Question

How do you find the solution to `x - y > 3` and `x + y < 3`?

Answer 1

A solution to a system of two inequalities with two variables `x` and `y` is a set of pairs `(x,y)` that satisfy both inequalities.

If you are looking for a solution to these two inequalities in terms of two separate inequalities, one for `x` and another for `y`, you will not find them, there are no such solutions. Variables `x` and `y` are related and we cannot separate them.

The best approach to "solve" this system of two inequalities is to better represent all the pairs `(x,y)` that satisfy them.

Let's represent the solutions to this system of inequalities (that is, all pairs `(x,y)` that satisfy them) graphically.
First of all, let's transform both inequalities in a more graph-friendly representation.
Add the same number `y-3` to the first inequality obtaining
`x-y+y-3 > 3+y-3` or
`x-3 > y` or
`y < x-3`
Add the same number `-x` to the second inequality obtaining
`x+y-x < 3-x` or
`y < -x+3`

Now we have two inequalities:
`y < x-3` and
`y < -x+3`

Let's graph them now.
The graph of `y=x-3` is
graph{x-3 [-6, 6, -4, 2]}

All pairs `(x,y)` that satisfy the equality `y = x-3` are on the line of this graph.
All pairs `(x,y)` that satisfy the inequality `y > x-3` are above the line of this graph.
All pairs `(x,y)` that satisfy the inequality `y < x-3` are below the line of this graph.
So, all the solutions to the first inequality are those pairs `(x,y)` that lie below the line of this graph.

Similarly, the graph of `y=-x+3` is
graph{-x+3 [-6, 6, -4, 2]}

All pairs `(x,y)` that satisfy the equality `y = -x+3` are on the line of this graph.
All pairs `(x,y)` that satisfy the inequality `y > -x+3` are above the line of this graph.
All pairs `(x,y)` that satisfy the inequality `y < -x+3` are below the line of this graph.
So, all the solutions to the second inequality are those pairs `(x,y)` that lie below the line of this graph.

Notice that both graphs intersect at a point `(0,3)`.

Now we have to imagine an area on the coordinate plane that is both below the first graph AND below the second graph. This area represents a solution to a system of two inequalities we have.

You can consider this area as a quarter of a plain with a vertex at a point `(0,3)` and borders diagonally going SE and SW from it.
Here is a graphical representation of this area with solutions to our system of inequalities being all the pairs `(x,y)` representing all points below the following graph line.
graph{-|x-3| [-6, 6, -3, 3]}

Incidentally, you can express this with another inequality:
`y < |x-3|`