# Why does this system of equations result in an always-positive output?

I have two variables, M and P, whose relationship is described by the following two equations:

[1] P = 50.5 * M / ( M - 50)

[2] M = C * P


where C is a positive constant. I've arbitrarily designated M to be the output.

In equation 2, M will be positive as long as P is positive. If P is negative, M will be negative. However, if we substitute in P from equation 1 into equation 2 we get:

[3] M = 50.5 * C + 50  OR  M = 0


Now M is guaranteed to be non-negative.

Why is this? I've graphed the equations and noticed that they only intersect in quadrant I. It makes sense that equation 3 would reflect this. But I still feel like there's something mysterious about the whole thing. The phenomenon (M suddenly being only positive) just seems to pop out randomly when looking at it as equations rather than graphically. Is there a fact or concept that will help me demystify the equation side of things?

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The second equation requires both $M$ and $P$ have the same sign. So your solution must be on first or third quadrant. Then... – Sigur Nov 19 '12 at 23:04
The substitution actually give something slightly different: either $M=0$, or $M=50.5C+50$. – Brian M. Scott Nov 19 '12 at 23:06
@Sigur: Sure, and equation 1 excludes the third quadrant, leaving only the first left. That much I understand, but for whatever reason this explanation doesn't satisfy me :/ – SharpHawk Nov 19 '12 at 23:07
@BrianM.Scott: good catch, thanks. – SharpHawk Nov 19 '12 at 23:13

Given \begin{align} P&=\frac{50.5M}{M-50}\\ M&=CP, \end{align}

we have $$M=C\frac{50.5M}{M-50}.$$ This results in \begin{align} M(M-50)&=C(50.5M)\\ M^2-50M&=50.5CM\\ M^2-50M-50.5CM&=0\\ M^2+(-50-50.5C)M&=0 \end{align} Given that this is a quadratic equation, the solutions are: \begin{align} M_1&=\frac{50+50.5C+\sqrt{(-50-50.5C)^2}}{2}=0\\ \text{and } M_2&=\frac{50+50.5C-\sqrt{(-50-50.5C)^2}}{2}=50+50.5C \end{align} This indicates definitively that $M$ is either 0 or $50+50.5C$.

What confuses you, precisely?

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Nothing pertaining to the solving of the equations. It's just, more of a feeling, as if my intuition failed to predict that M would always be positive. Maybe I should just accept that it's arbitrary, and there's no deeper truth left to find. – SharpHawk Nov 19 '12 at 23:17
It isn't arbitrary. It is simply the fact that you usually cannot foresee properties of variables in equations. You shouldn't have expected yourself to do so, either. Intuition is great for a lot of things, but it can do little in this case. By the way, if you feel I've met the question and answered it precisely, accept this answer. – 000 Nov 19 '12 at 23:20
I wasn't looking for a derivation, since I'd done it myself. However, since you forced me to confront my unrealistic expectations about my mathematical intuition (which I think was the key issue here), I'll go ahead and accept it. – SharpHawk Nov 19 '12 at 23:38
In the end, I'm just glad I helped! – 000 Nov 19 '12 at 23:39