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This one's a little harder:

Graph      y > ( 1 / 3 )x

* Remember -- to graph this guy use y = mx + b.  And we'll use a dashed line since there is no =.

the graph of y > ( 1 / 3 )x, which passes through the points ( -3 , - 1 ) , ( 0 , 0 ) , and ( 3 , 1 ) ... remember the line is dashed What side do we shade?

The only glitch on this one is that we can't try
( 0 , 0 )
because it's ON the line.

Let's try ( 1 , 2 ) :

y > ( 1 / 3 ) x ... 2 > ( 1 / 3 ) ( 1 ) ... 2 > ( 1 / 3 )

the graph of the line y = ( 1 / 3 )x ... test the point ( 1 , 2 )

Yep!  It works!  So, shade that side:

 
the graph of y > ( 1 / 3 )x ... the portion of the graph above the line is shaded All the ( x , y ) points in the shaded region work in

y > ( 1 / 3 )x

 


YOUR TURN:

Graph

y is less than or equal to 2x