - 12=-16 \). Since \( - 16\leq16 \), this point is a solution (Yes).
• For \( (0, 0) \): Substitute \( x = 0 \), \( y = 0 \). \( 4(0)-3(0)=0 \). Since \( 0\leq16 \), this point is a solution (Yes).
• For \( (1, - 4) \): Substitute \( x = 1 \), \( y=-4 \). \( 4(1)-3(-4)=4 + 12 = 16 \). Since \( 16\leq16 \), this point is a solution (Yes).
• For \( (3, 3) \): Substitute \( x = 3 \), \( y = 3 \). \( 4(3)-3(3)=12 - 9 = 3 \). Since \( 3\leq16 \), this point is a solution (Yes).

(For the graphing part, plot the solid line \( y = -\frac{2}{3}x - 3 \) and shade above it for \( y\geq-\frac{2}{3}x - 3 \), and plot the solid line \( 4x-3y = 16 \) (or \( y=\frac{4}{3}x-\frac{16}{3}\)) and shade above the line for \( 4x - 3y\leq16 \) (since \( y\geq\frac{4}{3}x-\frac{16}{3}\)) )

(For writing the inequality from the graph, we need to analyze the slope and the y - intercept of the boundary line and the direction of shading. But since the question is mainly about graphing and testing points for the given inequalities, we focus on the point - testing part as above)

Answer:

For the inequality \( y\geq-\frac{2}{3}x - 3 \):

• \( (-3, 3) \): Yes
• \( (-1, 4) \): Yes
• \( (0, 0) \): Yes
• \( (0, - 3) \): Yes
• \( (3, 3) \): Yes

For the inequality \( 4x - 3y\leq16 \):

• \( (-3, 3) \): Yes
• \( (-1, 4) \): Yes
• \( (0, 0) \): Yes
• \( (1, - 4) \): Yes
• \( (3, 3) \): Yes

(The graphing part: For \( y\geq-\frac{2}{3}x - 3 \), draw a solid line through \( (0, - 3) \) and \( (-4.5, 0) \) and shade above the line. For \( 4x - 3y\leq16 \), draw a solid line through \( (0,-\frac{16}{3}) \) and \( (4, 0) \) and shade above the line)