Question

§9.3: sketch each system shown below.
19.) \

$$\begin{cases} x + 2y \\leq 6 \\\\ 2x - 3y \\geq 12 \\\\ x > -8 \\end{cases}$$

20.) \

$$\begin{cases} 2x - y \\leq 4 \\\\ x > 1 \\\\ y > 2 \\\\ 3x + 3y \\leq 12 \\end{cases}$$

21.) \

$$\begin{cases} y < x + 3 \\\\ x^2 + y^2 \\geq 9 \\\\ y \\geq -3 \\end{cases}$$

Answer:

Problem 19: System \(

$$\begin{cases} x + 2y \leq 6 \\ 2x - 3y \geq 12 \\ x > -8 \end{cases}$$

\)

Step 1: Analyze \(x + 2y \leq 6\)

Rewrite as \(y \leq -\frac{1}{2}x + 3\). This is a line with slope \(-\frac{1}{2}\), y - intercept \(3\). Shade below the line (since \(\leq\)), solid line.

Step 2: Analyze \(2x - 3y \geq 12\)

Rewrite as \(y \leq \frac{2}{3}x - 4\) (wait, no: \(2x - 3y \geq 12 \implies -3y \geq -2x + 12 \implies y \leq \frac{2}{3}x - 4\)? Wait, no: dividing by negative flips inequality. Correct: \(2x - 3y \geq 12 \implies -3y \geq -2x + 12 \implies y \leq \frac{2}{3}x - 4\)? Wait, no: \( -3y \geq -2x + 12 \implies y \leq \frac{2}{3}x - 4\) (since dividing by -3, inequality flips). Wait, no: \(2x - 3y \geq 12 \implies -3y \geq -2x + 12 \implies y \leq \frac{2}{3}x - 4\)? Wait, no, let's do it again: \(2x - 3y \geq 12\)
Subtract \(2x\): \(-3y \geq -2x + 12\)
Divide by \(-3\) (reverse inequality): \(y \leq \frac{2}{3}x - 4\). Line with slope \(\frac{2}{3}\), y - intercept \(-4\). Shade below, solid line.

Step 3: Analyze \(x > -8\)

Vertical line \(x = -8\), dashed (since \(>\)), shade to the right of \(x = -8\).

Step 4: Find Intersection

Find intersection of \(x + 2y = 6\) and \(2x - 3y = 12\):
Multiply first equation by 2: \(2x + 4y = 12\)
Subtract second equation: \((2x + 4y) - (2x - 3y) = 12 - 12 \implies 7y = 0 \implies y = 0\). Then \(x = 6\). So intersection at \((6, 0)\).
Intersection of \(x + 2y = 6\) and \(x = -8\): \( -8 + 2y = 6 \implies 2y = 14 \implies y = 7\). Point \((-8, 7)\).
Intersection of \(2x - 3y = 12\) and \(x = -8\): \(2(-8) - 3y = 12 \implies -16 - 3y = 12 \implies -3y = 28 \implies y = -\frac{28}{3} \approx -9.33\). Point \((-8, -\frac{28}{3})\).
Shade the region that satisfies all three: below \(y = -\frac{1}{2}x + 3\), below \(y = \frac{2}{3}x - 4\), and right of \(x = -8\).

Problem 20: System \(

$$\begin{cases} 2x - y \leq 4 \\ x > 1 \\ y > 2 \\ 3x + 3y \leq 12 \end{cases}$$

\)

Step 1: Analyze \(2x - y \leq 4\)

Rewrite as \(y \geq 2x - 4\) (divide by -1, reverse inequality). Line \(y = 2x - 4\), slope 2, y - intercept -4. Shade above, solid line.

Step 2: Analyze \(x > 1\)

Vertical line \(x = 1\), dashed, shade right.

Step 3: Analyze \(y > 2\)

Horizontal line \(y = 2\), dashed, shade above.

Step 4: Analyze \(3x + 3y \leq 12\)

Simplify: \(x + y \leq 4 \implies y \leq -x + 4\). Line slope -1, y - intercept 4. Shade below, solid line.

Step 5: Find Intersections

• Intersection of \(2x - y = 4\) and \(x + y = 4\): Add equations: \(3x = 8 \implies x = \frac{8}{3} \approx 2.67\), \(y = 4 - \frac{8}{3} = \frac{4}{3} \approx 1.33\) (but \(y > 2\), so not in region).
• Intersection of \(x = 1\) and \(y = 2\): \((1, 2)\) (dashed lines, so open point).
• Intersection of \(x = 1\) and \(y = 2x - 4\): \(y = 2(1) - 4 = -2\) (below \(y = 2\), so not in region).
• Intersection of \(x = 1\) and \(y = -x + 4\): \(y = 3\). Point \((1, 3)\) (solid line \(y = -x + 4\), dashed \(x = 1\), so open at \(x = 1\), closed at \(y = -x + 4\)).
• Intersection of \(y = 2\) and \(y = -x + 4\): \(2 = -x + 4 \implies x = 2\). Point \((2, 2)\) (dashed \(y = 2\), solid \(y = -x + 4\), so open at \(y = 2\), closed at \(y = -x + 4\)).
• Intersection of \(y = 2\) and \(y = 2x - 4\): \(2 = 2x - 4 \implies x = 3\). Point \((3, 2)\) (dashed \(y = 2\), solid \(y = 2x - 4\), open at \(y = 2\), closed at \(y = 2x - 4\)).
• Intersection of \(y = 2x - 4\) and \(y = -x + 4\): As before, \((\frac{8}{3}, \frac{4}{3})\) (below \(y = 2\), not in region).

