Question

text description for graph
a. $y > \frac{2}{3}x + 1$
b. $y < \frac{2}{3}x + 1$
c. $y \geq \frac{2}{3}x + 1$
d. $y \leq \frac{2}{3}x + 1$

Explanation:

Step1: Find the slope of the line

The two points on the line are \((3, 3)\) and \((-3, -1)\). The slope \(m\) is calculated as \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{3 - (-1)}{3 - (-3)}=\frac{4}{6}=\frac{2}{3}\).

Step2: Find the equation of the line

Using the point - slope form \(y - y_1=m(x - x_1)\) with the point \((3,3)\) and \(m = \frac{2}{3}\), we have \(y - 3=\frac{2}{3}(x - 3)\). Simplifying, \(y-3=\frac{2}{3}x - 2\), so \(y=\frac{2}{3}x + 1\). The line is dashed, so the inequality is either \(y>\frac{2}{3}x + 1\) or \(y<\frac{2}{3}x + 1\).

Step3: Test a point in the shaded region

Let's test the point \((0,5)\) (which is in the shaded region). Substitute \(x = 0\) and \(y = 5\) into the inequality \(y>\frac{2}{3}x + 1\): \(5>\frac{2}{3}(0)+1\), \(5 > 1\) which is true. Substitute into \(y<\frac{2}{3}x + 1\): \(5<\frac{2}{3}(0)+1\), \(5 < 1\) which is false. Also, test the point \((-3,-1)\) (on the line, but the line is dashed, so the points on the line are not included). The shaded region is above the line, so the inequality is \(y>\frac{2}{3}x + 1\).

Answer:

A. \(y>\frac{2}{3}x + 1\)