Question

1. solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.

$y < -\frac{3}{4}x + 8$
$y \geq \frac{1}{2}x + 3$

1. solve the following system of inequalities graphically on the set of axes below. state the coordinates of a point in the solution set.

$y > x - 6$
$y \leq -\frac{1}{3}x - 2$

Explanation:

For Problem 2:

Step1: Graph $y < -\frac{3}{4}x + 8$

1. Identify the line $y = -\frac{3}{4}x + 8$:

• y-intercept: $(0, 8)$
• x-intercept: Set $y=0$: $0 = -\frac{3}{4}x + 8 \implies \frac{3}{4}x = 8 \implies x = \frac{32}{3} \approx 10.67$

1. Draw a dashed line (since the inequality is $<$) through these points.
2. Shade the region below the line (since $y$ is less than the expression).

Step2: Graph $y \geq \frac{1}{2}x + 3$

1. Identify the line $y = \frac{1}{2}x + 3$:

• y-intercept: $(0, 3)$
• x-intercept: Set $y=0$: $0 = \frac{1}{2}x + 3 \implies \frac{1}{2}x = -3 \implies x = -6$

1. Draw a solid line (since the inequality is $\geq$) through these points.
2. Shade the region above the line (since $y$ is greater than or equal to the expression).

Step3: Find intersection point

Set the two line equations equal to find where they cross:
$$-\frac{3}{4}x + 8 = \frac{1}{2}x + 3$$
Multiply all terms by 4 to eliminate denominators:
$$-3x + 32 = 2x + 12$$
$$32 - 12 = 2x + 3x$$
$$20 = 5x \implies x=4$$
Substitute $x=4$ into $y = \frac{1}{2}x + 3$:
$$y = \frac{1}{2}(4) + 3 = 2 + 3 = 5$$
Intersection point: $(4, 5)$

Step4: Identify solution region

The solution set is the overlapping shaded region (below the dashed line, above the solid line).

Step1: Graph $y > x - 6$

1. Identify the line $y = x - 6$:

• y-intercept: $(0, -6)$
• x-intercept: Set $y=0$: $0 = x - 6 \implies x=6$

1. Draw a dashed line (since the inequality is $>$) through these points.
2. Shade the region above the line (since $y$ is greater than the expression).

Step2: Graph $y \leq -\frac{1}{3}x - 2$

1. Identify the line $y = -\frac{1}{3}x - 2$:

• y-intercept: $(0, -2)$
• x-intercept: Set $y=0$: $0 = -\frac{1}{3}x - 2 \implies \frac{1}{3}x = -2 \implies x=-6$

1. Draw a solid line (since the inequality is $\leq$) through these points.
2. Shade the region below the line (since $y$ is less than or equal to the expression).

Step3: Find intersection point

Set the two line equations equal to find where they cross:
$$x - 6 = -\frac{1}{3}x - 2$$
Multiply all terms by 3 to eliminate denominators:
$$3x - 18 = -x - 6$$
$$3x + x = 18 - 6$$
$$4x = 12 \implies x=3$$
Substitute $x=3$ into $y = x - 6$:
$$y = 3 - 6 = -3$$
Intersection point: $(3, -3)$

Step4: Identify solution region

The solution set is the overlapping shaded region (above the dashed line, below the solid line).

Answer:

A point in the solution set is $\boldsymbol{(4, 5)}$ (or any point in the overlapping shaded region, e.g., $(0, 4)$)

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For Problem 4: