<s>[INST] Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find  and . Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find $$ and $$.
Can anyone see a slick way to do this? The solution is $x=10$ and $y=2$, but I am struggling with the method.
 [/INST] Hint: The first relation gives \begin{align}\dbinom{x}{y}=\frac38\dbinom{x}{y+1}\implies& \frac{x!}{y!(x-y)!}=\frac38\frac{x!}{(y+1)!(x-y-1)!}\\[0.3cm]\implies&\frac{x!}{y!(x-y)(x-y-1)!}=\frac38\frac{x!}{(y+1)y!(x-y-1)!}\\[0.3cm]\implies&\frac{1}{(x-y)}=\frac38\frac{1}{(y+1)}\implies -3x+11y=-8\end{align} For the second, recall that $$\dbinom{x}{y}=\dbinom{x}{x-y}$$ which gives that $x-y=y+6$.
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