# Is this equation with two unknowns solvable?

Can this equation be solved? If so how?

I would like to find both $X$ and $Z$ .

$4.33=\dfrac{0.4397-Z}{X-0.4397}$

where $Z$ is known to be in the range of $0.1931$ to $0.2352$

and $X$ is known to be in the range of $0.3549$ to $0.5576$

Ps. I am new here so if I could phrase my title better or format my question better please tell me. Also I don't know what type of equation this would be, that is if it is a legitimate equation, so I may have tagged it incorrectly.

Kind Regards

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In general, it is not possible to solve for both variables when there are two variables in a single equation (certain rare cases do exist though, such as in the equation, $x^2+y^2=0$, where $x$ and $y$ are both real).

In more generality, it is typically not possible to solve for all the variables in a system of $m$ equations in $n$ variables if $n > m$.

Hope this helps!

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Thanks Shaun, this is really the answer I was expecting, I just wasn't sure if the given ranges of values made any difference. – Duncan Gravill Nov 9 '13 at 16:59
Right, the ranges do not have much bearing on whether there is a unique solution. – Shaun Ault Nov 10 '13 at 0:01

Solve for $X$:

$$X = 0.5412 - 0.2309 Z \, \text{ if } Z \ne 0.4397$$

So $0.3549 \le 0.5412 - 0.2309 Z \le 0.5576$ which means $-0.8069 \le -Z \le 0.0710$ so $-0.0710 \le Z \le 0.8069$. That's not very helpful though because it has a wider range of values than what you already know.

You can do the same by solving for $Z$:

$$Z = 2.3436-4.33 X \, \text{ if } X \ne 0.4349$$

So in the end you'll get $0.4869 \le X \le 0.4967$.

All you can do, then, is narrow the possible range of values to $0.4869 \le X \le 0.4967$ and $0.1931 \le Z \le 0.2352$.

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Thanks MasterOfBinary, this is definitely interesting. Though I don't follow how you developed the first line from the equation I gave. Could you please explain? – Duncan Gravill Nov 9 '13 at 17:11
@FunkyFresh84 If you solve your first equation for $X$ with respect to $Z$ that's what you get. – MasterOfBinary Nov 10 '13 at 2:01