Let's start with the first inequality: #x+y<3#. We can add a value #a# to the left-hand side of this inequality as long as we add another value #b≥a# to the right-hand side of this inequality. After looking at the second inequality, we notice that we can set #a=x-y# and #b=4# as #x-y<4# from the second inequality.
This becomes #x+y+x-y<3+4#. The #y#'s cancel out, and we get #2x<7#. We can divide both sides by #2# to get #x<7/2#.
Now, #x+y<3# and #x-y<4#. Since #x<7/2#, #y# can have any value. To see this, solve for #y# to get #y<3-x# and #y>x-4#. In the first inequality, as #x->-oo# (allowed since #x<7/2#), #y < oo#. In the second inequality, as #x->-oo#, #y> -oo#.
So our solution is #{x, y|x<7/2nnyinRR}#, or #x in (-oo, 7/2)# and #y in (-oo,oo)#.
