Summary
Highlights
Graphing a system of linear inequalities involves graphing each inequality individually and then finding the region that satisfies all inequalities, which is the intersection of their solution sets.
A system of linear inequalities in two variables consists of two or more linear inequalities considered together. The solution to such a system is an ordered pair that satisfies all inequalities in the system.
To determine if an ordered pair is a solution, substitute the x and y values into each inequality. If both inequalities result in true statements, then the ordered pair is a solution to the system.
The video demonstrates checking if (6, 3) is a solution for x + 2y > 11 and -2x + y <= -7. Both inequalities are true for (6, 3), making it a solution. Conversely, (5, 3) is shown not to be a solution as it fails to satisfy the first inequality.
The example checks if (1, 4) is a solution for 3x + y < 4 and y <= -4x + 3. Both inequalities are false for (1, 4), meaning it is not a solution. However, (-4, 6) is verified as a solution because it satisfies both inequalities.
The first inequality, 4x + 2y < 12, is graphed by finding its x and y-intercepts (3, 0) and (0, 6) and drawing a dashed line. A test point (0, 0) shows that the region containing the origin is the solution set, so it is shaded.
The second inequality, y <= 5x + 4, is graphed using its y-intercept (0, 4) and a solid line due to 'less than or equal to'. The test point (0, 0) indicates that the region below the line is the solution set and is shaded green.
The solution to the system is the overlapping region where both shaded areas intersect. Any point within this darkest shaded region satisfies both inequalities.
Another system, x + y < 3 and y >= (2/5)x + 3, is graphed. The first inequality has a dashed line and is shaded below the line. The second has a solid line and is shaded above the line (indicated by the test point (0,0) being false).
The video demonstrates graphing a system including x-2, y<3, and y > x-1. A vertical line for x=-2, a horizontal line for y=3, and a diagonal line for y=x-1 are drawn, with appropriate shading based on the inequalities. The solution is the region where all three shaded areas overlap.
The final section works backward, showing how to determine the linear inequalities given a graph of their solution region. It involves identifying the boundary lines (slope, y-intercept) and using test points to determine the inequality signs (greater than, less than, or equal to).