Inequality system problem Say there is a cat that in three days eat 12 fishes.
Say that each day the cat eats more than the day before.
Say that the last day the cat has eaten less than the addition of the two previous days.
x+y+z = 12
z>y>x
z<x+y

I assume x -> first day, y -> second day, z -> third day.
How am I supposed to solve this? I have been taught to solve inequality systems with two variables (doing graphically), but I don't know how to do this one.
Thanks.
 A: To do a graphical solution, you can use the equality $z=12-x-y\ \ $ in the inequalities to get $12 \lt 2x+2y ,\ y \gt x ,\ 12 \gt 2y+x.\ \ $  You can plot these three as you are used to, shading the proper side of the lines.  Then as you are looking for integer solutions you have to look for lattice points in the allowable region.
A: Sometimes, unfortunately, the method you are "supposed" to use is gross overkill. But I will write out a linear inequalities approach, under the assumption that the 
number of fish eaten on any day is a non-negative integer.
Also, when your text said less than and your formula said $\le\;$, I assume it should be $\lt$.  Similarly, if your text says more and your formula said $\ge\;$, I assume it should be $\gt$.  So when there is conflict, I give priority to the words.  
We have three variables, but one can be eliminated, using say $z=12-(x+y)$.
Since $x<y$, we have
(1) $x+1 \le y$.
Since $y \lt z$, we have $y+1 \le z$, and therefore $y+1 \le 12-(x+y)$. This can be rewritten as 
(2) $x+2y \le 11$.
We have $z \lt x+y$, so $z+1 \le x+y$, and therefore $12-(x+y)+1 \le x+y$, giving
(3) $13 \le 2x+2y$.
Finally, we need to remember that $x$, $y$, and $12-(x+y)$ are all $\ge 0$.
That gives the inequalities
(4) $x \ge 0$, $y \ge 0$, $12 \ge x+y$.
Now we can draw lines as usual. 
Much too much work, since it is clear that $x=3$, $y=4$, $z=5$ is the only solution!  This is because if $x\lt y \lt z$ and $x+y+z=12$, we must have $z\gt 12/3$, so $z \ge 5$.  If $z\ge 6$, then $x+y \le 6$, but $x+y \le z$ is not allowed. So $z=5$.  Now we have $7$ fish, to be distributed among the two days. But the second day must be $\le 4$. To get a total of $7$ we need $y=4$, $x=3$.
Note: If everywhere we use your formula inequalities, not your text inequalities, all the $+1$'s need to be removed. That changes the inequalities, and we get a few more solutions. Which ones are OK depends on interpretation of the wording. The solutions could include some or all of  $(0,6,6)$, $(1,5,6)$, $(2,4,6)$, $(2,5,5)$, $(3,3,6)$, $(3,4,5)$, and $(4,4,4)$.      
A: A general solution of a system of inequalities is very difficult (see e.g. mathoverfow.net). Your special case can be solved graphically as answered by @Ross. You can also argue as follows. 
All variables must obviously be non-negative and less or equal 12. Combining the first equality and the last inequality we get $z \le 6$. $z$ must also satisfy $z \ge 4 $ 
because $z \lt 4$ would imply $y \lt 4$, $x \lt 4$, $x+y+z \lt 12$.
Similarly, $x$ must satisfy $x \le 4 $ because $x > 4$ would imply $y \gt 4$, $z \gt 4$, $x+y+z \gt 12$.
So, all solutions are obtained by choosing 
$$x \in [0, 4]$$
$$z \in [4, 6]$$
$$y = 12 - (x+z)$$
If only positive solutions are allowed, then $x \in (0, 4]$.
