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[I had trouble deciding how to tag this question, so please edit if there's a better classification.]

Some special types of indeterminate equations are solvable. for example, consider this example from the textbook:

Solve:

$(x+y)(x+z)=30$

$(y+z)(y+x)=15$

$(z+x)(z+y)=18$

Put $y+z = u, z+x = v, x+y = w$. Thus,

$vw = 30, wu = 15, uv = 18$

Multiplying them together yields $u^2v^2w^2 = 90^2$, etc., from where the following solution sets are obtained: $\{4,1,2\},\{-4,-1,-2\}$.

That was the textbook method. But when I had attempted the problem, I had noticed the common factor in any two equations and proceeded thus:

Divide first equation by second, second by third, and first by third, we get the following three equations:

$x-2y-z = 0$

$x + 6y - 5z = 0$

$3x-2y-5z = 0$

Unfortunately, all that these equations yield is the simple ratio that $x:y:z = 4:1:2$. Even if I assume $x=4k, y=k, z=2k$, etc., plugging this in any of the equations gives $k =0$. It looks like the equations are indeterminate in the true sense of the word!

Now my question is: How come some jugglery can yield nice little integral solutions whereas straightforward efforts leave us going round in circles? Perhaps what I'm asking is more on the philosophical side, but I'd like some light to be please thrown on this.

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ankush981
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1 Answers1

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The linear system is indeterminate because, for example, the third equation can be obtained from the first two. If you couple the latter with any equation of the original system, you are done.$$\begin{cases} x-2y-z = 0\\x+6y-5z = 0\\(x+y)(x+z)=30 \end{cases}$$

Tony Piccolo
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  • Thanks, but I don't see how the third linear equation may be obtained from the first two. – ankush981 Aug 20 '13 at 08:13
  • Multiplying side by side the first two, before eliminating the denominators of course $\frac {a}{c}=\frac {b}{d} \land \frac {c}{e}=\frac {d}{f} \rightarrow \frac {a}{e}=\frac {b}{f} ,$. – Tony Piccolo Aug 20 '13 at 08:36
  • Sorry, but I think something is wrong somewhere. Where does denominator come in $x−2y−z=0, x+6y−5z=0, 3x−2y−5z=0$? – ankush981 Aug 20 '13 at 09:59
  • Remember you started from $\frac {x+z}{y+z}=\frac {30}{15}$ and so on: I refer to those. – Tony Piccolo Aug 20 '13 at 10:36
  • Thanks, that makes sense. But I'm still wondering how an indeterminate system can yield a solution using an equation derived from the system itself. Or perhaps it's a stupid question to ask ... – ankush981 Aug 20 '13 at 11:09
  • The system, I am speaking of, is not linear: it contains two equations obtained by division and an equation of the original system. You must not consider an indeterminate linear system. – Tony Piccolo Aug 20 '13 at 13:46