<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] $$\overbrace{\frac 38 \binom x{y+1}=\binom xy}^\text{given}=\frac {y+1}{x-y}\binom x{y+1}\\
\frac {y+1}{x-y}=\frac 38\\
3x=11y+8\qquad\cdots (1)\\\\
\binom x{\color{orange}{y}}=\binom x{\color{blue}{y+6}}\\
\binom x{\color{blue}{x-y}}=\binom x{\color{orange}{x-y-6}}\\
x-y=y+6\\
x=2y+6\qquad\cdots (2)\\
\text {From }(1),(2), \hspace{6cm}\\
\color{red}{x=10, y=2}$$
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