Maybe it's worthwhile to talk through where the dual comes from. This will take a while, but hopefully the dual won't seem so mysterious when we're done.
Suppose we want to use the primal's constraints as a way to find an upper bound on the optimal value of the primal. If we multiply the first constraint by $9$, the second constraint by $1$, and add them together, we get $9(2x_1 - x_2) + 1(x_1 +3 x_2)$ for the left-hand side and $9(1) + 1(9)$ for the right-hand side. Since the first constraint is an equality and the second is an inequality, this implies $$19x_1 - 6x_2 \leq 18.$$
But since $x_1 \geq 0$, it's also true that $5x_1 \leq 19x_1$, and so $$5x_1 - 6x_2 \leq 19x_1 - 6x_2 \leq 18.$$
Therefore, $18$ is an upper-bound on the optimal value of the primal problem.
Surely we can do better than that, though. Instead of just guessing $9$ and $1$ as the multipliers, let's let them be variables. Thus we're looking for multipliers $y_1$ and $y_2$ to force $$5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9).$$
Now, in order for this pair of inequalities to hold, what has to be true about $y_1$ and $y_2$? Let's take the two inequalities one at a time.
The first inequality: $5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2)$
We have to track the coefficients of the $x_1$ and $x_2$ variables separately. First, we need the total $x_1$ coefficient on the right-hand side to be at least $5$. Getting exactly $5$ would be great, but since $x_1 \geq 0$, anything larger than $5$ would also satisfy the inequality for $x_1$. Mathematically speaking, this means that we need $2y_1 + y_2 \geq 5$.
On the other hand, to ensure the inequality for the $x_2$ variable we need the total $x_2$ coefficient on the right-hand side to be exactly $-6$. Since $x_2$ could be positive, we can't go lower than $-6$, and since $x_2$ could be negative, we can't go higher than $-6$ (as the negative value for $x_2$ would flip the direction of the inequality). So for the first inequality to work for the $x_2$ variable, we've got to have $-y_1 + 3y_2 = -6$.
The second inequality: $y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9)$
Here we have to track the $y_1$ and $y_2$ variables separately. The $y_1$ variables come from the first constraint, which is an equality constraint. It doesn't matter if $y_1$ is positive or negative, the equality constraint still holds. Thus $y_1$ is unrestricted in sign. However, the $y_2$ variable comes from the second constraint, which is a less-than-or-equal to constraint. If we were to multiply the second constraint by a negative number that would flip its direction and change it to a greater-than-or-equal constraint. To keep with our goal of upper-bounding the primal objective, we can't let that happen. So the $y_2$ variable can't be negative. Thus we must have $y_2 \geq 0$.
Finally, we want to make the right-hand side of the second inequality as small as possible, as we want the tightest upper-bound possible on the primal objective. So we want to minimize $y_1 + 9y_2$.
Putting all of these restrictions on $y_1$ and $y_2$ together we find that the problem of using the primal's constraints to find the best upper-bound on the optimal primal objective entails solving the following linear program:
$$\begin{align*}
\text{Minimize }\:\:\:\:\: y_1 + 9y_2& \\
\text{subject to }\:\:\:\:\: 2y_1 + y_2& \geq 5 \\
-y_1 + 3y_2& = -6\\
y_2 & \geq 0.
\end{align*}$$
And that's the dual.
It's probably worth summarizing the implications of this argument for all possible forms of the primal and dual. The following table is taken from p. 214 of
Introduction to Operations Research, 8th edition, by Hillier and Lieberman. They refer to this as the SOB method, where SOB stands for Sensible, Odd, or Bizarre, depending on how likely one would find that particular constraint or variable restriction in a maximization or minimization problem.
Primal Problem Dual Problem
(or Dual Problem) (or Primal Problem)
Maximization Minimization
Sensible <= constraint paired with nonnegative variable
Odd = constraint paired with unconstrained variable
Bizarre >= constraint paired with nonpositive variable
Sensible nonnegative variable paired with >= constraint
Odd unconstrained variable paired with = constraint
Bizarre nonpositive variable paired with <= constraint
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.
Best Answer
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 $[\color{red}{x_1, x_2, x_3, \bbox[2px, border: solid 1px]{x_4}}, \color{blue}{t_1,t_2}]^T = [\color{red}{0,0,0, \bbox[2px, border: solid 1px]{8}}, \color{blue}{86,0}]$.