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).
