Question

How do you graph the system `y >= - 2x - 3` and `3x - y > 2`?

Answer 1

A graph of a particular equation or inequality that involves two unknown variables `x` and `y` is a set of all points `(x,y)` on a Cartesian plane where values of `x` and `y` satisfy the given equation or inequality.

A graph of a system of equations or inequalities that involve two unknown variables `x` and `y` is a set of all points `(x,y)` on a Cartesian plane where values of `x` and `y` satisfy each and every given equation or inequality.

The above definition means that the graph of a system of equations or inequalities is an intersection of graphs representing each one of these equations or inequalities.

Draw a graph of the first inequality
`y >= -2x-3` (border line is included)
graph{y>= -2x-3 [-10, 10, -5, 5]}

Draw a graph of the second inequality
`3x-y > 2` or, equivalently, `y < 3x-2` (border line is not included)
graph{y < 3x-2 [-10, 10, -5, 5]}

All we need to do now is to intersect these two areas, and that would be a graph of a system of two inequalities.
To be precise, let's find the point of intersection of the left boundaries of both graphs (there is no right boundaries, the corresponding areas of the graphs infinitely continue as an argument `x` increases).
The point of intersection is, obviously, a solution of a system of equations:
`y=-2x-3` and
`3x-y=2`

To solve it, let's summarize left and right parts of both equations and get an equation for `x` only since `y` would be canceled.
`y+3x-y=-2x-3+2`
`3x=-2x-1`
`5x=-1`
`x=-1/5`
Substituting this into the first equation gives `y`:
`y=-2*(-1/5)-3=2/5-3=-13/5`

Therefore, the left boundary of the intersection of two areas represented by two graphs above has a key point
`(x,y)=(-1/5,-13/5)`
from which two lines are going - up along the line `y=3x-2` (the line itself is not included, only the area to the right of it) and down along the line `y=-2x-3` (including the line itself and the area to the right of it).
The area of intersection would approximately look like this (except the lower part should be solid line which implies that the border is included):
graph{x+1/5>|y+13/5|/2 [-10, 10, -5, 5]}