We are asked to graph the solution set for the following system of linear inequalities:

`{([2x+4y>16],[-x+3y>=7])`

Each linear inequality in two variables divides the plane into half-planes. One half-plane contains all of the solutions for the inequality. Typically you would graph the underlying line (e.g. for 2x+4y>16 we would graph the line 2x+4y=16.) ((If the inequality is strict , i.e. < or >, we draw the line as a dashed or dotted line. If the inequality is inclusive we draw a solid line.))

After drawing the line we would shade the side of the line that has all of the solutions. One way is to pick a random point that does not lie on the line; if the point satisfies the inequality we would shade on that side of the line.

A system of two linear inequalities consists of two intersecting lines or two parallel lines. If the lines are parallel then there are three cases. (1) There are no solutions (e.g. the shaded area is "above" one line and "below" the other),(2) The solution of one inequality is all of the solutions and (3) the solution set (shaded area) is between the parallel lines.

If the two lines intersect the plane is divided into four regions, one of which is the solution set.

Considering only linear inequalities the intersection of three or more lines can create polygonal regions. If the solution set is the interior of a polygon we say that the solution set is bounded. If the solution set is the exterior of all of the polygons then the set is unbounded.

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For this problem graph the line 2x+4y=16 (dotted since no point on the line satisfies the inequality.) We choose a convenient point not on the line; say (0,0). Now 2(0)+4(0)<16 so (0,0) is not a solution so we shade the side of the line away from (0,0). Then graph the line -x+3y=7 (solid since the inequality is inclusive.) Checking (0,0) we see that -(0)+3(0)<7 so (0,0) is not a solution.

The solution set is the region that is shaded by both constraints (inequalities.) The point (0,5) lies in the solution region. 2(0)+4(5)>16 and -(0)+3(5)>7. (See attachment.)

The solution set is unbounded -- the lines do not create a polygon with the solution set in the interior.

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