Shade region: right of \(x = 1\), above \(y = 2\), above \(y = 2x - 4\), and below \(y = -x + 4\). The feasible region is a polygon with vertices (open/closed) around \((2, 2)\), \((1, 3)\), and intersection of \(y = 2\) and \(y = -x + 4\) (but check bounds: \(x > 1\), \(y > 2\), so the region is bounded by \(x = 1\) (right), \(y = 2\) (above), \(y = 2x - 4\) (above), and \(y = -x + 4\) (below)).

Problem 21: System \(

$$\begin{cases} y < x + 3 \\ x^2 + y^2 \geq 9 \\ y \geq -3 \end{cases}$$

\)

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

Line \(y = x + 3\), slope 1, y - intercept 3. Shade below, dashed line.

Step 2: Analyze \(x^2 + y^2 \geq 9\)

Circle with radius 3, center at origin. Shade outside (since \(\geq\)), solid line.

Step 3: Analyze \(y \geq -3\)

Horizontal line \(y = -3\), solid line, shade above.

Step 4: Find Intersections

• Intersection of \(y = x + 3\) and \(x^2 + y^2 = 9\): Substitute \(y = x + 3\) into circle: \(x^2 + (x + 3)^2 = 9 \implies x^2 + x^2 + 6x + 9 = 9 \implies 2x^2 + 6x = 0 \implies 2x(x + 3) = 0 \implies x = 0\) or \(x = -3\).
• \(x = 0\): \(y = 3\) (point \((0, 3)\)).
• \(x = -3\): \(y = 0\) (point \((-3, 0)\)).
• Intersection of \(y = -3\) and \(x^2 + y^2 = 9\): \(x^2 + 9 = 9 \implies x = 0\). Point \((0, -3)\).
• Intersection of \(y = -3\) and \(y = x + 3\): \(x = -6\). Point \((-6, -3)\).

Shade region: above \(y = -3\), below \(y = x + 3\) (dashed), and outside the circle \(x^2 + y^2 = 9\) (solid). The region includes the area above \(y = -3\), below the line \(y = x + 3\), and outside the circle (e.g., to the right of the circle, above \(y = -3\), and below \(y = x + 3\); also to the left of the circle but above \(y = -3\) and below \(y = x + 3\) (but check if inside circle: at \(x = -3\), \(y = 0\) is on circle, \(x = -4\), \(y = -1\): \(x^2 + y^2 = 16 + 1 = 17 \geq 9\), so left of \(x = -3\) is outside circle, above \(y = -3\), below \(y = x + 3\) (e.g., \(x = -5\), \(y = -2\): \(y = -2 < x + 3 = -2\)? No, \(y = -2 = x + 3\) when \(x = -5\), so \(y < x + 3\) would be \(y < -2\), but \(y \geq -3\), so left region is between \(y = -3\) and \(y = x + 3\), outside circle.

Sketching Instructions (General)

For each system:

1. Graph each inequality (solid/dashed line, shade region).
2. Identify intersection points of boundary lines.
3. The solution is the overlapping shaded region (if any).

For example, in Problem 19:

• Solid line \(x + 2y = 6\) (shade below), solid line \(2x - 3y = 12\) (shade below), dashed line \(x = -8\) (shade right).
• The overlapping region is where all three shadings meet: right of \(x = -8\), below both lines.

In Problem 20:

• Dashed \(x = 1\) (right), dashed \(y = 2\) (above), solid \(y = 2x - 4\) (above), solid \(x + y = 4\) (below).
• Overlap: right of \(x = 1\), above \(y = 2\), above \(y = 2x - 4\), below \(y = -x + 4\) (a polygon with vertices near \((1, 3)\), \((2, 2)\), and intersection of \(y = 2x - 4\) and \(y = -x + 4\) (but adjusted for \(y > 2\))).

In Problem 21:

• Dashed \(y = x + 3\) (below), solid \(y = -3\) (above), solid circle \(x^2 + y^2 = 9\) (outside).
• Overlap: above \(y = -3\), below \(y = x + 3\), outside the circle (two regions: left of the circle (between \(y = -3\) and \(y = x + 3\)) and right of the circle (between \(y = -3\), below \(y = x + 3\), and outside circle)).

(Note: Since the question is to sketch, the final answer is the graphical representation as described. For each system, plot the lines, shade the regions, and identify the overlapping area.)