Questions: Graph the system of inequalities and shade the portion on the graph represented by the solution set. x-4y ≤ -3 x+4y ≥

Solution Steps

The given inequalities are:

\(x - 4y \le -3\) \(x + 4y \ge 3\)

We rewrite these inequalities in slope-intercept form (i.e., \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept).

For the first inequality: \(x - 4y \le -3\) \(-4y \le -x - 3\) \(4y \ge x + 3\) \(y \ge \frac{1}{4}x + \frac{3}{4}\)

For the second inequality: \(x + 4y \ge 3\) \(4y \ge -x + 3\) \(y \ge -\frac{1}{4}x + \frac{3}{4}\)

We graph the lines \(y = \frac{1}{4}x + \frac{3}{4}\) and \(y = -\frac{1}{4}x + \frac{3}{4}\).

The first line has a y-intercept of \(\frac{3}{4}\) and a slope of \(\frac{1}{4}\). This means that for every 1 unit increase in \(x\), \(y\) increases by \(\frac{1}{4}\). The second line has a y-intercept of \(\frac{3}{4}\) and a slope of \(-\frac{1}{4}\). This means that for every 1 unit increase in \(x\), \(y\) decreases by \(\frac{1}{4}\).

Since the inequalities are greater than or equal to, we use solid lines to represent the boundaries.

For the inequality \(y \ge \frac{1}{4}x + \frac{3}{4}\), we shade the region above the line \(y = \frac{1}{4}x + \frac{3}{4}\). For the inequality \(y \ge -\frac{1}{4}x + \frac{3}{4}\), we shade the region above the line \(y = -\frac{1}{4}x + \frac{3}{4}\).

The solution set is the intersection of the shaded regions, representing the area where both inequalities are satisfied.

Final Answer