Solve binomial coefficient equation

The formula should be $$\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}.$$ It is the fundamental recurrence of the binomial coefficients.

Hence $$\binom{6}{2}+\binom{6}{x}=\binom{7}{x}=\binom{6}{x-1}+\binom{6}{x}\implies \binom{6}{2}= \binom{6}{x-1}.$$ and, by symmetry, it follows that we have TWO solutions: $x-1=2$ OR $x-1=6-2=4$.

We have

$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{2!4!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$ $$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$

therefore since $(n)!=(m)!$ implies $n=m$ only when $n,m > 1$ we see that

$$(x-1)! = 2! \implies x-1=2 \implies x=3$$ $$(6-(x-1))!=4! \implies 6 -(x-1)=4 \implies -x=-2-1\implies x=3$$

which gives us our first solution of $x=3$. Next, because $$\begin{pmatrix} 6\\2 \end{pmatrix}=\begin{pmatrix} 6\\4 \end{pmatrix}$$ we have

$$\begin{pmatrix} 6\\4 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{4!2!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$ $$\begin{pmatrix} 6\\{x-1} \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\frac{6!}{(x-1)!(6-(x-1))!}+\frac{6!}{x!(6-x)!}=\frac{7!}{x!(7-x)!}=\begin{pmatrix} 7\\x \end{pmatrix}$$

hence

$$(x-1)! = 4! \implies x-1=4 \implies x=5$$ $$(6-(x-1))!=2! \implies 6 -(x-1)=2 \implies -x=-4-1\implies x=5$$ which gives us our second solution of $x=5$.