Does adding a new variable to this optimization problem alter the solution? Here is an optimal tableau where the constraints are of $\leq$ type:

If I added a new activity $x_9$ with coefficients $(2,0,3)^t$ in the constraints and a price of $5$ (in the objective function), would the solution be altered?
Do I have to do out the entire tableau with the new variable, or is there an easier way?
 A: Consider a primal problem $\cal{P}$
\begin{align*}
\min  \ \ & c^{\top}x\\
&Ax \leq b\\
&x\geq 0
\end{align*}
and put it in standard form $\cal{P}_{SF}$
\begin{align*}
\min  \ \ & c^{\top}x + 0^{\top}x^s\\
&Ax + Ix^s= b\\
&x, x^s\geq 0 
\end{align*}
Then the Dual problem $\cal{D}$ is
\begin{align*}
\max  \ \ & b^{\top}y \\
&A^{\top}y \leq c\\
&y\leq 0 
\end{align*}
It is well-known that for every  primal-feasible basis $B$ the $m$-vector $y^{\top}=c_{B}^{\top}B^{-1}$, together with the primal feasible solution 
\begin{align*}
x=\left[
\begin{array}{c}
 B^{-1}b\\
0
\end{array}
\right]
\end{align*}
 satisfies the complementary slackness conditions. Therefore $x$ and $y$ are optimal solutions if $y$ is feasible for the dual problem. The feasibility of $y$ can be checked implicitly in the primal Tableu by checking the reduced costs 
\begin{align*}
&\hat{c}_{B}^{\top}=0_m^{\top}\\
&\hat{c}_{N}^{\top}=c_N^{\top} -c_{B}^{\top}B^{-1}N
\end{align*}
since they represents the slack variables of the dual constraints. If all the reduced costs are non-negative, with respect the basis $B$, then $B$ is an optimal basis. Observe that in the example you report the reduced costs of the primal slack variables represent the slack variables of the dual constraints $y \leq 0$, namely
\begin{align*}
&y_1 + \hat{c}_6=0\\
&y_2 + \hat{c}_7=0\\
&y_3 + \hat{c}_8=0
\end{align*}
Therefore 
\begin{align*}
y^{\top}=[y_1,y_2,y_3]=c_{B}^{\top}B^{-1}=[-\hat{c}_6, -\hat{c}_7, -\hat{c}_8]=[-2,-1/10, -2]
\end{align*}
Now to analyse how the optimal solution changes when a new activity is added, you have just to compute the reduced cost of the new activity with respect to the optimal basis. It is given by
\begin{align*}
\hat{c}_9=c_9-c_{B}^{\top}B^{-1}A_9=5-[-2,-1/10, -2]\left[\begin{array}{c} 2\\ 0\\ 3\end{array}\right]=5+10>0
\end{align*} 
meaning that the current optimal basis is still optimal even if the new activity is added.
