How to graph inequality on a plane

May 2010
52
0
The part that gets me is the plane thing.

X + Y < 1

If x is 0 then Y is 0, -1, -2, -3 and so on right?

If y is 0 then the same applies for X =0, -1,-2, -3 and so on?

Please correct me if I am wrong.

Therefore How do I graph this? And what sets of numbers should I use?

00 for both won't work???
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Yes, the values you give are correct but the graph has to include non integers, rational and irrational numbers as well.

The graph of the equation, y= x+1, is a straight line. On that line, when x= 0, y= 1 and when y= 0, x= -1. The graph of y= x+1 is the straight line through (0, 1) and (-1, 0).

The reason you should graph that line first is that it forms the boundary of the points that satisfy y< x+ 1. Obviously that line divides all of \(\displaystyle R^2\) into two subsets. The points that satisfy y< x+ 1 is one of the two subsets on either side of the line. Choose a point on each side of the line to determine which side satifies the inequality.

Also, because this is y< x+ 1, rather than \(\displaystyle y\le x+1\), you don't want to include the line itself in the graph. One way to indicate that is to use a dashed line rather than a full line.
 
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