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

1 Answer
Apr 5, 2015

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|#