Question

graph each inequality in the coordinate plane.\
see examples 1, 2, and 4\

1. $y \geq -2x + 3$\
2. $y < x - 6$\
3. $y \leq \frac{2}{3}x - 1$\
4. $y > x - 2$\
5. $y = -0.5x + 2$\
6. $y \geq 1.5x - 4$\
7. $2x > 12$\
8. $-2y \leq 6$

Explanation:

Step1: Analyze the inequality type

For linear inequalities in two variables (like \(y < x - 6\), \(y\geq - 2x + 3\) etc.), we first consider the boundary line. If the inequality is strict (\(<\) or \(>\)), the boundary line is dashed; if it is non - strict (\(\leq\) or \(\geq\)), the boundary line is solid. Then we determine which side of the line to shade by testing a point (usually the origin \((0,0)\) if it is not on the boundary line).

Step2: Graph for \(y\geq - 2x + 3\) (Problem 16)

- **Boundary Line**: The equation of the boundary line is \(y=-2x + 3\). Since the inequality is \(\geq\), the line is solid.
- Find two points on the line. When \(x = 0\), \(y=-2(0)+3=3\), so the point is \((0,3)\). When \(y = 0\), \(0=-2x + 3\), \(2x=3\), \(x=\frac{3}{2}=1.5\), so the point is \((1.5,0)\).
- **Shading**: Test the origin \((0,0)\) in the inequality \(y\geq - 2x+3\). Substitute \(x = 0\), \(y = 0\): \(0\geq-2(0)+3\) which is \(0\geq3\), false. So we shade the side of the line that does not contain the origin.

Step3: Graph for \(y < x - 6\) (Problem 17)

- **Boundary Line**: The equation of the boundary line is \(y=x - 6\). Since the inequality is \(<\), the line is dashed.
- When \(x = 0\), \(y=0 - 6=-6\), so the point is \((0,-6)\). When \(y = 0\), \(0=x - 6\), \(x = 6\), so the point is \((6,0)\).
- **Shading**: Test the origin \((0,0)\) in the inequality \(y < x - 6\). Substitute \(x = 0\), \(y = 0\): \(0<0 - 6\) which is \(0<-6\), false. So we shade the side of the line that does not contain the origin.

Step4: Graph for \(y\leq\frac{2}{3}x - 1\) (Problem 18)

- **Boundary Line**: The equation of the boundary line is \(y=\frac{2}{3}x-1\). Since the inequality is \(\leq\), the line is solid.
- When \(x = 0\), \(y=\frac{2}{3}(0)-1=-1\), so the point is \((0,-1)\). When \(y = 0\), \(0=\frac{2}{3}x-1\), \(\frac{2}{3}x = 1\), \(x=\frac{3}{2}=1.5\), so the point is \((1.5,0)\).
- **Shading**: Test the origin \((0,0)\) in the inequality \(y\leq\frac{2}{3}x - 1\). Substitute \(x = 0\), \(y = 0\): \(0\leq\frac{2}{3}(0)-1\) which is \(0\leq - 1\), false. So we shade the side of the line that does not contain the origin.

Step5: Graph for \(y=-0.5x + 2\) (Wait, this is an equation, maybe a typo? If it's \(y\leq - 0.5x+2\) or \(y\geq - 0.5x + 2\) or \(y < - 0.5x+2\) or \(y > - 0.5x+2\))

Assuming it's \(y\leq - 0.5x + 2\) (since it's in the same set of inequality problems)
- **Boundary Line**: The equation of the boundary line is \(y=-0.5x + 2\). Since the inequality is \(\leq\) (assuming), the line is solid.
- When \(x = 0\), \(y = 2\), so the point is \((0,2)\). When \(y = 0\), \(0=-0.5x+2\), \(0.5x = 2\), \(x = 4\), so the point is \((4,0)\).
- **Shading**: Test the origin \((0,0)\) in the inequality \(y\leq - 0.5x + 2\). Substitute \(x = 0\), \(y = 0\): \(0\leq-0.5(0)+2\) which is \(0\leq2\), true. So we shade the side of the line that contains the origin.

Step6: Graph for \(y > x - 2\) (Problem 19)

- **Boundary Line**: The equation of the boundary line is \(y=x - 2\). Since the inequality is \(>\), the line is dashed.
- When \(x = 0\), \(y=0 - 2=-2\), so the point is \((0,-2)\). When \(y = 0\), \(0=x - 2\), \(x = 2\), so the point is \((2,0)\).
- **Shading**: Test the origin \((0,0)\) in the inequality \(y > x - 2\). Substitute \(x = 0\), \(y = 0\): \(0>0 - 2\) which is \(0>-2\), true. So we shade the side of the line that contains the origin.

Step7: Graph for \(y\geq1.5x - 4\) (Problem 21)

- **Boundary Line**: The equation of the boundary line is \(y = 1.5x-4\). Since the inequality is \(\geq\), the line is solid.
- When \(x = 0\), \(y=-4\), so the point is \((0,-4)\). When \(y = 0\), \(0 = 1.5x-4\), \(1.5x=4\), \(x=\frac{4}{1.5}=\frac{8}{3}\approx2.67\), so the point is \((\frac{8}{3},0)\).
- **Shading**: Test the origin \((0,0)\) in the inequality \(y\geq1.5x - 4\). Substitute \(x = 0\), \(y = 0\): \(0\geq1.5(0)-4\) which is \(0\geq - 4\), true. So we shade the side of the line that contains the origin.

Step8: Graph for \(2x>12\) (Problem 22)

- First, solve the inequality for \(x\): \(x > 6\). This is a vertical line. The boundary line is \(x = 6\), and since the inequality is \(>\), the line is dashed. We shade the region to the right of the line \(x = 6\) (all points with \(x\) - coordinate greater than 6).

Step9: Graph for \(-2y\leq6\) (Problem 23)

- First, solve the inequality for \(y\). Divide both sides by \(- 2\). Remember that when we divide or multiply an inequality by a negative number, the direction of the inequality sign changes. So we get \(y\geq - 3\). This is a horizontal line. The boundary line is \(y=-3\), and since the inequality is \(\geq\), the line is solid. We shade the region above the line \(y=-3\) (all points with \(y\) - coordinate greater than or equal to - 3).

(Note: Since the problem says "Graph each inequality", the final answer would be the set of graphs constructed as per the above steps for each inequality. But since we can't draw the graphs here, we can describe the process of graphing each one as above.)

Answer:

The graphs are constructed by first determining the boundary line (dashed for strict inequalities, solid for non - strict), finding two points on the boundary line, testing a point (usually the origin) to determine the shading region, and then shading the appropriate region. For example:

• For \(y\geq - 2x + 3\): Solid line through \((0,3)\) and \((1.5,0)\), shade above the line.
• For \(y < x - 6\): Dashed line through \((0,-6)\) and \((6,0)\), shade below the line.
• For \(y\leq\frac{2}{3}x - 1\): Solid line through \((0,-1)\) and \((1.5,0)\), shade below the line.
• For \(y\leq - 0.5x + 2\) (assuming): Solid line through \((0,2)\) and \((4,0)\), shade below the line.
• For \(y > x - 2\): Dashed line through \((0,-2)\) and \((2,0)\), shade above the line.
• For \(y\geq1.5x - 4\): Solid line through \((0,-4)\) and \((\frac{8}{3},0)\), shade above the line.
• For \(2x>12\): Dashed vertical line \(x = 6\), shade to the right.
• For \(-2y\leq6\) (i.e., \(y\geq - 3\)): Solid horizontal line \(y=-3\), shade above.