Question

1. which region represents the solution set to the following system of inequalities:

a. region 1 \\( y \geq \frac{2}{3}x + 1 \\)
b. region 2 \\( 5x + 6y \leq - 30 \\)
c. region 3
d. region 4

1. given the linear equation, \\( y = \frac{3}{4}x - 2 \\), find the following:

a) the equation of a line through the point (2, 3), parallel to the given line.

Explanation:

Question 20

Step1: Analyze \( y \geq \frac{2}{3}x + 1 \)

The line \( y=\frac{2}{3}x + 1 \) has a positive slope. The inequality \( y\geq\frac{2}{3}x + 1 \) means we shade above this line.

Step2: Analyze \( 5x + 6y \leq - 30 \)

Rewrite it as \( y\leq-\frac{5}{6}x - 5 \). The line \( y = -\frac{5}{6}x - 5 \) has a negative slope. The inequality \( y\leq-\frac{5}{6}x - 5 \) means we shade below this line.

Step3: Find the intersection of the two shaded regions

The region that is above \( y=\frac{2}{3}x + 1 \) and below \( y = -\frac{5}{6}x - 5 \) (by checking the graph's regions) corresponds to Region 2.

Step1: Recall parallel line slope

Parallel lines have the same slope. The given line \( y=\frac{3}{4}x - 2 \) has a slope \( m=\frac{3}{4} \). So the new line also has \( m = \frac{3}{4} \).

Step2: Use point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(2,3) \) and \( m=\frac{3}{4} \).
Substitute the values: \( y - 3=\frac{3}{4}(x - 2) \).

Step3: Simplify the equation

Expand the right - hand side: \( y - 3=\frac{3}{4}x-\frac{3}{2} \).
Add 3 to both sides: \( y=\frac{3}{4}x-\frac{3}{2}+3=\frac{3}{4}x+\frac{3}{2} \).

Answer:

B. Region 2

Question 21a