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After solving a system of equations I get this:

\begin{cases} x + 0.03125z = 0.242813 \\ y + 0.40625z = 0.456563\end{cases}

How can I filter solutions that are only positive for $x$, $y$ and $z$

Ted

3 Answers 3

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For equations that look complicated, it helps to artificially simplify them by guessing really boring values that you might solve for.

For example let $y$ and $z$ be $0$ (yes, that's not positive but it's not particularly negative either (unless you're speaking French)..actually just consider $x,y,z$ as non-negative...makes things easier all around). At least $y$ and $z$ are as small as possible. Then you'll notice that $x$ is as big as your RHS (right hand side), which really means it is as big as possible. And notice that $x>0$.

Now do the same for $y$ (set $x$ and $z$ to $0$) and separately for $z$ (set $x$ and $y$ to $0$).

Notice now that if you subtract a little from any one of them, you'll be able to add a little more to the others (not the same amount for each).

So the smallest value for each could be $0$, and the largest value will be what your solutions are for the above simplifications. And by the above 'notice', all values in between will work.

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Perhaps you could glean something from this.

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Are your constants really reals? It looks like it may be rounded from $x+\frac{z}{32}=\frac{777y}{3200}+\frac{13z}{32}=\frac{1461}{3200}$ If not, you can use the decimals in the following. Break it into $3200x+100z=1461$ and $777y+1300z=1461$. You can see that $x \le \frac {1461}{3200}, y \le \frac {1461}{777}, z\le \frac {1461}{1300}$ and as any one gets larger, it drives down the others.