why is it that in a linear inequality, each point on a dashed boundary line is not a solution?
suppose you have the graph of
y > 2x + 5
the graph would consist of all the points above the line y = 2x + 5 , but exclude all the points which lie on the line itself
The equation acts like a boundary, but the points on the equation are not part of the region.
to show that the line is only the boundary, we draw the line as a dotted line, if the points are included we make it a solid line
if you want all the points above the line and all the points which lie on the line we used the notation
y ≥ 2x + 5
This is a universally accepted notation.
thanks
In a linear inequality, the dashed boundary line represents the boundary between the two regions of the coordinate plane: the solution region and the non-solution region.
To understand why each point on a dashed boundary line is not a solution, we need to consider the definition of a linear inequality. A linear inequality is an inequality that can be represented by a linear equation, such as "y > 2x + 1" or "3y - 4x ≤ 10."
For example, let's take the linear inequality "y > 2x + 1." The dashed line that represents this inequality can be obtained by first graphing the related linear equation "y = 2x + 1" (which is the boundary line) and then making it dashed to indicate that the points on the line are not included in the solution set.
Now, consider a point on the dashed boundary line. It can be expressed as (x, y) with x and y satisfying the equation of the line, y = 2x + 1. However, in the case of a strict inequality like "y > 2x + 1," the points on the line do not satisfy the inequality because the inequality excludes the points exactly on the line. The points above the line are the actual solutions, which form the solution region.
To determine the solutions, it is necessary to test points in both regions, above and below the dashed line. For example, choose a point below the line, substitute its coordinates into the inequality, and verify if the inequality holds true. Similarly, choose a point above the line and check if it satisfies the inequality. Points on the dashed line are not tested, as they are not part of the solution.
In conclusion, each point on a dashed boundary line in a linear inequality is not a solution because the inequality only includes the points in the solution region, which are either above or below the dashed line.