Question

which linear inequality will not have a shared solution set with the graphed linear inequality?
○ $y < \frac{5}{3}x - 2$
○ $y < -\frac{5}{3}x + 1$
○ $y > \frac{5}{3}x + 2$
○ $y > -\frac{5}{3}x + 2$
(graph: a coordinate plane with a dashed line $y < \frac{5}{3}x + 1$ and the region above the line shaded.)

Explanation:

Step1: Analyze the graphed inequality

The graphed inequality is \( y < \frac{5}{3}x + 1 \). It has a slope of \( \frac{5}{3} \) and a y - intercept of 1, and the region below the dashed line is shaded.

Step2: Analyze each option

• Option 1: \( y < \frac{5}{3}x - 2 \)
• The slope is the same as the graphed inequality (\( \frac{5}{3} \)). The region below this line (since \( y < \)) will overlap with the region of \( y < \frac{5}{3}x + 1 \) for some values of \( x \) and \( y \). For example, when \( x = 0 \), the graphed inequality has \( y < 1 \) and this inequality has \( y < - 2 \). But for larger \( x \) values, there will be overlap.
• Option 2: \( y < -\frac{5}{3}x + 1 \)
• The slope is negative (\( -\frac{5}{3} \)) and the y - intercept is 1. The region below this line will intersect with the region of \( y < \frac{5}{3}x + 1 \) at some points. For example, at the intersection of the two lines \( \frac{5}{3}x+1=-\frac{5}{3}x + 1\), we get \( \frac{10}{3}x=0\), \( x = 0 \), \( y = 1 \). The two regions will overlap around this point.
• Option 3: \( y > \frac{5}{3}x + 2 \)
• The slope is \( \frac{5}{3} \) (same as the graphed inequality) and the y - intercept is 2, which is above the y - intercept of the graphed inequality (1). The region above this line (\( y >\)) and the region below the graphed line (\( y < \frac{5}{3}x + 1\)) will never overlap because for any \( x \), \( \frac{5}{3}x + 2>\frac{5}{3}x + 1\). So \( y > \frac{5}{3}x + 2 \) and \( y < \frac{5}{3}x + 1 \) have no shared solutions.
• Option 4: \( y > -\frac{5}{3}x + 2 \)
• The slope is negative (\( -\frac{5}{3} \)) and the y - intercept is 2. The region above this line will intersect with the region of \( y < \frac{5}{3}x + 1 \) at some points. For example, solve \( \frac{5}{3}x + 1=-\frac{5}{3}x+2\), \( \frac{10}{3}x = 1\), \( x=\frac{3}{10}\), \( y=\frac{5}{3}\times\frac{3}{10}+1=\frac{1}{2}+1=\frac{3}{2}\). The two regions will overlap around this point.

Answer:

\( y > \frac{5}{3}x + 2 \) (the third option: \( y > \frac{5}{3}x + 2 \))