Question

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

Answer 1

See below.

Explanation:

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