The solution set of inequality $3x + 5y < 4$ is

The inequality $3x + 5y < 4$ represents a region in the xy-plane. The related equation $3x + 5y = 4$ is a straight line that acts as a boundary.

The inequality has a strict inequality sign "<" (not "≤"), which means the boundary line is not included in the solution set. This makes it an open region.

This eliminates the option of a closed half plane.

The inequality $3x + 5y < 4$ divides the plane into two parts:

• Points where $3x + 5y < 4$ (one half)
• Points where $3x + 5y > 4$ (other half)

The solution is a half plane, not the whole plane.

This eliminates the option of the whole xy-plane.

To determine which half plane, test the origin $(0, 0)$ in the inequality:

$3(0) + 5(0) < 4$

$0 < 4$

This is true, so the origin satisfies the inequality and is in the solution set.

The solution set is an open half plane containing the origin.