A student proposed the following method of sketching the solution to a system of two inequalities. “First, I shade the region of points that satisfies the first inequality. Then, I shade the...
A student proposed the following method of sketching the solution to a system of two inequalities. “First, I shade the region of points that satisfies the first inequality. Then, I shade the region of points that satisfy the second inequality. The solution consists of all points that have been shaded.” Do you agree? Explain. Support your answer using the system `5x-3y>1`
Yes, this is an acceptable way to solve the problem.
The solution to an inequality is the set of all points that satisfy the inequality. The solution to a system of two linear inequalities is also the set of all points that satisfy the system -- that is the set of all points that satisfy both inequalities.
There are three cases in the case of two lines in a plane:
(1) They are actually the same line. In this case any solution to the first inequality is a solution to the second inequality.
(2) The lines are parallel. Multiple cases arise from this situation: there might be no solutions to the system, the solutions could be all of the points between the lines or the solution could be all of the points to one side of one of the lines.
(3) The lines intersect as is the case in the example. The line will divide the plane into four quadrants. One of these quadrants is the solution set.
See the attached graph:
Here we graph 5x-3y>1. (The points (-1,-2) and (2,3) lie on the line 5x-3y=1). Note that we graph with a dotted line as points on the line are not solutions. Then check a point not on the line;say (0,0). Since 0 is not greater than 1, (0,0) and all other points on the same side of the line are not solutions to the inequality so we shade the other side in blue.
Now graph 3x+4y<=18 in the same coordinate system. The points (2,3) and (6,0) lie on the line and we graph with a solid line as points on the line are solutions to the inequality. Now check a point not on the line; say (0,0). Since 0<18 (0,0) is a solution as are all points on the same side of the line so shade this side red.
The solution to the system is the area shaded twice -- in this case purple. Note that all points in the region are solutions, as are all points along the line 3x+4y=18 where x>2. The point (2,3) is not a solution as it satisfies only one inequality. None of the points on the line 5x-3y=1 are solutions.