Given the system of inequalities, list all the points included in the solution: \[x + y > 4\] \[-2x + y < 3\]

To solve the system of inequalities \(x + y > 4\) and \(-2x + y < 3\), it's not practical to list all the points because there are infinitely many points that satisfy both inequalities. Instead, one can describe the region that represents the solution set.

1. Graph the first inequality \(x + y > 4\): - To graph the related equation, \(x + y = 4\), find two points that lie on the line by choosing values for \(x\) and solving for \(y\) or vice versa. For example: - If \(x = 0\), then \(y = 4\). - If \(y = 0\), then \(x = 4\). - Draw a dashed line passing through the points (0, 4) and (4, 0) because the inequality is strictly greater than (not equal to). - Choose a test point not on the line, such as the origin (0,0), to determine which side of the line represents the inequality: - Plugging (0,0) into \(x + y > 4\) gives \(0 + 0 > 4\), which is false, so the region that does not include (0,0) is the solution to the inequality. This is the region above the line.

2. Graph the second inequality \(-2x + y < 3\): - Graph the related equation, \(-2x + y = 3\), using a similar method to find two points. For example: - If \(x = 0\), then \(y = 3\). - If \(y = 0\), then \(x = -1.5\). - Draw another dashed line through the points (0, 3) and (-1.5, 0). - To determine which half-plane to shade for the inequality, use a test point again. Plugging (0,0) into \(-2x + y < 3\) gives \(0 < 3\), which is true, so the region that includes (0,0) is the solution to this inequality. This is the region below the line.

The solution to the system of inequalities includes all the points that are in the region above the line \(x + y = 4\) and below the line \(-2x + y = 3\). To visualize the solution set, draw both dashed lines on the same coordinate plane and shade the area which overlaps that satisfies both inequalities.