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I did $$\frac{y!}{(y-2)!2}+10y=108\\$$

I don't really know how to go from here...

This equation is actually part of a system:

$$\begin{cases}{}^xC_2 = 45 \\ {}^yC_2 + xy = 108\end{cases}$$

Solving the first part of the system I find that x = 10.

My questions:

  • How do I solve $\frac{y!}{(y-2)!2!}+10y=108$
  • Would it be simpler to solve x and y by solving the system, instead of the isolated equations?
  • How do I solve the system?

NOTE: C here is the C from the formula of combinations, as in: $${}^nC_k = \frac{n!}{(n-k)!k!}$$

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Assuming you meant to solve $\frac{y!}{(y-2)!2!}+10y=108$,

$$\frac{y!}{(y-2)!}=y(y-1)=y^2-y$$

$$\frac{y!}{(y-2)!2!}+10y=0.5y^2+9.5y=108$$

$$y^2+19y=216$$

Quadratic, with solution given as

$$y=8,-27$$

Usually $y>0$, so then $y=8$.

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    $\begingroup$ Only the positive root is strictly valid. The negative one fails to provide well-defined factorials in the original equation. $\endgroup$ – Oscar Lanzi Oct 15 '16 at 0:58
  • $\begingroup$ @OscarLanzi Yeah, I guess that's true too. $\endgroup$ – Simply Beautiful Art Oct 15 '16 at 1:04
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    $\begingroup$ @OscarLanzi: Good observation, +1. Much as $x=0$ is a solution of the equation $x^2=0$ but not of the equation $\frac{1/x}{1/x^3}=0$. $\endgroup$ – MPW Oct 15 '16 at 1:05
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$$y^2+19y=216$$ $$y(y+19)=(8)(27)$$ so the $$y=8$$

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