How do you graph the inequality #3-x>0# and #y+x< -6#?

First we write the inequalities as linear equations, and the graph these. Remember to use a dashed line, as these are not equal to inequalities, so the line itself will not be an included region.

#3-x>0#

#3-x=0# , #x=3color(white)(88888888888)# First equation.

#y+x<-6#

#y+x=-6# , #y=-x-6color(white)(88)# Second equation.

We now graph these:

With graph plotted, we can see that there are four possible regions.

A , B , C and D

The required region has to satisfy both inequalities, so we test a set of coordinates in each region. We can save some work by realising that if a coordinate in a region fails for the first inequality we test, then the region can't be the required region and it is not necessary to test the other inequality as well.

Region A

Coordinates: #(-2 ,2)#

#color(blue)(3-x>0)#

#3-(-2)>0color(white)(88)# , #5>2color(white)(88888888)# TRUE

#color(blue)(y+x<-6)#

#2+(-2)<-6color(white)(88)# ,#0<-6color(white)(888)# FALSE

Region A is not the required region.

Region B

Coordinates: #(2 ,2)#

#color(blue)(3-x>0)#

#3-(2)>0color(white)(88)# , #1>0color(white)(888888888)# TRUE

#color(blue)(y+x<-6)#

#2+2<-6color(white)(88)# , #4<-6color(white)(888)color(white)(88)# FALSE

Region B is not the required region.

Region C

Coordinates: #(-2 , -6)#

#color(blue)(3-x>0)#

#3-(-2)>0color(white)(88)# , #5>0color(white)(888888888888888)# TRUE

#color(blue)(y+x<-6)#

#(-6)+(-2)<-6color(white)(88)# , #-8<-6color(white)(8888)# TRUE

Region C is a required region.

We do not need to test region D. As can be seen from the graph above if C is an included region and is to the left of the line #x=3#, then to the right of this line is an excluded region.

Shade the required region C