Prove that  $V={(𝑥_1,𝑥_2,⋯,𝑥_n) \in ℝ^n ∣ 𝑥_1+𝑥_2+...+𝑥_{𝑛−1}−2𝑥_𝑛=0}\}$ is a subspace of $\R^n$.

Let  $t,s \in \R$ and $x,y\in V$. Then $x=(x_1, \dots, x_{n-1}, x_n)$ and $y=(y_1, \dots, y_{n-1}, y_n)$ with
\[x_1 ​+x _2 ​ +...+x_{n−1} ​−2x_ n ​ =0\]
and 
\[y_1 ​+y _2 ​ +...+y_{n−1} ​−2y_ n ​ =0.\]
Hence
\[z=tx+sy=t(x_1, \dots, x_{n-1}, x_n)+s(y_1, \dots, y_{n-1}, y_n)\]
\[=(tx_1+sy_1, tx_2+sy_2, \dots, tx_{n}+sy_{n}).\]
Thus we have
\[z_1 ​+z _2 ​ +...+z_{n−1} ​−2z_ n=\]
\[(tx_1+sy_1)+(tx_2+sy_2)+\dots +(tx_{n-1}+sy_{n-1})-2 (tx_{n}+sy_{n})\]
\[=t(x_1 ​+x _2 ​ +...+x_{n−1} ​−2x_ n)+s(y_1 ​+y _2 ​ +...+y_{n−1} ​−2y_ n )=t\times 0+s\times 0=0.\]
Therefore $tx+sy \in V$ and hence $V$ is a vector space.