# 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 $50+50.5C$.