Understanding Linear Inequalities and their Graphs

The question asks us to determine the graphical representation of the linear in-equation \(2x - 5y \le 5\) in the Cartesian plane. Graphing a linear inequality involves two main steps: first, graphing the corresponding linear equation, and second, determining the region that satisfies the inequality.

Graphing the Boundary Line

The given in-equation is \(2x - 5y \le 5\). The boundary line for this inequality is the graph of the corresponding linear equation:

\(2x - 5y = 5\)

To graph this line, we can find two points that lie on the line. A common method is to find the intercepts:

• x-intercept: Set \(y = 0\) and solve for \(x\).

  \(2x - 5(0) = 5\)

  \(2x = 5\)

  \(x = \frac{5}{2} = 2.5\)

  So, the x-intercept is \((2.5, 0)\).
• y-intercept: Set \(x = 0\) and solve for \(y\).

  \(2(0) - 5y = 5\)

  \(-5y = 5\)

  \(y = -1\)

  So, the y-intercept is \((0, -1)\).

Plotting these two points \((2.5, 0)\) and \((0, -1)\) and drawing a line through them gives us the graph of \(2x - 5y = 5\). Since the inequality is \(2x - 5y \le 5\), which includes "equal to" (\(\le\)), the line \(2x - 5y = 5\) is part of the solution set and should be drawn as a solid line.

Determining the Solution Region

The inequality \(2x - 5y \le 5\) divides the Cartesian plane into two regions: one above the line \(2x - 5y = 5\) and one below it. We need to determine which region satisfies the inequality. We can do this by picking a test point that is not on the line. The origin \((0, 0)\) is usually the easiest test point, provided it is not on the line \(2x - 5y = 5\). Substituting \((0, 0)\) into the equation \(2x - 5y = 5\):

\(2(0) - 5(0) = 0\)

Since \(0 \ne 5\), the origin is not on the line. Now, substitute the test point \((0, 0)\) into the original inequality \(2x - 5y \le 5\):

\(2(0) - 5(0) \le 5\)

\(0 \le 5\)

This inequality \(0 \le 5\) is a true statement. This means that the region containing the test point \((0, 0)\) is the solution region for the inequality \(2x - 5y \le 5\).

Locating the Origin Relative to the Line

To determine if \((0, 0)\) is above or below the line \(2x - 5y = 5\), we can rewrite the equation in slope-intercept form \(y = mx + c\):

\(2x - 5y = 5\)

\(-5y = -2x + 5\)

\(y = \frac{-2}{-5}x + \frac{5}{-5}\)

\(y = \frac{2}{5}x - 1\)

The y-intercept is \(-1\). The line crosses the y-axis at \((0, -1)\). The origin \((0, 0)\) is located directly above the point \((0, -1)\) on the y-axis. Since the line has a positive slope (\(2/5\)), it goes upwards as \(x\) increases. Therefore, the origin \((0, 0)\) is located above the line \(y = \frac{2}{5}x - 1\), which is equivalent to \(2x - 5y = 5\).

Since the test point \((0, 0)\) satisfies the inequality \(2x - 5y \le 5\), and \((0, 0)\) is located above the line, the solution region is the area above the line \(2x - 5y = 5\). The "equal to" part of the inequality (\(\le\)) means that the line itself is included in the solution.

Thus, the graph of the in-equation \(2x - 5y \le 5\) is the region on and above the line \(2x - 5y = 5\).

Revision Table: Graphing Linear Inequalities

Additional Information: The Test Point Method

The test point method is a reliable way to determine which side of the boundary line represents the solution set of a linear inequality. Once the boundary line is graphed, pick any point that is not on the line. Substitute the coordinates of this point into the original inequality. If the inequality holds true for the test point, then the region containing that point is the solution region. If the inequality is false for the test point, then the solution region is the area on the other side of the line. The line itself is included in the solution set if the inequality symbol is \(\le\) or \(\ge\), and it is excluded (drawn as a dashed line) if the symbol is < or >.