How do you graph #5x + 3y > -6# and #2y + x < 6#?

First, graph Line (1) -> 5x + 3y + 6 = 0 by its 2 intercepts. Make x = 0 --> y = - 2. Make y = 0 --> x = -6/5. Use the origin O as test point. Replace x = 0 and y = 0 into inequality (1). We get: 6 > 0. True!. Then, the solution set of inequality (1) is the area containing O. Color it. Next, graph Line (2) -> 2y + x - 6 = 0 by its 2 intercepts. Make x = 0 --> y = 3. Make y = 0 --> x = 6. Use O as test point. The solution set of (2) is the area containing O. Color it. The compound solution set is the area that is shared by both graphs. graph{5x + 3y + 6 = 0 [-10, 10, -5, 5]} graph{2y + x - 6 = 0 [-10, 10, -5, 5]}

To graph (5x + 3y > -6) and (2y + x < 6), follow these steps:

1. Graph the boundary lines for each inequality by first replacing the inequality symbol with an equal sign. For (5x + 3y > -6), the boundary line is (5x + 3y = -6), and for (2y + x < 6), the boundary line is (2y + x = 6).
2. To graph each boundary line, solve for (y) in terms of (x). For (5x + 3y = -6), (y = (-5/3)x - 2), and for (2y + x = 6), (y = (-1/2)x + 3).
3. Plot each boundary line on the coordinate plane.
4. Choose a test point not on the boundary line and substitute its coordinates into the original inequality. If the inequality is true, shade the region containing the test point. If false, shade the opposite region.
5. For (5x + 3y > -6), if you test the origin (0,0), you find that (5(0) + 3(0) > -6), which is true. So, shade the region above the line (5x + 3y = -6).
6. For (2y + x < 6), if you test the origin (0,0), you find that (2(0) + (0) < 6), which is true. So, shade the region below the line (2y + x = 6).
7. The region where both shaded areas overlap represents the solution to the system of inequalities.