Question 150765
First graph the line {{{y=3x}}}



{{{ graph( 500, 500, -10, 10, -10, 10,3x+0) }}} Graph of {{{y=3x}}}



Now plug in a test point (0,1) into the inequality {{{y<3x}}}. Note: the test point should <font size=4><b>not</b></font> be on the line.



{{{y<3x}}} Start with the given inequality.



{{{1<3*(0)}}} Plug in {{{x=0}}} and {{{y=1}}}



{{{1<0}}} Evaluate and simplify.



Since the inequality is false, this means that we shade the entire region that does <b>NOT</b> contain the point (0,1)



In other words, we simply shade the <b>entire</b> region that is below the line.



{{{drawing( 500, 500, -10, 10, -10, 10,
graph( 500, 500, -10, 10, -10, 10,3x),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-3),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-6),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-9),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-12),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-15),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-18),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-21),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-24),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-27),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-30),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-33),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-36),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-39),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-42),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-45),
graph( 500, 500, -10, 10, -10, 10,3x,3x+-48))}}} Graph of {{{y<3x}}} with the shaded region in green


Note: since the graph has a less than sign "<", this means that the boundary line is a dotted line.