Clearly, one can solve a system of linear equations by adding the constituent equations together or subtracting them from one another (via the so-called "elimination" method), but can one do the same with a system of linear inequalities. In other words, is an operation of the following form valid?

   x > y
+  a > b

   Therefore, x + a > y + b.

Many thanks!


Yes, it is valid. You can prove it's validness in two steps: First, if $a<b$ then $a+x<b+x$.
Secondly, if $x<y$ then $0<y-x$, by adding $-x$ to both sides. Hence $b+x<b+x+y-x=b+y$.
Finally, $a+x<b+y$

  • $\begingroup$ Perfect -- thank you very much for the clear explanation! May I ask a followup question? What about multiplying inequalities? $\endgroup$ – James Evans Oct 2 '12 at 12:53
  • $\begingroup$ Multiplying equations is problematic, since it depends on signs. If $a<b$ and $x>0$ then $ax<bx$, but if $x<0$ then $ax>bx$. $\endgroup$ – Dennis Gulko Oct 2 '12 at 16:21
  • $\begingroup$ For example, $-2<-1$ and $-1<3$, but is $2<-3$? $\endgroup$ – Dennis Gulko Oct 2 '12 at 17:24

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