I was working with a student today who was having problems graphing a system of linear equations. So, I am posting a small part of my book explaining the technique to graph such problems.
Systems of Linear Inequalities
Solving systems of linear inequalities graphically is very similar to solving systems of linear equations. We
graph the system inequalities on the same rectangular coordinate system. Shade the area which represents the intersection of the graphs. To check, we can pick a point from the shaded region and make sure the coordinates satisfy both inequalities in the system. We will start by graphing both separately and then graphing both on the same rectangular coordinate system.
Example: Solve the system by graphing
y ≥ 2x + 3
3x + y < 6
The graph of y ≥ 2x + 3.
Find the shaded region by picking a point on either side of the line to see if it satisfies the inequality. The
easiest to choose is (0, 0). Substitute into y ≥ 2x + 3 to get 0 ≥ 2(0) + 3. This is a false statement so the solution is on the other side of the line. We can check to make sure the shaded region is correct by choosing a point in that region and see if it satisfies the inequality. We'll choose the point (-3, 0). Therefore 0 ≥ 2(-3) + 3, 0 ≥ -3 is a true statement, so our solution is correct.
Next we will graph 3x + y < 6.
If we choose any point in the unshaded region and substitute into both inequalities, we will notice that the
inequalities will not be satisfied. The shaded area in the graph below is the region that only satisfies 3x + y < 6. We find the shaded region by picking a point on either side of the line to see if it satisfies the inequality. Choose (0, 0) since it's the easiest to work with. Therefore 3(0) + 0 < 6 is a true statement and is part of the solution set. All points on the same side of the dotted line satisfy the inequality.
By graphing both inequalities on the same rectangular coordinate system, we will see where the two solutions
The large shaded area where the inequality y ≥ 2x + 3 is located satisfies both equations. Therefore it is the
solution to the system of inequalities.
•Note that we show that the green area is the solution by picking a point in the region and testing it in
both inequalities. Any point in that region will satisfy both inequalities