First of all the optimal solution of the dual problem is $(y_1^*,y_2^*)=(10,0)$ and $z^*=250$.
First Option
Use the following equation to evaluate the optimal values of $\mathbf x$:
$\left[c-A^Ty^* \right]^T\cdot x^*=0$
Therefore the equations are
$(8-20)x_1=0 \Rightarrow x_1=0$
$(6-10)x_2=0 \Rightarrow x_2=0$
$(-10+10)x_3=0 \ \Rightarrow x_3 \ \text{ firstly cannot be determined.}$
$(20-20)x_4=0 \ \Rightarrow x_4 \ \text{firstly cannot be determined.}$
$(2-10)x_5=0 \Rightarrow x_5=0$
From the primal problem we get the two equations
$-x_3+2x_4=25$
$2x_3-x_4=20$
Now solve for $x_3$ and $x_4$.
Second Option
Or you can conclude directly from your dual program that $s_3=s_4=0$ and $s_1,s_2$ and $s_5$ are unequal to $0$. Just plug in the values of the optimal solution.
Using
$s_j\cdot x_j=0 \ \ \forall \ j\in \{1,2,3,4,5 \}$
it follows that $x_1=x_2=x_5=0$
It remains to calculate the values of $x_3$ and $x_4$. The equations a equal to the equations of the first option.
- $[y_1, y_2]^T = [2,1]$ is not a feasible solution to the dual since it doesn't satisfy the first constraint $4y_1 -y_2 = 7 > -7$.
- $[y_1, y_2]^T = [2,1]$ is not a basic solution. You may say "by the same calculations", but not "by the same arguments".
\begin{align}
4y_1 -y_2 = 7 > -7 &\implies [y_1, y_2]^T \text{ infeasible} \\
4y_1 -y_2 = 7 \ne -7 &\implies [y_1, y_2]^T \text{ non-basic}
\end{align}
The second implication follows since we actually transform the LP into
\begin{array}{lrr}
\text{max} & w = -22y_1 -8y_2 &\\
\text{s.t.} & -2y_1 - y_2 +s_1 =& -4 \\
& 4y_1 -y_2 +s_2 =& -7\\
&-y_1 -3y_2 +s_3 =& 5\\
&8y_1 -y_2 +s_4 =& -14\\
& y_1, y_2, s_1, s_2, s_3, s_4 \geq 0
\end{array}
The calculations in the question body actually shows that $s_i$'s are nonzero, so the solution $[y_1,y_2]^T$ has more than two nonzero components, so it's not a basic solution.
- You have the right choice of method. To use the dual simplex method, we need a basic (infeasible) solution with the optimality satisfied. Note that feasibility and optimality are two independent concepts. It's easy to observe that in the simplex tableau, the $w$-row ($[22, 8, 0,\dots,0]$) is nonnegative, so the optimality condition is satisfied. As a result, the most obvious choice for a basic variable would be $[y_1,y_2]^T=[0,0]$.
Before continuing the dual simplex method, in response to OP's comment corcerning $[y_1,y_2]^T = [2, -1]$, we have $s_1 = -4 + 2(2) + (-1) = -1 \ne 0$, so $[y_1,y_2]^T = [2,-1]$ has more than two nonzero components, so it's nonbasic and infeasible. We can't use this to solve the dual.
Current basis: $s_1, s_2, s_3, s_4$
\begin{array}{rrrrrrr|r}
& y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \\ \hline
s_1 & -2 & -1 & 1 & 0 & 0 & 0 & -4 \\
s_2 & 4 & -1 & 0 & 1 & 0 & 0 & -7 \\
s_3 & -1 & -3 & 0 & 0 & 1 & 0 & 5 \\
s_4 & 8 & -1 & 0 & 0 & 0 & 1 & -14 \\ \hline
& 22 & 8 & 0 & 0 & 0 & 0 & 0 \\
\text{ratio} & 11/4 & -8 & & & & 0 &
\end{array}
Leaving variable: $s_4$, entering variable: $y_2$
Current basis: $s_1, s_2, s_3, y_2$
\begin{array}{rrrrrrr|r}
& y_1 & y_2 & s_1 & s_2 & s_3 & s_4 & \\ \hline
s_1 & -10 & 0 & 1 & 0 & 0 & -1 & 10 \\
s_2 & -4 & 0 & 0 & 1 & 0 & -1 & 7 \\
s_3 & -25 & 0 & 0 & 0 & 1 & -3 & 47 \\
y_2 & -8 & 1 & 0 & 0 & 0 & -1 & 14 \\ \hline
& \color{blue}{86} & \color{blue}{0} & \color{red}{0} & \color{red}{0} & \color{red}{0} & \color{red}{\bbox[2px, border: solid 1px]{8}} & -112 \\
\end{array}
Hence the optimal solution is $[y_1,y_2]^T = [0,14]$ with $w = -112$.
By adding slack variabes $t_1,t_2$ to the primal
\begin{array}{rlr}
\min z = & \color{red}{-4x_1 - 7x_2 + 5x_3 -14x_4} &\\
\text{s.t.} & 2x_1 -4x_2 + x_3 -8x_4 +\color{blue}{t_1} &= 22 \\
& x_1 + x_2 + 3x_3 + x_4 +\color{blue}{t_2} &= 8 \\
& \color{red}{x_1, x_2, x_3, \bbox[2px, border: solid 1px]{x_4}}, \color{blue}{t_1,t_2} \geq 0,
\end{array}
we can read the solution of the primal from the simplex tableau for the dual: the optimal solution for the primal is $[\color{red}{x_1, x_2, x_3, \bbox[2px, border: solid 1px]{x_4}}, \color{blue}{t_1,t_2}]^T = [\color{red}{0,0,0, \bbox[2px, border: solid 1px]{8}}, \color{blue}{86,0}]$.
Best Answer
Your formulation of the dual Problem is right-you forgot the negative sign at the third constraint only. As I said, you can eleminate the third constraint (primal problem).In this case the dual problem is: $$ \text{min } 8y_1+10y_2\\ 2y_1-y_2\ge 5\\ 4y_1+y_2\ge 7\\ -2y_1+2y_2\ge -3\\ y_1,y_2\ge0 $$
The solution of this problem is $y^T=(y_1,y_2)=(3.5;2)$ Your solution is right. The solution is $y_3=0$, because the third constraint is not necessary.
The compementary slackness condition is $X^TC^*=b^TY^*$
This gives: $\begin{pmatrix}5 & 7 & -3\end{pmatrix}\cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 8 & 10 \end{pmatrix} \cdot \begin{pmatrix} 3.5 \\ 2 \end{pmatrix} \Rightarrow \begin{pmatrix}5 & 7 & -3 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=48$
Additional
The following condition must hold $x_j \cdot z_j=0 \ \ \forall n$
$z_j$ are the slack variables of the dual problem. If you insert the solution for the dual problem, then you will see, that $z_1$ and $z_3$ are zero and $z_2$ is not zero. Thus the equations are (for all n):
$x_1\cdot 0=0 $
$x_2\cdot z_2=0$
$x_3\cdot 0=0$
Thus you know for sure, that $x_2=0$
And secondly this equation must hold: $s_i\cdot y_i=0 \ \ \forall m$
$s_i$ are the slack variables of the primal problem. Thus $s_1$ and $s_2$ are zero and so we have the equations:
$2x_1-2x_3=8$
$-x_1+2x_3=10 $
Remember, that $x_2=0$
This two equations you can solve.