Example

Graph:      a) 2x + 3y > 6      and       x  1

Solution:

First we’ll graph the linear equations from the inequalities which are the boundaries for our shaded areas on the graph

2x + 3y > 6

Graphing      2x + 3y = 6 
Substitute x = 0 then y = 2   ð  (0, 2)
Substitute y = 0 then x = 3   ð  (3, 0) 
Graph the line on your xy-plane as a broke
line because there is no “=” sign in your 
linear inequality.               
Choose (0, 0) as your test point.  
Substitute the test point in the original Inequality:  2(0) + 3(0) > 6    
                0 > 6                False
Then you’ll shade the area where your test
point does not belong to.   

X

Y

0

2

1

0

x  1 

 

 Graphing      X = 1
Substitute y = 0 then x = 1   ð  (1, 0)
Substitute y = 3 then x = 1   ð  (1, 3)
Graph the line on your xy-plane as a connected line because there is “=” sign in your linear inequality.
Choose (0, 0) as our test point Substitute the test point in the original  Inequality:  
0  1                   True 
Then we’ll shade the area where our our test point is at or belong to.

X

Y

1

0

1

3


 If you notice on the graph our shaded area for both inequalities together is the overlap (intersection) of the graph of each inequality, which is the dark shaded area between the tow lines

 


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