Given the primal $$\max z= 5x_1-4x_2+3x_3$$ subject to
$2x_1+x_2-6x_3=20$
$ 6x_1+5x_2+10x_3\leq 76$
$8x_1-3x_2+6x_3\geq 50$
with $x_1\in \mathbb{R}, x_2\geq 0,x_3\leq 0$.
The question is to construct the dual and find the dual solution using complementary slackness theorem if at the primal's optimal table the basic variables are $s_3$ (slack in the 3rd constarint),$s_2$ (slack in the 2nd constarint), $x_1$.
My attempt is since $x_2,x_3$ are not in the optimal table this implies $x_2=x_3=0$ at the optimal. So this also should satisfy the constraints. This gives $x_1=10$ and hence the optimum value as $z=50$. Now the dual constructed is $$\min w=20y_1+76y_2+50y_3$$ subject to
$2y_1+6y_2+8y_3=5$
$y_1+5y_2-3y_3\geq -4$
$6y_1-10y_2-6y_3\geq -3$ with $y_1\in \mathbb{R}, y_2\geq 0, y_3\leq 0$
Now I know since the first constraint is an equation in the primal$\implies~y_1=0$, after that I am stuck. can somebody help.