Section 3-3:  Polynomial Inequalities in Two Variables
 

Demo: Linear Inequalities (Larry Green)
 
The graphs of inequalities in two variables consist of points in the x-y graphing plane.  Two find the graph, first graph the equation.  This is the boundary line between what makes the inequality greater than or less than.  The line can be solid (points on the line make the statement true) or dotted (does not include the points on the line).  You must decide which side of the line to shade.  The easiest way to tell is to pick a point on one side of the line.  If it makes the statement true, you have the correct side.  Go ahead and shade it.  If it makes the statement false, you have the wrong side.  Shade the other side.
 
 
Sample Problems
 
1)  Graph  the inequality y < 2x + 1
The boundary is a line.  The y-intercept is (0,1), with slope of 2.  Graph the line!
Notice the line is dotted because the line is not included ( no or equal to in the statement).
Now pick a point not on the line.  Use (0,0).  Is 0 < 2(0) + 1?  Sure is, so shade this side of the line!
 

 
2)  Graph the solution of  y > x + 3 and y < 9 - x2
 
The first has a boundary line.  The second boundary is a parabola.  The line graphs with slope 1 and y-intercept (0, 3).  The parabola has a vertex point at (0, 9) and crosses the x-axis at 3 and -3.  Look back to graphing parabolas if you forgot about that!!
 Now pick a point.  Try (0, 0).  Try the line.  0 > 0 + 3.  This is false.  Shade the other direction, above the line.  Now try (0, 0) in the parabola.
0 < 9 - 0.  This is true.  Shade inside the parabola.  Viola!!  You get:
The solution is the cross-hatched area!!

 
3)  Find the solution for |y| > 1 and |x| < 3 in two variables.
Both are absolute value graphs  The first can be written as y > 1 or y < -1.  The second can be written as -3 < x < 3.
The first absolute value graph means we want values above one and below -1 including the equality.
The second absolute value means x must be between -3 and 3 including these points.  Add that graph!!
The solution is the crossed-hatched area!!

4)  Find the solution for y < x + 1, y > -x + 4 and y > 2
 
Each of these has a boundary which is a line.  Two are slanted and the other is horizontal.  The first has slope 1 and y-intercept 1.  The second has slope -1 and y-intercept 4.  The last is a horizontal line at 2.
You can try (0,0) for all three lines.  In the first line 0 < 0 + 1 is true.  Shade below the blue line.  In the second line, 0 > 0 + 4 is false.  Shade above the green line.  In the third line, 0 > 2 is false.  Shade above the purple line.  The result is:

 
 On to linear programming: 
 
Back it up dude!! I'm lost!!