Solving for two unknowns in an inequality

This is probably one of the more embarrassing moments of my life, but I will soldier on. Having not dealt with inequalities for quite some time, I was rather befuddled by a question from a friend of mine regarding solving for two variables given a pair of inequalities. The problem goes like this:

Given $x + y \lt 11$ and $x \gt 6$, what is the lower bound of $x - y$ ?

I'm fairly sure she told me the answer was $4$, but I've exhausted my bag of tricks trying to get there. If these were equalities, it would be fairly easy, of course, but would someone be so kind as to help nudge me in the right direction?

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take x=6.1 and y=4.8, then you have x+y=10.9 x-y=1.3 – anonymous Dec 6 '10 at 4:07
Are $x$ and $y$ positive integers? – Aryabhata Dec 6 '10 at 4:09
Sorry guys...yes the problem does say they are integers. @Chandru...sorry I'm having a brain fart. Could you perhaps expound on that just a little bit? – treeface Dec 6 '10 at 4:11
so lower bound of x-y is not 4 – anonymous Dec 6 '10 at 4:12
@treeface: Integers, or positive integers? Big difference. Anyway, based on the answer, I am pretty sure it will be positive integers. – Aryabhata Dec 6 '10 at 4:14

Are $x$ and $y$ required to be integers? Because if they are not, then the answer cannot be $4$: take $x=6.5$ and $y=4$. Then $x+y = 10.5\lt 11$, but $x-y = 2.5 \lt 4$. In fact, taking $x=6+\epsilon$ and $y=5-(\epsilon+\delta)$ with $\epsilon\gt 0$ and $\delta\gt 0$ arbitrary but positive, and this gives you $x+y = 11-\delta\lt 11$, but $x-y = 1+2\epsilon+\delta$, so we can make $x-y$ as close to $1$ as we like (without it ever being equal to $1$); on the other hand, since $x\gt 6$ and $x+y\lt 11$, then $y\lt 11-x \lt 11-6 = 5$, so $x-y \gt 6-5 = 1$. Thus, the difference is always greater than $1$; but since we can make it as close to $1$ as we want, the best possible lower bound for $x-y$ is $1$ (without it actually being achieved).

However, if $x$ and $y$ are required to be integers, then the smallest value of $x-y$ will occur when $x$ is as small as possible and $y$ as large as it possibly can be given that.

Added: To see why, first suppose that $x+y\lt 11$ with $x\gt 6$, but $y$ is not as large as it can be; that is, that $x+(y+1)$ is also less than $11$. Setting $y'=y+1$, you have that $x-y' = x-y-1 \lt x-y$, so you get a smaller value with $y'$. That means that in order to find the smallest possible value of $x-y$ for a given $x$, you may assume $y$ is as large as it possibly can be. Once you have that, you can check that larger values of $x$ will give you larger differences, so you also want $x$ to be as small as it possibly can be.

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Thank you, Arturo. This helped me enormously! – treeface Dec 6 '10 at 4:35

The answer is $x-y>1$. What you can do is to graph the set $A$: $x+y<11, x>6$ and note that $x-y=1$ passes through the intersection of the two lines $x+y=11$ and $x=6$. The set $A$ is to the right of $x-y=1$, which is $x-y>1$.

If $x,y$ are integers, then the smallest integer $k$ such that the set $x-y\ge k$ includes all the points in $A$ is 4 ( $(7,3)$ lies on the line $x-y=4$). This is clear from the graph.

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"what you can / could do..."? (-; – Arturo Magidin Dec 6 '10 at 4:21
@Arturo. Corrected. – TCL Dec 6 '10 at 6:22